Chapter test · nothing is marked until you submit

Factors and Multiples: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 132 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. One numeral, five tests . 11 points. Question 1 of 10.

A single whole number can be put past several divisibility tests in one pass, and each test reads a different part of the numeral: some read one digit, some read a block of digits, and some read every digit there is.
1. Part A.
  Decide which of $2$ , $4$ , $5$ , $8$ and $10$ divide $7{,}416$ , and say for each test which part of the numeral it reads.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  The four-digit number $6{,}3\square 0$ has one digit missing. Give every digit that makes it divisible by $9$ , every digit that makes it divisible by $4$ , and every digit that does both.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A student writes: " $7{,}416$ is divisible by $8$ , because its last two digits form $16$ and $16$ is a multiple of $8$ ." Decide whether the verdict is right, decide whether the reasoning is right, and give a four-digit number on which the student's rule and the true verdict disagree, if one exists.
  
  Carry your own answer forward Judge the student's verdict against your own answer for $7{,}416$ in part A, whatever it came to.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
The answer

Part A

$2$ , $4$ and $8$ divide $7{,}416$ ; $5$ and $10$ do not. The tests for $2$ , $5$ and $10$ read the last digit, the test for $4$ the last two digits, and the test for $8$ the last three.

Part B

For $9$ the digit is $0$ or $9$ . For $4$ it is $0$ , $2$ , $4$ , $6$ or $8$ . Only $0$ does both.

Part C

The verdict is right, since $7{,}416 = 8 \times 927$ , but the reasoning is not: the test for $8$ reads the last three digits, because $1000$ is a multiple of $8$ while $100$ is not. The same reasoning passes $2{,}932$ , whose last two digits give $32$ , and $2{,}932$ is not a multiple of $8$ .

Worked solution

Part A

Three of the tests read the last digit alone, because every column above the ones is built from tens. The last digit here is $6$ , which is even, so $2$ divides the number, while $5$ and $10$ need a last digit of $0$ or $5$ and do not get one.

$7{,}416 = (741 \times 10) + 6$

The test for $4$ reads one column further in, since $100 = 4 \times 25$ makes everything above the last two digits a multiple of $4$ . The test for $8$ reads one further still, since $1000 = 8 \times 125$ .

$16 = 4 \times 4, \qquad 416 = 8 \times 52$

Both of those leftovers are clean, so $4$ and $8$ divide $7{,}416$ as well.

Part B

For $9$ , add the digits and set the sum against the multiples of $9$ . The known digits give $6 + 3 + 0 = 9$ , so the blank digit $d$ has to make $9 + d$ a multiple of $9$ .

$9 + d = 9 \Rightarrow d = 0, \qquad 9 + d = 18 \Rightarrow d = 9$

For $4$ , only the last two digits matter, and they read $d$ then $0$ , which is the number $10 \times d$ . Run through the ten digits and pick out the two-digit numbers that are multiples of $4$ .

$00, \; 20, \; 40, \; 60, \; 80 \text{ are multiples of } 4; \qquad 10, \; 30, \; 50, \; 70, \; 90 \text{ are not}$

So the even digits pass the test for $4$ . The only digit on both lists is $0$ , giving $6{,}300$ , which is $9 \times 700$ and $4 \times 1{,}575$ .

Part C

Take the two questions apart. The verdict happens to be right, and dividing settles that.

$7{,}416 = 8 \times 927$

The reasoning is still wrong, and the place-value split shows why. Splitting at the hundreds column leaves a piece that is a multiple of $100$ , and $100$ is not a multiple of $8$ : it is $8 \times 12 + 4$ . So the hundreds column can still carry a remainder into the answer, and the last two digits cannot settle the question on their own. Splitting one column further out fixes it, because $1000 = 8 \times 125$ .

A number where the student's rule breaks is one whose last two digits form a multiple of $8$ while the last three do not.

$2{,}932: \quad 32 = 8 \times 4, \qquad 932 = (8 \times 116) + 4$

The student's rule calls $2{,}932$ divisible by $8$ , and it is not.

In one line

Of $2$ , $4$ , $5$ , $8$ and $10$ , the divisors of $7{,}416$ are $2$ , $4$ and $8$ , found by reading the last digit, the last two digits and the last three digits. In $6{,}3\square 0$ the blank must be $0$ or $9$ for divisibility by $9$ and any even digit for divisibility by $4$ , so only $0$ does both. The student's verdict on $7{,}416$ is right and the reasoning is not: the test for $8$ reads three digits, not two, and on $2{,}932$ the same reasoning gives the wrong answer.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Applies each test to the block of digits that test reads, and gives the right verdict for all five divisors. . Worth 2 points.

Names which part of the numeral each test reads: the last digit for $2$ , $5$ and $10$ , the last two for $4$ , the last three for $8$ . . Worth 1 point.

Part B 3 points

Finds every digit making the digit sum a multiple of $9$ , and every digit for which the last two digits form a multiple of $4$ . . Worth 2 points.

Identifies the single digit that satisfies both conditions at once. . Worth 1 point.

Part C 5 points

Separates the verdict from the reasoning, accepting the first and rejecting the second. . Worth 2 points. needs an explanation, not just an answer

Says that the test for $8$ reads the last three digits because $1000$ is a multiple of $8$ while $100$ is not. . Worth 2 points. needs an explanation, not just an answer

Supplies a four-digit number on which the student's two-digit rule and the true verdict disagree. . Worth 1 point.
2. What the factor count decides . 12 points. Question 2 of 10.

Prime and composite are settled by a count and nothing else: how many distinct whole numbers divide the number leaving no remainder. Exactly two makes it prime, more than two makes it composite.
1. Part A.
  List every factor of $92$ and every factor of $301$ . Give each count, and classify each number.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Do the same for $1$ and for $2$ : list every factor of each, give the counts, and classify each number by the same rule.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A classmate offers a shorter rule: a whole number is prime exactly when no whole number below it divides it except $1$ , and composite exactly when it is even or ends in $5$ . Test each half of that rule against the four numbers above, and say for each half whether it is sound, correcting anything it gets wrong.
  
  Carry your own answer forward Test the classmate's rule against the four numbers you classified in parts A and B, whatever you decided about each.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$92$ has six factors: $1$ , $2$ , $4$ , $23$ , $46$ , $92$ . $301$ has four: $1$ , $7$ , $43$ , $301$ . Both counts pass two, so both numbers are composite.

Part B

$1$ has one factor, $1$ itself, so it is neither prime nor composite. $2$ has two factors, $1$ and $2$ , so it is prime.

Part C

Both halves fail. The first calls $1$ prime, since no whole number below $1$ divides it, although its single factor puts it outside both groups. The second calls $2$ composite although $2$ is prime, and it misses $301$ , which is odd, does not end in $5$ , and is $7 \times 43$ .

Worked solution

Part A

Work up through the candidate divisors, taking each factor together with its partner. For $92$ , the tests for $2$ and $4$ both pass, and $92 \div 2 = 46$ , $92 \div 4 = 23$ .

$92 = 1 \times 92 = 2 \times 46 = 4 \times 23$

The candidates $5$ , $6$ , $7$ , $8$ and $9$ each leave a remainder, and $10 \times 10 = 100$ has passed $92$ , so no factor pair is left to find and the list is complete: $1, 2, 4, 23, 46, 92$ , six factors in all.

For $301$ the quick tests fail, since it is odd, its digits add to $4$ and it does not end in $0$ or $5$ . The next candidate is $7$ .

$301 = 7 \times 43$

That gives the factor list $1, 7, 43, 301$ , four factors. Six and four are both more than two, so each number is composite.

Part B

A factor of $1$ is a whole number dividing $1$ with no remainder, and only $1$ does, since anything larger leaves a remainder. The " $1$ " and the "itself" are the same number here, so they are one entry, not two.

$\text{factors of } 1: \; 1 \qquad \text{factors of } 2: \; 1, \, 2$

One factor is neither exactly two nor more than two, so $1$ meets neither definition and stands outside both groups. The number $2$ has exactly two distinct factors, so it is prime, and it stays prime although it is even: the factor $2$ is the number itself, not a third factor.

Part C

Take the halves one at a time, and test both directions of each.

The first half agrees with the definition for every number greater than $1$ , but it says nothing about how many factors there are, so it slips at the bottom of the list. No whole number below $1$ divides $1$ at all, so the classmate's rule calls $1$ prime.

$\text{factors of } 1: \; 1 \quad \text{(one factor, so neither prime nor composite)}$

The second half fails in both directions at once. Going one way, $2$ is even, so the rule calls it composite, and $2$ is prime. Going the other way, a composite need not be even and need not end in $5$ .

$301 = 7 \times 43$

That number is odd and does not end in $5$ , so the rule's second half never reaches it and a composite slips straight through. Being even or ending in $5$ is a quick way to CATCH some composites, never a definition of one: it names two divisors out of all there are.

In one line

$92$ has the six factors $1, 2, 4, 23, 46, 92$ and $301$ has the four factors $1, 7, 43, 301$ , so both are composite; $1$ has a single factor and is neither prime nor composite, while $2$ has exactly two and is prime. The classmate's shorter rule fails at both halves: the first calls $1$ prime, and the second calls the prime $2$ composite while letting the composite $301 = 7 \times 43$ through, since being even or ending in $5$ names only two of the divisors a number might have.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Lists all six factors of $92$ and all four factors of $301$ , with no repeats and none missing. . Worth 2 points.

Classifies each number from its count of distinct factors rather than from its appearance. . Worth 2 points.

Part B 3 points

Gives one factor for $1$ and two for $2$ , counting the number itself once only. . Worth 2 points.

Reads each classification off the count, placing $1$ outside both groups and $2$ among the primes. . Worth 1 point.

Part C 5 points

Rejects the first half by pointing at $1$ , and says that the definition counts factors rather than searching below the number. . Worth 2 points. needs an explanation, not just an answer

Rejects the second half in both directions, with a prime it wrongly calls composite and a composite it wrongly lets through. . Worth 3 points. needs an explanation, not just an answer
3. A tree and a column . 12 points. Question 3 of 10.

A factor tree splits a number in whatever way you notice first and branches out; repeated division works down a single column, always taking the smallest prime that fits. Both stop for the same reason.
1. Part A.
  Build a factor tree for $588$ , taking $588 = 12 \times 49$ as the first split. Carry every branch as far as it will go, then give the factorization in prime-power form.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Find the prime factorization of $1{,}452$ by repeated division, taking the smallest prime that fits at every step, and name the test that told you each divisor would fit.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Say how you can tell a factor tree has gone as far as it can without multiplying anything back, and say what the small raised number in a prime-power form counts. Then decide whether $2^2 \times 25 \times 11$ is a prime factorization of $1{,}100$ , and repair it if it is not.
  
  Carry your own answer forward Judge that form by the same standard you applied to your own tree in part A.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

$588 = 2 \times 2 \times 3 \times 7 \times 7 = 2^2 \times 3 \times 7^2$ .

Part B

$1{,}452 = 2^2 \times 3 \times 11^2$ , from the divisors $2$ , $2$ , $3$ , $11$ , $11$ : an even last digit for each $2$ , the digit sum $12$ for the $3$ , and division for each $11$ .

Part C

A tree is finished when every leaf is prime, since only a composite can be split again. The raised number counts copies of that prime. The form $2^2 \times 25 \times 11$ multiplies back to $1{,}100$ , but $25 = 5^2$ is composite, so repaired it reads $2^2 \times 5^2 \times 11$ .

Worked solution

Part A

Neither $12$ nor $49$ is prime, so each branch splits again. The $12$ goes to $4 \times 3$ and then the $4$ goes to $2 \times 2$ ; the $49$ goes to $7 \times 7$ .

$588 = 12 \times 49 = (2 \times 2 \times 3) \times (7 \times 7)$

Every leaf is now prime, so the tree is finished. Collecting the repeats into powers and listing the primes from smallest to largest,

$588 = 2^2 \times 3 \times 7^2$

and multiplying back gives $4 \times 3 \times 49 = 588$ .

Part B

The last digit is even, so $2$ fits, and the quotient is even again.

$1{,}452 \div 2 = 726, \qquad 726 \div 2 = 363$

Now $363$ is odd, so $2$ is finished. Its digits add to $12$ , a multiple of $3$ , so $3$ fits.

$363 \div 3 = 121$

The quotient $121$ is odd, its digits add to $4$ and it does not end in $0$ or $5$ , so $3$ and $5$ are out; $7$ leaves a remainder, and $11$ divides it exactly, twice over.

$121 \div 11 = 11, \qquad 11 \div 11 = 1$

The quotient has reached $1$ , so the divisors used are the factorization: $1{,}452 = 2^2 \times 3 \times 11^2$ .

Part C

The stopping condition is read off the leaves, not off a multiplication. A composite number has a factor other than $1$ and itself, so it can always be split again; a prime has none, so it cannot. A tree is therefore finished exactly when every leaf is prime, and no check by multiplication is needed to see it.

The raised number is shorthand for repeated multiplication, and it counts copies of the base.

$5^2 = 5 \times 5 = 25, \qquad 2^2 = 2 \times 2 = 4$

So it says how many times the prime appears, never what the prime is multiplied by.

Now test the form offered. Its arithmetic is correct.

$2^2 \times 25 \times 11 = 4 \times 25 \times 11 = 1{,}100$

But a prime factorization is a product of PRIMES, and $25$ is not one, so this is a tree with a branch left unfinished rather than a slip in the multiplication. Splitting that leaf gives $5 \times 5$ and the finished form.

$1{,}100 = 2^2 \times 5^2 \times 11$

In one line

The tree for $588$ finishes at $2^2 \times 3 \times 7^2$ . Repeated division on $1{,}452$ uses the divisors $2$ , $2$ , $3$ , $11$ , $11$ , justified by an even last digit for each $2$ , the digit sum $12$ for the $3$ , and division for each $11$ , giving $2^2 \times 3 \times 11^2$ . A tree has gone as far as it can exactly when every leaf is prime, and the raised number counts how many copies of that prime are multiplied together. The form $2^2 \times 25 \times 11$ multiplies back to $1{,}100$ but is not a prime factorization, since $25 = 5 \times 5$ is composite; repaired it reads $2^2 \times 5^2 \times 11$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Splits every composite branch again, so that all five leaves are prime. . Worth 2 points.

Collects the repeated primes into powers and lists the primes from smallest to largest. . Worth 1 point.

Part B 4 points

Divides by the smallest prime that fits at each step and carries on until the quotient reaches $1$ . . Worth 2 points.

Names the test behind each divisor: an even last digit for $2$ , a digit sum of $12$ for $3$ , and division for $11$ . . Worth 1 point.

Gives the answer in prime-power form with both exponents right. . Worth 1 point.

Part C 5 points

States the stopping condition as every leaf being prime, and says why a composite leaf can always be split again. . Worth 2 points. needs an explanation, not just an answer

Says the exponent counts copies of the prime that are multiplied together. . Worth 1 point.

Rejects the offered form because $25$ is composite, grants that its multiplication is correct, and repairs it to $2^2 \times 5^2 \times 11$ . . Worth 2 points. needs an explanation, not just an answer
4. Kits and reorders . 13 points. Question 4 of 10.

A school supply room holds $110$ pens and $154$ notebooks. Pens are reordered every $12$ days and notebooks every $27$ days, and both orders were placed today.
1. Part A.
  The pens and notebooks are to be made up into identical kits, every kit holding the same number of pens and the same number of notebooks, with none left over and as many kits as possible. Give the number of kits and what one kit holds.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Give the number of days that pass before the two reorders next fall on the same day.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Each part above was settled by one of the two answers a pair of prime factorizations can give. Say which, and what in the wording chose it. Then give the value part A would have produced if the other one had been taken, and say why a supply room could not use it.
  
  Carry your own answer forward Argue from the two answers you produced in parts A and B, whatever they came to.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer

Part A

There are $22$ kits, each holding $5$ pens and $7$ notebooks.

Part B

$108$ days pass, which is $9$ pen orders and $4$ notebook orders.

Part C

Part A asks for the largest size fitting both stocks exactly, which takes the lowest power of each shared prime; part B asks when two cycles first coincide, which takes the highest power of every prime. The highest powers in part A give $770$ , more kits than there are pens, so no kit could be filled.

Worked solution

Part A

Each kit takes an equal share of both stocks, so the number of kits divides $110$ and also divides $154$ , and as many kits as possible means the greatest number that does both.

$110 = 2 \times 5 \times 11, \qquad 154 = 2 \times 7 \times 11$

The primes in both are $2$ and $11$ , each once in each number, and the lowest power of a shared prime is what a common factor may carry.

$\operatorname{GCF}(110, 154) = 2 \times 11 = 22$

So there are $22$ kits, and each one holds $110 \div 22 = 5$ pens and $154 \div 22 = 7$ notebooks, with nothing left over.

Part B

Pen orders land on the multiples of $12$ and notebook orders on the multiples of $27$ , so they fall together on a common multiple and NEXT fall together on the smallest one.

$12 = 2^2 \times 3, \qquad 27 = 3^3$

A common multiple must hold enough copies of each prime for either cycle on its own, so take the highest power of each.

$\operatorname{LCM}(12, 27) = 2^2 \times 3^3 = 108$

Checking, $108 \div 12 = 9$ and $108 \div 27 = 4$ , both whole, so day $108$ really is a day both orders are placed.

Part C

The two questions pull in opposite directions, and the wording says which way each pulls.

Part A hands out a fixed stock, so its answer has to DIVIDE both totals, which caps every prime at the count held by the poorer number. That is the lowest power of each shared prime.

$\operatorname{GCF}(110, 154) = 22$

Part B waits for two repeating cycles, so its answer has to be a MULTIPLE of both cycle lengths, which demands at least the count held by the richer number. That is the highest power of each prime.

$\operatorname{LCM}(12, 27) = 108$

Taking the highest powers in part A instead would give a different number altogether.

$\operatorname{LCM}(110, 154) = 2 \times 5 \times 7 \times 11 = 770$

A supply room cannot use $770$ kits: there are only $110$ pens, so most of the kits would stand empty. Whenever the answer to a sharing question comes out larger than the stock being shared, the two rules have been swapped.

In one line

The stocks make $\operatorname{GCF}(110, 154) = 22$ kits, each holding $5$ pens and $7$ notebooks, and the two reorders next coincide after $\operatorname{LCM}(12, 27) = 108$ days. Sharing a fixed stock asks for a number that divides both totals, so it takes the lowest power of each shared prime; waiting for two cycles to meet asks for a number both cycle lengths divide, so it takes the highest power of every prime. Taking the highest powers in part A would give $770$ , far more kits than there are pens to fill them.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Sets the number of kits up as a number dividing both stocks, and takes the greatest such number. . Worth 2 points.

Takes the lowest power of each shared prime and reaches $22$ . . Worth 1 point.

Reports what one kit holds, in pens and in notebooks. . Worth 1 point.

Part B 4 points

Recognizes the meeting day as a common multiple of the two cycles and asks for the least one. . Worth 2 points.

Takes the highest power of each prime, including all three copies of $3$ from $27$ . . Worth 1 point.

Reports the answer as a number of days. . Worth 1 point.

Part C 5 points

Attaches the lowest powers to the sharing question and the highest powers to the coinciding question, quoting the wording that decides it. . Worth 3 points. needs an explanation, not just an answer

Produces $770$ as the value the other rule would have given in part A, and says why a count of kits larger than the stock is impossible. . Worth 2 points. needs an explanation, not just an answer
5. Where a test may be split . 14 points. Question 5 of 10.

The number under test throughout is $7{,}848$ . Some divisors have a rule of their own, read off one block of digits, and some have no standalone rule taught in this chapter, so here they are tested by combining two taught rules.
1. Part A.
  Write $7{,}848$ as a multiple of $100$ plus the number made by its last two digits, and again as a multiple of $1000$ plus the number made by its last three. Use those two splits to say why the test for $4$ needs only two digits and the test for $8$ only three.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
2. Part B.
  Say which of $4$ , $6$ , $8$ and $9$ divide $7{,}848$ , naming what each test reads.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  A student proposes a general move: to test for any divisor, split it into two factors and run the tests for those two. Decide whether the move always works, taking $24$ split as $4 \times 6$ as your case, and state the condition on the two factors that makes it sound.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$7{,}848 = (78 \times 100) + 48 = (7 \times 1000) + 848$ . Since $100 = 4 \times 25$ , the first piece is a multiple of $4$ whatever stands above the last two digits, so only $48$ can decide that test; since $1000 = 8 \times 125$ , only $848$ can decide the test for $8$ .

Part B

All four divide $7{,}848$ : the last two digits give $48$ , the last three give $848$ , the digits add to $27$ , and the number is even with that digit sum a multiple of $3$ .

Part C

The move is not always sound. $12$ passes the tests for $4$ and for $6$ and is not a multiple of $24$ , because the two factors share a $2$ that then gets counted twice. It is sound exactly when the two factors share no factor above $1$ , as $3$ and $8$ do not.

Worked solution

Part A

Cut the numeral at the hundreds column, then at the thousands column.

$7{,}848 = (78 \times 100) + 48, \qquad 7{,}848 = (7 \times 1000) + 848$

In the first split the leading piece counts whole hundreds, and $100 = 4 \times 25$ , so that piece is $4 \times 25 \times 78$ : a multiple of $4$ , contributing remainder $0$ however large the number is. If a sum of two pieces is being tested for $4$ and one piece is already a multiple of $4$ , the verdict rests entirely on the other piece.

$48 = 4 \times 12$

So $7{,}848$ is divisible by $4$ , and the digits above the last two never entered the argument. The second split works the same way one column out, because $1000 = 8 \times 125$ makes the leading piece a multiple of $8$ .

$848 = 8 \times 106$

So $7{,}848$ is divisible by $8$ as well, decided by three digits. The reason the tests differ in length is exactly that $100$ is a multiple of $4$ but not of $8$ , while $1000$ is a multiple of both.

Part B

Run the four tests, each on the part of the numeral it reads.

$48 = 4 \times 12, \qquad 848 = 8 \times 106, \qquad 7 + 8 + 4 + 8 = 27 = 9 \times 3$

The first two settle $4$ and $8$ . The digit sum $27$ is a multiple of $9$ , so $9$ divides the number, and since $27$ is also a multiple of $3$ and the last digit $8$ is even, the number passes both halves of the test for $6$ as well.

$7{,}848 \div 6 = 1{,}308$

So every one of $4$ , $6$ , $8$ and $9$ divides $7{,}848$ .

Part C

Try the proposed split on a small number.

$12 \div 4 = 3, \qquad 12 \div 6 = 2, \qquad 12 \div 24 \text{ leaves a remainder}$

So $12$ passes both halves of the student's test for $24$ and is not a multiple of $24$ : one counterexample is enough to sink the move as a general rule.

The reason is that $4$ and $6$ overlap. Passing both tests puts two copies of $2$ into the number (from the $4$ ) and one copy of $2$ with one copy of $3$ (from the $6$ ), but the single copy of $2$ that the $6$ demands may be one the $4$ has already supplied. Nothing forces a third copy, and $24 = 2^3 \times 3$ needs three.

Split $24$ the other way and the overlap disappears.

$24 = 3 \times 8, \qquad 3 \text{ and } 8 \text{ share no factor above } 1$

Now the copies cannot be shared: the $8$ supplies the three $2$ s and the $3$ supplies the $3$ . So the move is sound exactly when the two factors have no common factor above $1$ , and $7{,}848$ , which passes the tests for $3$ and $8$ , is indeed a multiple of $24$ .

In one line

$7{,}848 = (78 \times 100) + 48 = (7 \times 1000) + 848$ , and because $100 = 4 \times 25$ and $1000 = 8 \times 125$ the leading piece of each split contributes no remainder, so two digits settle the test for $4$ and three settle the test for $8$ . All of $4$ , $6$ , $8$ and $9$ divide $7{,}848$ , its digit sum being $27$ . Splitting a divisor into two factors and running their tests is sound only when those factors share no factor above $1$ : $12$ passes the tests for $4$ and $6$ without being a multiple of $24$ , while the split $24 = 3 \times 8$ carries no such overlap.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Writes both splits correctly, as a multiple of $100$ plus $48$ and as a multiple of $1000$ plus $848$ . . Worth 2 points.

Argues that the leading piece is a multiple of the divisor, quoting $100 = 4 \times 25$ and $1000 = 8 \times 125$ , so the verdict rests on the leftover alone. . Worth 3 points. needs an explanation, not just an answer

Part B 4 points

Gives the right verdict for all four divisors. . Worth 2 points.

Names what each test reads, and treats $6$ as the two tests for $2$ and $3$ run together. . Worth 2 points.

Part C 5 points

Rejects the move as a general rule and backs the verdict with a number passing both tests and failing the test for $24$ . . Worth 3 points. needs an explanation, not just an answer

States the condition, that the two factors share no factor above $1$ , and explains what goes wrong when they do. . Worth 2 points. needs an explanation, not just an answer
6. One answer inside the other . 13 points. Question 6 of 10.

For two whole numbers the greatest common factor and the least common multiple are not free of one another. Fix the two numbers and either answer settles the other.
1. Part A.
  Find $\operatorname{GCF}(40, 56)$ and $\operatorname{LCM}(40, 56)$ from their prime factorizations. Then multiply your two answers together, multiply $40$ by $56$ , and compare the results.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Two whole numbers have a greatest common factor of $10$ and a least common multiple of $120$ . Give their product, and give a pair of numbers that fits, with neither of them equal to $10$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Suppose one of two whole numbers divides the other. Say what the greatest common factor and the least common multiple must then be, and check the product rule against that case.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$\operatorname{GCF}(40, 56) = 8$ and $\operatorname{LCM}(40, 56) = 280$ . Both $8 \times 280$ and $40 \times 56$ come to $2{,}240$ .

Part B

Their product is $1{,}200$ , and the pair $30$ and $40$ fits.

Part C

The greatest common factor is the number that divides and the least common multiple is the other one. Multiplying the two answers is then multiplying the two numbers in the other order, so the rule holds with nothing left to check.

Worked solution

Part A

Write both numbers in prime-power form.

$40 = 2^3 \times 5, \qquad 56 = 2^3 \times 7$

The only shared prime is $2$ , at three copies in each, so the lowest power is $2^3$ and the greatest common factor is $8$ . For the least common multiple take the highest power of every prime that appears at all.

$\operatorname{LCM}(40, 56) = 2^3 \times 5 \times 7 = 280$

Now multiply the two answers, and multiply the two numbers.

$8 \times 280 = 2{,}240, \qquad 40 \times 56 = 2{,}240$

The two agree, as they must: between them the answers take the lower and the higher count of every prime, which is the pair of counts over again.

Part B

The two answers multiply back to the two numbers multiplied together, so the product is fixed even though the numbers are not yet known.

$10 \times 120 = 1{,}200$

To find a pair, note that both numbers are multiples of $10$ , say $10 \times u$ and $10 \times v$ , and their product must be $1{,}200$ , so $u \times v = 12$ . Take $u$ and $v$ sharing no factor above $1$ , or the greatest common factor would climb past $10$ .

$u = 3, \; v = 4 \;\Rightarrow\; 30 \text{ and } 40$

Checking, $30 = 2 \times 3 \times 5$ and $40 = 2^3 \times 5$ , so the greatest common factor is $2 \times 5 = 10$ and the least common multiple is $2^3 \times 3 \times 5 = 120$ .

Part C

Let the smaller number divide the larger. The smaller divides itself and it divides the larger, so it is a common factor, and no common factor can exceed the smaller number, so it is the greatest one. The larger is a multiple of itself and a multiple of the smaller, so it is a common multiple, and no common multiple can be below the larger number, so it is the least one.

$\operatorname{GCF}(40, 120) = 40, \qquad \operatorname{LCM}(40, 120) = 120$

The two answers are therefore the two numbers over again, and the product rule becomes an identity.

$40 \times 120 = 40 \times 120$

The prime powers say the same thing: if one number divides the other, then at every prime its count is the lower one, so one number supplies every lowest power and the other supplies every highest power.

In one line

$\operatorname{GCF}(40, 56) = 8$ and $\operatorname{LCM}(40, 56) = 280$ , and $8 \times 280 = 40 \times 56 = 2{,}240$ . Two numbers with a greatest common factor of $10$ and a least common multiple of $120$ must have product $1{,}200$ , and $30$ and $40$ is such a pair. When one number divides the other, the greatest common factor is the number that divides and the least common multiple is the other one, so the product rule reduces to multiplying the two numbers in the other order.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Takes the lowest power of each shared prime for one answer and the highest power of every prime for the other. . Worth 2 points.

Carries out both multiplications and states that the two results agree. . Worth 2 points.

Part B 4 points

Gives the product as the two answers multiplied together. . Worth 2 points.

Produces a pair of numbers, neither equal to $10$ , whose greatest common factor and least common multiple are the ones given. . Worth 2 points.

Part C 5 points

Identifies the greatest common factor as the dividing number and the least common multiple as the other, with a reason for each. . Worth 3 points. needs an explanation, not just an answer

Checks the product rule in this case, observing that the two answers are the two numbers again. . Worth 2 points. needs an explanation, not just an answer
7. A number given only by its primes . 13 points. Question 7 of 10.

Two whole numbers are written out as prime powers: $N = 2^3 \times 3^2 \times 11$ and $M = 2^2 \times 5 \times 7$ . Nothing here needs multiplying out, because the exponents are the whole record of what each number is made of.
1. Part A.
  Decide for each of $18$ , $33$ and $20$ whether it is a factor of $N$ , giving your reason in terms of the copies of each prime.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Give the prime factorization of $N \times M$ in prime-power form, and say how many copies of the prime $2$ it holds.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Two students argue about $N$ . One says every whole number greater than $1$ that divides $N$ has to be one of $2$ , $3$ and $11$ . The other says no prime outside $2$ , $3$ and $11$ can divide $N$ . Decide each claim.
  
  Carry your own answer forward You may use your conclusions from part A, whatever they were.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$18$ and $33$ are factors of $N$ . $20$ is not, because it asks for a copy of the prime $5$ and $N$ holds none.

Part B

$N \times M = 2^5 \times 3^2 \times 5 \times 7 \times 11$ , which holds five copies of the prime $2$ .

Part C

The first is wrong: $6$ divides $N$ and is none of the three, since a factor may be built from several of the primes at once. The second is right: a prime dividing $N$ would have to appear in the single factorization $N$ has, and only $2$ , $3$ and $11$ appear there.

Worked solution

Part A

A number divides $N$ exactly when every prime it asks for is present in $N$ in at least the quantity it asks for. So factor each candidate and compare tallies.

$18 = 2 \times 3^2, \qquad 33 = 3 \times 11, \qquad 20 = 2^2 \times 5$

Against $N = 2^3 \times 3^2 \times 11$ : the first asks for one $2$ and two $3$ s, and $N$ has three $2$ s and two $3$ s, so it fits; the second asks for one $3$ and one $11$ , and both are there, so it fits too.

The third asks for a copy of $5$ . Nothing in $N$ can supply it, because $N$ has exactly one prime factorization and $5$ does not appear in it, so $20$ is not a factor.

Part B

Multiplying two numbers sets their factorizations side by side, so counting a prime inside the product means adding the two counts. The prime $2$ appears three times in $N$ and twice in $M$ .

$N \times M = (2^3 \times 3^2 \times 11) \times (2^2 \times 5 \times 7)$

Gathering equal primes and listing them from smallest to largest,

$N \times M = 2^5 \times 3^2 \times 5 \times 7 \times 11$

so the product holds five copies of $2$ . Every other prime appears in only one of the two numbers, so it enters the product at the power it already had.

Part C

The two claims sound alike and are not, because one is about factors and the other about PRIME factors.

The first is false, and one factor settles it.

$6 = 2 \times 3, \qquad N = 6 \times (2^2 \times 3 \times 11)$

So $6$ divides $N$ without being one of the three primes. A factor may be assembled from any of the copies $N$ owns, and $4$ , $9$ , $22$ and $33$ divide $N$ for the same reason.

The second is true, and uniqueness is what makes it true. Suppose a prime $p$ divided $N$ , so that $N = p \times k$ for some whole number $k$ . Factoring $k$ into primes and putting $p$ in front produces a prime factorization of $N$ that contains $p$ . But $N$ has exactly one prime factorization apart from order, and the one it has is $2^3 \times 3^2 \times 11$ , so $p$ has to be $2$ , $3$ or $11$ .

That is the whole use of the theorem here: a prime cannot hide inside a number without showing up in its factorization.

In one line

Of the three candidates, $18 = 2 \times 3^2$ and $33 = 3 \times 11$ divide $N$ , while $20 = 2^2 \times 5$ does not, since $N$ holds no copy of $5$ . Multiplying adds the prime counts, so $N \times M = 2^5 \times 3^2 \times 5 \times 7 \times 11$ , with five copies of $2$ . The first student is wrong, because $6$ divides $N$ and is not one of its primes; the second is right, because a prime dividing $N$ would have to appear in the one prime factorization $N$ has.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Factors each candidate and compares its prime counts against those of $N$ . . Worth 2 points.

Gives the right verdict for all three, rejecting $20$ because the prime $5$ is absent from $N$ . . Worth 2 points.

Part B 4 points

Adds the counts of each prime across the two numbers rather than multiplying or listing them twice. . Worth 2 points.

Writes the result in prime-power form and reports five copies of the prime $2$ . . Worth 2 points.

Part C 5 points

Rejects the first claim with a factor of $N$ that is not one of its primes. . Worth 2 points. needs an explanation, not just an answer

Accepts the second claim and argues from the uniqueness of the prime factorization, not from having checked the small primes. . Worth 3 points. needs an explanation, not just an answer
8. What a third number changes . 14 points. Question 8 of 10.

Three whole numbers are on the table: $30$ , $42$ and $70$ . The parts below ask for the two answers a set of prime factorizations gives, and then for what becomes of them when the list gets shorter.
1. Part A.
  Write each of the three numbers in prime-power form, then give $\operatorname{GCF}(30, 42, 70)$ and $\operatorname{LCM}(30, 42, 70)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Now drop $70$ from the list. Give the greatest common factor and the least common multiple of the two numbers left, and say for each answer whether it rose, fell or stayed.
  
  Carry your own answer forward Compare the two answers here with the pair you produced in part A, whatever they came to.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain why dropping a number can never lower a greatest common factor or raise a least common multiple, and account for what each of your two answers did.
  
  Carry your own answer forward Account for the change, or the lack of one, between your own answers in parts A and B.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

$30 = 2 \times 3 \times 5$ , $42 = 2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$ , so $\operatorname{GCF} = 2$ and $\operatorname{LCM} = 210$ .

Part B

For $30$ and $42$ the greatest common factor is $6$ , which has risen, and the least common multiple is $210$ , which has stayed where it was.

Part C

A common factor of fewer numbers has fewer conditions to meet, so the greatest common factor can only rise or stay; a common multiple of fewer numbers faces fewer demands, so the least common multiple can only fall or stay. It stayed at $210$ because $42$ still supplies the $7$ and $30$ still supplies the $5$ .

Worked solution

Part A

Take the three factorizations first.

$30 = 2 \times 3 \times 5, \qquad 42 = 2 \times 3 \times 7, \qquad 70 = 2 \times 5 \times 7$

For the greatest common factor, keep only the primes every one of the three owns, each at its lowest power. The prime $3$ is missing from $70$ , the prime $5$ from $42$ and the prime $7$ from $30$ , so all three drop out and only $2$ survives.

$\operatorname{GCF}(30, 42, 70) = 2$

For the least common multiple, take the highest power of every prime that appears anywhere.

$\operatorname{LCM}(30, 42, 70) = 2 \times 3 \times 5 \times 7 = 210$

Both ends check: $2$ divides all three, and $210 \div 30 = 7$ , $210 \div 42 = 5$ , $210 \div 70 = 3$ .

Part B

With $70$ gone, only $30$ and $42$ have a say.

$30 = 2 \times 3 \times 5, \qquad 42 = 2 \times 3 \times 7$

The primes in both are now $2$ and $3$ , each once in each number, so the greatest common factor picks up a factor it could not have before.

$\operatorname{GCF}(30, 42) = 2 \times 3 = 6$

That has risen from $2$ , because $70$ was the number holding no copy of $3$ . For the least common multiple, the highest powers are unchanged.

$\operatorname{LCM}(30, 42) = 2 \times 3 \times 5 \times 7 = 210$

That is where it already was.

Part C

Think of each number in the list as imposing conditions, prime by prime.

A common factor must fit inside every number on the list, so each number caps how many copies of a prime the factor may carry. Removing a number removes caps and never adds one, so everything that was a common factor still is, and there may be more besides. The greatest of a larger collection cannot be smaller, so the answer rises or stays.

A common multiple must cover every number on the list, so each number sets a floor under how many copies of a prime the multiple must hold. Removing a number removes a floor and never adds one, so everything that was a common multiple still is, and the least of a larger collection cannot be larger.

Here the caps were what moved. The number $70$ holds no copy of $3$ , so while it was on the list no common factor could carry a $3$ ; once it went, the $3$ that $30$ and $42$ both own became available.

$2 \longrightarrow 2 \times 3 = 6$

The floors did not move at all, because no prime had its highest count held by $70$ alone.

$5 \text{ sits in } 30, \qquad 7 \text{ sits in } 42$

Every prime that $70$ demanded was already demanded by one of the other two, at the same power, so the least common multiple had nothing to give up and stayed at $210$ .

In one line

From $30 = 2 \times 3 \times 5$ , $42 = 2 \times 3 \times 7$ and $70 = 2 \times 5 \times 7$ , the greatest common factor is $2$ and the least common multiple is $210$ . Dropping $70$ raises the greatest common factor to $6$ and leaves the least common multiple at $210$ . A shorter list caps a common factor less and demands less of a common multiple, so the first answer can only rise or stay and the second can only fall or stay; here the second stayed because $30$ still supplies the $5$ and $42$ still supplies the $7$ that $70$ had been contributing.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Writes all three prime-power forms correctly. . Worth 1 point.

Keeps for the greatest common factor only the primes every one of the three owns, so that $3$ , $5$ and $7$ all drop out. . Worth 2 points.

Takes for the least common multiple the highest power of every prime appearing anywhere, and reaches $210$ . . Worth 2 points.

Part B 4 points

Recomputes both answers for the remaining pair. . Worth 2 points.

Says of each answer whether it rose, fell or stayed, against the answers from part A. . Worth 2 points.

Part C 5 points

Argues the direction of each change from what a shorter list does to the conditions: fewer caps for a common factor, fewer floors for a common multiple. . Worth 3 points. needs an explanation, not just an answer

Explains the unchanged least common multiple by noting that both primes $70$ contributed are still held, at the same power, by one of the remaining numbers. . Worth 2 points. needs an explanation, not just an answer
9. Two numbers under trial division . 15 points. Question 9 of 10.

Trial division settles a number in one of two ways: it turns up a factor, or it runs out of candidates. Where it runs out is fixed by the number itself, not by how many divisions have already failed.
1. Part A.
  Decide whether $337$ is prime, saying which candidate divisors you tried and showing that your search went far enough to settle it.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Decide whether $611$ is prime, naming the divisor that settles it if there is one.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate tries $2$ , $3$ , $5$ , $7$ and $11$ on $611$ , finds no factor, and stops there, saying that five failed divisions in a row are enough to settle it. Say what the stopping rule actually permits, whether this stop was permitted, and what it would take to complete the search.
  
  Carry your own answer forward Judge the classmate against your own search in part B, whichever candidates you tried and whatever verdict you reached.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
The answer

Part A

$337$ is prime. The obliged candidates are $2$ , $3$ , $5$ , $7$ , $11$ , $13$ and $17$ , and $19$ is the first that may be skipped.

Part B

$611$ is composite. The divisor $13$ settles it, since $611 = 13 \times 47$ .

Part C

Five failed divisions settle nothing. The search may stop only once a candidate times itself passes the number, and $13 \times 13 = 169$ is far below $611$ , so $13$ was still owed. It divides, giving $611 = 13 \times 47$ , so the classmate has called a composite number prime.

Worked solution

Part A

Test candidates while the candidate times itself does not pass the number.

$17 \times 17 = 289 \le 337, \qquad 19 \times 19 = 361 > 337$

So the obliged list runs $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , and $19$ is the first candidate the stopping rule excuses. The quick tests clear the first three: $337$ is odd, its digits add to $13$ , and it does not end in $0$ or $5$ . The rest need division.

$337 = (7 \times 48) + 1 = (11 \times 30) + 7 = (13 \times 25) + 12 = (17 \times 19) + 14$

Every one leaves a remainder, so no factor pair exists and $337$ is prime.

Part B

The quick tests all fail: $611$ is odd, its digits add to $8$ , and it does not end in $0$ or $5$ . So the search moves up through the candidates.

$611 = (7 \times 87) + 2, \qquad 611 = (11 \times 55) + 6$

Neither fits, and the next candidate is $13$ .

$611 \div 13 = 47$

That is a third factor besides $1$ and $611$ , so the number is composite and the search can stop the moment it appears.

Part C

The rule is not about how many divisions have failed. It is about where the factor pairs can still be hiding.

$24 \times 24 = 576 \le 611, \qquad 25 \times 25 = 625 > 611$

So every candidate up to $24$ is owed, which means the primes $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , $19$ and $23$ . The classmate stopped after five of the nine.

What the rule guarantees is that the smaller partner of any factor pair is small enough that multiplying it by itself does not pass the number. Stopping early throws that guarantee away: a factor pair whose smaller partner is $13$ , $17$ , $19$ or $23$ would go unseen.

$611 = 13 \times 47$

The very next candidate was the one that worked, so the stop cost the whole verdict: the classmate would report a composite number as prime. A run of failed divisions is evidence of nothing on its own, since every composite number fails a great many divisions before the one that succeeds.

In one line

$337$ is prime: the obliged candidates are $2$ , $3$ , $5$ , $7$ , $11$ , $13$ and $17$ , since $17 \times 17 = 289$ does not pass $337$ while $19 \times 19 = 361$ does, and every one of them leaves a remainder. $611$ is composite, settled by $13$ , since $611 = 13 \times 47$ . The classmate's stop after five failed divisions is not permitted: the rule ends the search only once a candidate times itself passes the number, which for $611$ means going as far as $23$ , and the very next candidate, $13$ , was the one that divides.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Tries the candidate divisors in order and fixes where the search may stop by comparing a candidate times itself with $337$ . . Worth 2 points.

Carries out or dismisses every division that is owed, and reaches a verdict. . Worth 2 points.

Shows the search was complete by naming the first candidate whose square passes $337$ . . Worth 1 point.

Part B 5 points

Runs the quick tests first and continues past them to the larger candidates. . Worth 2 points.

Finds the divisor $13$ and gives the factor pair $611 = 13 \times 47$ . . Worth 2 points.

Reads that factor pair back as a classification: a third factor besides $1$ and the number itself makes $611$ composite. . Worth 1 point.

Part C 5 points

States the stopping rule as a comparison of a candidate times itself with the number, not as a count of failed divisions. . Worth 2 points. needs an explanation, not just an answer

Names the candidates still owed when the classmate stopped, and identifies $13$ among them. . Worth 2 points.

Completes the search, or says what completing it would take, and reports the verdict that follows. . Worth 1 point. needs an explanation, not just an answer
10. Carrying a rule to a third number . 15 points. Question 10 of 10.

For two whole numbers, the greatest common factor and the least common multiple multiply back to the two numbers multiplied together. The parts below put that rule to work on a pair, and then on a set of three.
1. Part A.
  For $45$ and $60$ , give the greatest common factor and the least common multiple, and check that the two answers multiply to $45 \times 60$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Now bring in a third number, $50$ . Give $\operatorname{GCF}(45, 50, 60)$ and $\operatorname{LCM}(45, 50, 60)$ , multiply those two answers together, and compare the result with $45 \times 50 \times 60$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The rule for two numbers turns on a pair of counts holding nothing but a lower one and a higher one. Using the prime $5$ in part B as your case, account for the comparison you reached there. Then decide a classmate's claim that for three numbers the two answers can never multiply to the product.
  
  Carry your own answer forward Argue from the counts in the three factorizations you wrote in part B, whatever answers they gave you.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

$\operatorname{GCF}(45, 60) = 15$ and $\operatorname{LCM}(45, 60) = 180$ , and $15 \times 180 = 45 \times 60 = 2{,}700$ .

Part B

$\operatorname{GCF}(45, 50, 60) = 5$ and $\operatorname{LCM}(45, 50, 60) = 900$ , so the two answers give $4{,}500$ , while $45 \times 50 \times 60 = 135{,}000$ . The two are not equal.

Part C

A third count sits between the lowest and the highest, and neither answer reaches for it: at the prime $5$ the counts are $1$ , $2$ and $1$ , so the two answers collect three copies where the product holds four. The claim is still wrong: $5$ , $8$ and $9$ share no prime, and $1 \times 360 = 5 \times 8 \times 9$ .

Worked solution

Part A

Write both in prime-power form and go prime by prime.

$45 = 3^2 \times 5, \qquad 60 = 2^2 \times 3 \times 5$

The shared primes are $3$ and $5$ . Lowest powers give the greatest common factor, highest powers give the least common multiple.

$\operatorname{GCF}(45, 60) = 3 \times 5 = 15, \qquad \operatorname{LCM}(45, 60) = 2^2 \times 3^2 \times 5 = 180$

Now multiply each pair.

$15 \times 180 = 2{,}700, \qquad 45 \times 60 = 2{,}700$

They agree, because at every prime the two answers take the lower count and the higher count, and a pair of counts holds nothing else.

Part B

Add the third factorization and run both rules across all three columns.

$45 = 3^2 \times 5, \qquad 50 = 2 \times 5^2, \qquad 60 = 2^2 \times 3 \times 5$

Only $5$ sits in all three, at one copy in the poorest of them, so that is the whole of the greatest common factor. For the least common multiple take the highest power of each prime anywhere: $2^2$ from $60$ , $3^2$ from $45$ and $5^2$ from $50$ .

$\operatorname{GCF} = 5, \qquad \operatorname{LCM} = 2^2 \times 3^2 \times 5^2 = 900$

Now set the two products side by side.

$5 \times 900 = 4{,}500, \qquad 45 \times 50 \times 60 = 135{,}000$

They are not equal, and not even close: the product is $30$ times the other.

Part C

With two numbers, sorting a pair of counts changes nothing about what the pair holds, so the lower and the higher between them use the pair up, and the two answers hold exactly as many copies of each prime as the product does. Three counts have a middle one, and neither answer ever reaches for it.

Watch the prime $5$ across the three numbers of part B. It appears once in $45$ , twice in $50$ and once in $60$ .

$\text{lowest } 1, \quad \text{middle } 1, \quad \text{highest } 2$

The greatest common factor takes one copy and the least common multiple takes two, three in all, while the product $45 \times 50 \times 60$ holds $1 + 2 + 1 = 4$ . One copy of $5$ goes uncollected. The primes $2$ and $3$ each lose a copy the same way, which is exactly the factor of $30$ separating $4{,}500$ from $135{,}000$ .

The classmate's claim goes too far, though. Nothing is lost when there is no middle copy to lose, and that happens whenever no two of the three numbers share a prime.

$5 = 5, \qquad 8 = 2^3, \qquad 9 = 3^2$

Here the greatest common factor is $1$ and the least common multiple is $2^3 \times 3^2 \times 5 = 360$ , which is the product itself, so the two answers do multiply to $5 \times 8 \times 9 = 360$ . The honest statement is that there is no RULE of that shape for three numbers, not that it always fails.

In one line

For $45$ and $60$ the answers are $15$ and $180$ , and $15 \times 180 = 45 \times 60 = 2{,}700$ . Bringing in $50$ gives $5$ and $900$ , whose product $4{,}500$ falls well short of $45 \times 50 \times 60 = 135{,}000$ , because a third number contributes a middle count at each prime that neither answer collects: at the prime $5$ the counts are $1$ , $1$ and $2$ , so three copies are collected where the product holds four. The classmate goes too far all the same, since $5$ , $8$ and $9$ share no prime and give $1 \times 360 = 5 \times 8 \times 9$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Finds both answers from the prime powers, taking lowest powers for one and highest for the other. . Worth 2 points.

Carries out both multiplications and reports that they agree. . Worth 2 points.

Part B 5 points

Applies both rules across all three numbers, dropping from the greatest common factor every prime that any one of them lacks. . Worth 2 points.

Reaches $5$ and $900$ , and multiplies them out. . Worth 2 points.

Compares the two results and states plainly that they differ. . Worth 1 point.

Part C 6 points

Identifies the middle count as what neither answer collects, and tracks it at the prime $5$ through the counts $1$ , $1$ and $2$ . . Worth 3 points. needs an explanation, not just an answer

Rejects the classmate's claim and exhibits three numbers, no two sharing a prime, for which the two answers do multiply to the product. . Worth 3 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. One numeral, five tests 11 points

Part A 3

Applies each test to the block of digits that test reads, and gives the right verdict for all five divisors. 2
Names which part of the numeral each test reads: the last digit for $2$ , $5$ and $10$ , the last two for $4$ , the last three for $8$ . 1

Part B 3

Finds every digit making the digit sum a multiple of $9$ , and every digit for which the last two digits form a multiple of $4$ . 2
Identifies the single digit that satisfies both conditions at once. 1

Part C 5

Separates the verdict from the reasoning, accepting the first and rejecting the second. 2
Says that the test for $8$ reads the last three digits because $1000$ is a multiple of $8$ while $100$ is not. 2
Supplies a four-digit number on which the student's two-digit rule and the true verdict disagree. 1

2. What the factor count decides 12 points

Part A 4

Lists all six factors of $92$ and all four factors of $301$ , with no repeats and none missing. 2
Classifies each number from its count of distinct factors rather than from its appearance. 2

Part B 3

Gives one factor for $1$ and two for $2$ , counting the number itself once only. 2
Reads each classification off the count, placing $1$ outside both groups and $2$ among the primes. 1

Part C 5

Rejects the first half by pointing at $1$ , and says that the definition counts factors rather than searching below the number. 2
Rejects the second half in both directions, with a prime it wrongly calls composite and a composite it wrongly lets through. 3

3. A tree and a column 12 points

Part A 3

Splits every composite branch again, so that all five leaves are prime. 2
Collects the repeated primes into powers and lists the primes from smallest to largest. 1

Part B 4

Divides by the smallest prime that fits at each step and carries on until the quotient reaches $1$ . 2
Names the test behind each divisor: an even last digit for $2$ , a digit sum of $12$ for $3$ , and division for $11$ . 1
Gives the answer in prime-power form with both exponents right. 1

Part C 5

States the stopping condition as every leaf being prime, and says why a composite leaf can always be split again. 2
Says the exponent counts copies of the prime that are multiplied together. 1
Rejects the offered form because $25$ is composite, grants that its multiplication is correct, and repairs it to $2^2 \times 5^2 \times 11$ . 2

4. Kits and reorders 13 points

Part A 4

Sets the number of kits up as a number dividing both stocks, and takes the greatest such number. 2
Takes the lowest power of each shared prime and reaches $22$ . 1
Reports what one kit holds, in pens and in notebooks. 1

Part B 4

Recognizes the meeting day as a common multiple of the two cycles and asks for the least one. 2
Takes the highest power of each prime, including all three copies of $3$ from $27$ . 1
Reports the answer as a number of days. 1

Part C 5

Attaches the lowest powers to the sharing question and the highest powers to the coinciding question, quoting the wording that decides it. 3
Produces $770$ as the value the other rule would have given in part A, and says why a count of kits larger than the stock is impossible. 2

5. Where a test may be split 14 points

Part A 5

Writes both splits correctly, as a multiple of $100$ plus $48$ and as a multiple of $1000$ plus $848$ . 2
Argues that the leading piece is a multiple of the divisor, quoting $100 = 4 \times 25$ and $1000 = 8 \times 125$ , so the verdict rests on the leftover alone. 3

Part B 4

Gives the right verdict for all four divisors. 2
Names what each test reads, and treats $6$ as the two tests for $2$ and $3$ run together. 2

Part C 5

Rejects the move as a general rule and backs the verdict with a number passing both tests and failing the test for $24$ . 3
States the condition, that the two factors share no factor above $1$ , and explains what goes wrong when they do. 2

6. One answer inside the other 13 points

Part A 4

Takes the lowest power of each shared prime for one answer and the highest power of every prime for the other. 2
Carries out both multiplications and states that the two results agree. 2

Part B 4

Gives the product as the two answers multiplied together. 2
Produces a pair of numbers, neither equal to $10$ , whose greatest common factor and least common multiple are the ones given. 2

Part C 5

Identifies the greatest common factor as the dividing number and the least common multiple as the other, with a reason for each. 3
Checks the product rule in this case, observing that the two answers are the two numbers again. 2

7. A number given only by its primes 13 points

Part A 4

Factors each candidate and compares its prime counts against those of $N$ . 2
Gives the right verdict for all three, rejecting $20$ because the prime $5$ is absent from $N$ . 2

Part B 4

Adds the counts of each prime across the two numbers rather than multiplying or listing them twice. 2
Writes the result in prime-power form and reports five copies of the prime $2$ . 2

Part C 5

Rejects the first claim with a factor of $N$ that is not one of its primes. 2
Accepts the second claim and argues from the uniqueness of the prime factorization, not from having checked the small primes. 3

8. What a third number changes 14 points

Part A 5

Writes all three prime-power forms correctly. 1
Keeps for the greatest common factor only the primes every one of the three owns, so that $3$ , $5$ and $7$ all drop out. 2
Takes for the least common multiple the highest power of every prime appearing anywhere, and reaches $210$ . 2

Part B 4

Recomputes both answers for the remaining pair. 2
Says of each answer whether it rose, fell or stayed, against the answers from part A. 2

Part C 5

Argues the direction of each change from what a shorter list does to the conditions: fewer caps for a common factor, fewer floors for a common multiple. 3
Explains the unchanged least common multiple by noting that both primes $70$ contributed are still held, at the same power, by one of the remaining numbers. 2

9. Two numbers under trial division 15 points

Part A 5

Tries the candidate divisors in order and fixes where the search may stop by comparing a candidate times itself with $337$ . 2
Carries out or dismisses every division that is owed, and reaches a verdict. 2
Shows the search was complete by naming the first candidate whose square passes $337$ . 1

Part B 5

Runs the quick tests first and continues past them to the larger candidates. 2
Finds the divisor $13$ and gives the factor pair $611 = 13 \times 47$ . 2
Reads that factor pair back as a classification: a third factor besides $1$ and the number itself makes $611$ composite. 1

Part C 5

States the stopping rule as a comparison of a candidate times itself with the number, not as a count of failed divisions. 2
Names the candidates still owed when the classmate stopped, and identifies $13$ among them. 2
Completes the search, or says what completing it would take, and reports the verdict that follows. 1

10. Carrying a rule to a third number 15 points

Part A 4

Finds both answers from the prime powers, taking lowest powers for one and highest for the other. 2
Carries out both multiplications and reports that they agree. 2

Part B 5

Applies both rules across all three numbers, dropping from the greatest common factor every prime that any one of them lacks. 2
Reaches $5$ and $900$ , and multiplies them out. 2
Compares the two results and states plainly that they differ. 1

Part C 6

Identifies the middle count as what neither answer collects, and tracks it at the prime $5$ through the counts $1$ , $1$ and $2$ . 3
Rejects the classmate's claim and exhibits three numbers, no two sharing a prime, for which the two answers do multiply to the product. 3

Multiple choice

Free response