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Foundations of Arithmetic: 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, 120 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 figure, corrected twice . 11 points. Question 1 of 10.

A stock ledger records the figure $52{,}608$ , and two separate corrections to it are proposed. Each correction is added to the figure as it was originally recorded.
1. Part A.
  Adding $300$ to $52{,}608$ changes exactly one of its digits. Name the column that changes, give the digit that replaces the one standing there now, and write the corrected figure.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  The second proposal adds $500$ to $52{,}608$ instead. Write $52{,}608$ in expanded form, add the correction to the term it belongs with, and produce the corrected figure. More than one digit changes this time; say which ones.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain how a correction as small as $500$ can reach beyond the column it was added to, and say what the two corrections together show about one column and the column on its left.
  
  Carry your own answer forward Argue from the two corrected figures you produced above, whichever they were. What is marked here is the account of why one correction stayed inside a single column and the other did not.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

The hundreds column changes, its $6$ becoming a $9$ , and the corrected figure is $52{,}908$ .

Part B

$52{,}608 = 50{,}000 + 2{,}000 + 600 + 8$ , and $600 + 500 = 1{,}100$ , so the figure becomes $53{,}108$ . The thousands digit and the hundreds digit both change.

Part C

A column holds one digit, so it carries at most nine of its own units. Eleven hundreds will not fit, and ten of them are exactly one thousand, so the excess moves one column left. Each column is worth ten of its right-hand neighbour, which is why ten of anything becomes one of the next thing along.

Worked solution

Part A

The correction is three hundreds, so it belongs in the hundreds column and touches no other. That column holds a $6$ , and $6 + 3 = 9$ still fits inside a single column:

$52{,}608 + 300 = 50{,}000 + 2{,}000 + (600 + 300) + 8 = 50{,}000 + 2{,}000 + 900 + 8 = 52{,}908.$

Every other column is left exactly as it was, because nothing was added to any of them.

Part B

Expanded form is the sum of what each digit contributes, so every column gets its own term and the correction has an obvious home:

$52{,}608 = 50{,}000 + 2{,}000 + 600 + 8, \qquad 600 + 500 = 1{,}100.$

But $1{,}100$ is eleven hundreds, and a column holds a single digit, so it cannot hold eleven of anything. Ten of those hundreds are one thousand, and they move one column to the left, leaving one hundred behind:

$50{,}000 + 2{,}000 + 1{,}100 + 8 = 50{,}000 + 3{,}000 + 100 + 8 = 53{,}108.$

The thousands digit has gone from $2$ to $3$ and the hundreds digit from $6$ to $1$ . A correction made entirely in the hundreds column has reached the column to its left.

Part C

Nothing about the correction is large; what matters is whether the column it lands in can hold the result. A column holds a single digit, so it tops out at nine of its own units. The first correction left six hundreds as nine hundreds and fitted:

$6 + 3 = 9.$

The second asked the same column to hold eleven hundreds, which no column can do. Ten hundreds are exactly one thousand:

$10 \times 100 = 1{,}000,$

so ten of them leave as a single unit of the next column and one hundred stays. That is the ten-times relationship seen from the inside: each column is worth ten of the column on its right, so ten units of any column are one unit of its left-hand neighbour, and an overflow always lands exactly one column over.

In one line

Adding $300$ changes only the hundreds digit, giving $52{,}908$ . Adding $500$ instead makes eleven hundreds, and since a column holds a single digit, ten of them bundle into one thousand: $52{,}608 + 500 = 53{,}108$ , where the thousands and the hundreds digits have both changed. A column cannot hold ten of its own units, and ten of them are exactly one unit of the column to its left.

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

Adds the correction in the hundreds column alone, leaving every other column as it was. . Worth 2 points.

Names the column that changes and the digit now standing in it, not merely the corrected figure. . Worth 1 point.

Part B 4 points

Writes the expanded form and adds the correction to the hundreds term rather than to the numeral as a whole. . Worth 2 points.

Bundles ten of the eleven hundreds into one thousand and reports the figure that results. . Worth 1 point.

Names both of the digits that change. . Worth 1 point.

Part C 4 points

Explains the overflow from a column holding only one digit, so ten of its units cannot remain in it. . Worth 2 points. needs an explanation, not just an answer

States the ten-times relationship between neighbouring columns and ties it to why the excess lands exactly one column to the left. . Worth 2 points. needs an explanation, not just an answer
2. What the rearrangement needed . 11 points. Question 2 of 10.

A cashier faces the sum $19 + 46 + 81 + 4$ and would rather not add it straight through.
1. Part A.
  Rearrange and regroup the sum so that it can be done in your head. Show the pairs you form and what each comes to, and give the total.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Give the value of $(29 \times 0) + (1 \times 36) + (0 + 45)$ without multiplying anything out, and name the property that settles each of the three brackets.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A classmate says the rearrangement in part A is licensed by the associative property, since the brackets moved. Decide whether that one property is enough to license everything you did there, and if it is not, name precisely what else was needed and what it did.
  
  Carry your own answer forward Judge the classmate against the rearrangement you actually made in part A. If you paired the numbers differently, name the freedoms your own pairing used.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

Pairing $19$ with $81$ gives $100$ , and pairing $46$ with $4$ gives $50$ , so the total is $100 + 50 = 150$ .

Part B

$0 + 36 + 45 = 81$ . The first bracket is settled by the zero property of multiplication, the second by the multiplicative identity, and the third by the additive identity.

Part C

It is not enough. Associativity chooses which pair is combined first and leaves every number where it stands, so it cannot bring the $81$ past the $46$ to sit beside the $19$ . That move is a change of order, which is the commutative property. Part A used both, and neither does the other's work.

Worked solution

Part A

Two pairs here fall on round numbers, and a sum permits both the moves needed to bring them together:

$19 + 46 + 81 + 4 = (19 + 81) + (46 + 4) = 100 + 50 = 150.$

Adding straight through reaches the same total by heavier steps, passing $65$ , then $146$ , then $150$ .

Part B

Each bracket is settled by a property rather than by arithmetic:

$29 \times 0 = 0, \qquad 1 \times 36 = 36, \qquad 0 + 45 = 45.$

Multiplying by $0$ collapses whatever it meets, multiplying by $1$ hands it back untouched, and adding $0$ hands it back untouched. Only the addition is left:

$0 + 36 + 45 = 81.$

The two identities are not one property with two faces: $0$ is the number that changes nothing under addition, $1$ is the number that changes nothing under multiplication, and $0$ under multiplication does the opposite of both.

Part C

The two freedoms are different, and part A used them in turn.

The associative property changes which pair is combined first and moves nothing:

$(19 + 46) + 81 = 19 + (46 + 81).$

Every number is still in its original position, and in that position the $46$ sits between the $19$ and the $81$ . So associativity on its own cannot produce the pairing part A wanted.

Bringing the $81$ across is a change of ORDER, which is the commutative property:

$19 + 46 + 81 + 4 = 19 + 81 + 46 + 4.$

Only once the numbers stand in that order does the grouping $(19 + 81) + (46 + 4)$ become available, and grouping it is the associative move. The classmate has named the second of the two and missed the first.

In one line

$19 + 46 + 81 + 4 = (19 + 81) + (46 + 4) = 100 + 50 = 150$ , and $(29 \times 0) + (1 \times 36) + (0 + 45) = 0 + 36 + 45 = 81$ , settled by the zero property, the multiplicative identity and the additive identity in turn. One property does not license part A: associativity chooses which pair goes first but moves nothing, so bringing the $81$ next to the $19$ needed the commutative property as well.

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

Forms pairs that fall on round numbers, using each of the four numbers exactly once. . Worth 2 points.

Reports the total the rearranged sum gives. . Worth 1 point.

Part B 3 points

Settles all three brackets correctly without carrying out a multiplication. . Worth 2 points.

Names a different property for each bracket, keeping the two identities apart from the zero property. . Worth 1 point.

Part C 5 points

Decides against the classmate and separates a change of grouping from a change of order, saying what each does to the written sum. . Worth 3 points. needs an explanation, not just an answer

Names the property the classmate left out and identifies the specific move in part A that needed it. . Worth 2 points. needs an explanation, not just an answer
3. One line, one operation at a time . 12 points. Question 3 of 10.

Each part below is about the single number a written line names, and about what changes when the writing changes.
1. Part A.
  Evaluate $40 - 2 \times 3^3 \div 6$ , settling one operation at a time.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Evaluate $\dfrac{60 - 4 \times 3}{4 + 8}$ .
  
  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 $40 - 2 \times 3^3 \div 6 = 40 - 2 \times 9 \div 6 = 37$ . Point to the earliest expression in that chain that is already wrong, account for how the student arrived at it, and finish the line correctly.
  
  Carry your own answer forward Compare the student's chain with your own part A line. Should part A have gone astray, settle the line again by any route you trust, then judge the chain against that.
  
  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

The expression names $31$ .

Part B

The expression names $4$ .

Part C

The second expression is the first wrong one: $3^3$ has been read as $3 \times 3$ . A power is repeated multiplication of the base, and the exponent counts the copies, so $3^3 = 27$ rather than $9$ . The tiers are then applied faultlessly to a wrong number, and the line should reach $31$ .

Worked solution

Part A

The highest tier present is the exponent, so the power is settled first. A power is repeated multiplication, and the exponent counts the copies of the base:

$40 - 2 \times 3^3 \div 6 = 40 - 2 \times 27 \div 6, \qquad \text{since } 3^3 = 3 \times 3 \times 3 = 27.$

Multiplication and division share the next tier down and run left to right, the multiplication first because it stands further left:

$40 - 2 \times 27 \div 6 = 40 - 54 \div 6 = 40 - 9 = 31.$

Subtraction is on the lowest tier of all, so it waits until everything above it has become a single number.

Part B

The bar gathers the whole of what stands above it and the whole of what stands below it, so each has to be settled before any dividing happens. Inside the numerator the multiplication comes first:

$\frac{60 - 4 \times 3}{4 + 8} = \frac{60 - 12}{12} = \frac{48}{12}.$

Only now does the bar act as a division:

$\frac{48}{12} = 4.$

Part C

Follow the chain one expression at a time. The tier order is not what fails here: the student settled the power first, which is right, then ran the multiplication and the division left to right and left the subtraction until last, which is also right.

The second expression is where it breaks. Its $9$ can only have come from multiplying the base by the exponent, $3 \times 3$ . But an exponent is not a factor; it is a count of how many copies of the base are multiplied:

$3^3 = 3 \times 3 \times 3 = 27.$

Working correctly from that line:

$40 - 2 \times 27 \div 6 = 40 - 54 \div 6 = 40 - 9 = 31.$

Everything after the slip is faultless arithmetic, which is exactly what makes it hard to catch: $2 \times 9 = 18$ , then $18 \div 6 = 3$ , then $40 - 3 = 37$ are each correct. The error is in what the notation was taken to mean, not in the working.

In one line

$40 - 2 \times 3^3 \div 6 = 31$ , and $\dfrac{60 - 4 \times 3}{4 + 8} = \dfrac{48}{12} = 4$ . The student's chain first fails at $40 - 2 \times 9 \div 6$ , where $3^3$ has been read as $3 \times 3$ ; an exponent counts copies of the base rather than multiplying it, so $3^3 = 27$ and the line reaches $31$ .

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

Settles the power before any of the operations below it, and treats it as repeated multiplication of the base rather than as a product of base and exponent. . Worth 1 point.

Runs the multiplication and the division left to right, and leaves the subtraction until both are done. . Worth 2 points.

Reports one number as the value of the whole line. . Worth 1 point.

Part B 3 points

Settles the whole numerator and the whole denominator before dividing, rather than dividing one term of the top by one term of the bottom. . Worth 2 points.

Reports one number as the value of the whole expression. . Worth 1 point.

Part C 5 points

Names the first expression in the chain that is wrong, rather than only reporting that the final value is wrong. . Worth 2 points. needs an explanation, not just an answer

Diagnoses the slip precisely, saying what the power was taken to mean and what an exponent actually counts, and notes that the tier order itself was sound. . Worth 2 points. needs an explanation, not just an answer

Gives the corrected value. . Worth 1 point.
4. Six kits, and a price the records lost . 13 points. Question 4 of 10.

A workshop makes up $6$ identical repair kits. Each kit holds a clamp costing $15$ dollars, a blade costing $8$ dollars, and a spool whose price the records do not give. The whole order came to $174$ dollars.
1. Part A.
  Priced by item, the records read $90 + 48$ and then a third term that has been smudged out, the first term being what all six clamps cost and the second what all six blades cost. Work out the smudged term, and from it the price of one spool.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Check that price by pricing a whole kit instead: give what one kit costs, multiply by the number of kits, and say whether the result matches the order.
  
  Carry your own answer forward Use the spool price you arrived at in part A, whatever it came to, and say honestly whether the check comes out.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  The blade price rises by $3$ dollars. Say what has to change in the item-priced form and what has to change in the kit-priced form, give the new total, and say which of the two shows the cost of the rise more directly and why.
  
  Carry your own answer forward Work from the spool price and the cost of a kit that you established above, whatever they were. What is marked here is the account of where the rise lands, not a particular pair of figures.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
The answer
Part A

The smudged term is $36$ dollars, which is what all six spools cost together, so one spool costs $6$ dollars.
- both amounts are wanted: the smudged term is what all six spools cost, and the price of one spool is that term divided by the number of kits, so reporting either alone leaves the other unstated
Part B

One kit costs $15 + 8 + 6 = 29$ dollars, and $6 \times 29 = 174$ dollars, which matches the order exactly.

Part C

In the kit-priced form the cost of a kit rises from $29$ to $32$ dollars; in the item-priced form only the blade term changes, from $48$ to $66$ . The new total is $192$ dollars. The item-priced form shows the rise most directly, because all of it sits in that one term, which grows by $18$ dollars, one rise for each kit.
Worked solution

Part A

The three terms have to account for the whole order between them, so the smudged one is whatever is left:

$90 + 48 + \text{smudged} = 174 \quad \Longrightarrow \quad \text{smudged} = 174 - 138 = 36.$

Every term of a form priced by item is six of one item, because the $6$ has been sent to each price in turn. The first two terms confirm that reading, since $6 \times 15 = 90$ and $6 \times 8 = 48$ . So the $36$ is six spools, and one spool costs

$36 \div 6 = 6.$

Part B

Pricing a kit first gathers the three item prices into one quantity and then takes six of it:

$6 \times (15 + 8 + 6) = 6 \times 29 = 174.$

That is the order as recorded, so the spool price stands. The check is worth making because the two routes reach the total through different quantities: this one through the cost of one kit, the other through the cost of all six of each item.

Part C

A rise on one item touches one place in each form. Priced by kit, the kit itself changes:

$15 + 11 + 6 = 32, \qquad 6 \times 32 = 192.$

Priced by item, only the blade term moves and the other two stand exactly as they were:

$90 + 66 + 36 = 192.$

The item-priced form is the one that shows what the rise costs. Its effect is confined to a single term, and the increase is visibly one rise for each kit:

$6 \times 3 = 18.$

In the kit-priced form that same $18$ dollars is just as real but buried, recoverable only by comparing $6 \times 29$ with $6 \times 32$ .

In one line

The smudged term is $36$ dollars, so one spool costs $6$ dollars, and pricing a kit first checks it: $6 \times (15 + 8 + 6) = 6 \times 29 = 174$ dollars. After a $3$ dollar rise on the blade the order comes to $192$ dollars. Only one place in each form changes, and the item-priced form shows the cost of the rise directly, because all of it sits in the blade term, which grows by $6 \times 3 = 18$ dollars.

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

Uses the whole order to recover the missing term, rather than guessing a price and testing it. . Worth 2 points.

Reads a term of the item-priced form as the number of kits times one item price, and divides by the number of kits to reach a single price. . Worth 2 points.

States both amounts in dollars, keeping what all six spools cost apart from what one spool costs. . Worth 1 point.

Part B 3 points

Prices one whole kit by adding the three item prices, then multiplies by the number of kits. . Worth 2 points.

Reports the total as an amount of money and sets it against the order. . Worth 1 point.

Part C 5 points

Locates the change in exactly one place in each of the two forms, rather than rebuilding either from scratch. . Worth 2 points.

Gives the new total. . Worth 1 point.

Says which form shows the cost of the rise more directly, with the reason it does. . Worth 2 points. needs an explanation, not just an answer
5. Three terms, and where the collecting stops . 13 points. Question 5 of 10.

Two three-term sums are collected below, and then a proposed rule about when the collecting stops is put to the test.
1. Part A.
  Take the greatest factor shared by all three terms out of $42 + 70 + 28$ , and check your answer by multiplying back out.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Do the same for $40 + 60 + 35$ . Then say why the factor you take out is smaller than the factor $40$ and $60$ share between them.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  A classmate looks at your collected form of $40 + 60 + 35$ and says it cannot be finished, because two of the three terms left inside the bracket still share a factor of $4$ . Decide whether the classmate is right, and state the test that settles when a collection is finished.
  
  Carry your own answer forward Judge the classmate against the collected form you produced in part B, whichever form that was.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$42 + 70 + 28 = 14 \times (3 + 5 + 2)$ , and multiplying back out returns $42 + 70 + 28$ .
- $(3 + 5 + 2) \times 14$ is the same product with the factors the other way round, and either side comes to $140$
Part B

$40 + 60 + 35 = 5 \times (8 + 12 + 7)$ . On their own $40$ and $60$ share $20$ , but $20$ does not divide $35$ , and a factor must divide every term, so the shared factor falls to $5$ .

Part C

The classmate is wrong. The test is whether some factor above $1$ divides every term inside the bracket, not whether one pair of them shares a factor. The $7$ shares nothing above $1$ with either of the other two terms, so nothing further can come out and the collection is finished.
Worked solution

Part A

A factor can be taken out only if it divides every term. The numbers dividing $42$ are $1, 2, 3, 6, 7, 14, 21, 42$ ; those dividing $70$ are $1, 2, 5, 7, 10, 14, 35, 70$ ; those dividing $28$ are $1, 2, 4, 7, 14, 28$ . The largest appearing on all three lists is $14$ :

$42 + 70 + 28 = 14 \times 3 + 14 \times 5 + 14 \times 2 = 14 \times (3 + 5 + 2).$

Multiplying back out recovers all three terms, and the total is $14 \times 10 = 140$ .

Part B

The first two terms share a great deal: $40$ and $60$ are both divisible by $4$ , by $10$ and so by $20$ . The third term is what constrains the answer. Since $35 = 5 \times 7$ , it is not even and shares nothing above $5$ with the other two:

$40 + 60 + 35 = 5 \times 8 + 5 \times 12 + 5 \times 7 = 5 \times (8 + 12 + 7).$

Checking, $5 \times 27 = 135$ and $40 + 60 + 35 = 135$ .

A common factor has to divide EVERY term, so one stubborn term can pull the shared factor a long way down, from $20$ to $5$ here.

Part C

The classmate has the right idea and is applying it to the wrong set of numbers. Taking a factor out of a bracket requires that factor to divide EVERY term inside, because what comes out has to come out of all of them at once.

The terms inside are $8$ , $12$ and $7$ . The first two do share a factor of $4$ , exactly as the classmate says. But $7$ is prime and divides neither of them, so

$8 + 12 + 7 = 4 \times 2 + 4 \times 3 + 7$

has no way to become $4$ times a whole number: the $7$ would be left behind.

That is the test worth carrying away, and it is a test over all the terms together. A collection is finished when the terms inside share no factor above $1$ , and a factor shared by some pair of them is not enough to continue.

In one line

$42 + 70 + 28 = 14 \times (3 + 5 + 2)$ , which totals $140$ , and $40 + 60 + 35 = 5 \times (8 + 12 + 7)$ , which totals $135$ . The second takes out only a $5$ because a common factor must divide every term and $35$ shares nothing larger. The classmate is wrong for the same reason read forwards: $8$ and $12$ do share a $4$ , but the $7$ does not, so nothing further can come out and the collection is finished.

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 the greatest factor common to all three terms rather than to two of them. . Worth 2 points.

Writes the sum as that factor multiplying one bracket holding all three remaining terms. . Worth 1 point.

Checks by multiplying back out and recovering all three original terms. . Worth 1 point.

Part B 4 points

Tests candidate factors against all three terms, so that the third term is what settles the answer. . Worth 2 points.

Explains that a factor shared by two of the terms is unavailable unless it divides the third as well. . Worth 2 points. needs an explanation, not just an answer

Part C 5 points

Rejects the classmate's conclusion and grounds the rejection in the term that shares nothing with the other two. . Worth 3 points. needs an explanation, not just an answer

States the finishing test over all the terms inside the bracket rather than over a pair of them. . Worth 2 points. needs an explanation, not just an answer
6. Five digits, and the codes they make . 11 points. Question 6 of 10.

A machine prints five-digit codes, using each of the digits $3$ , $0$ , $8$ , $1$ and $6$ exactly once in every code. No code may begin with $0$ .
1. Part A.
  Write the largest code the machine can print and the smallest, and say what the digit $8$ contributes in each of them.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  How many times larger is the $8$ 's contribution in the largest code than in the smallest? Say how that number could have been predicted from the two codes without working out either contribution.
  
  Carry your own answer forward Use the two codes and the two contributions you produced in part A, whatever they were.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  A colleague offers a rule for the smallest code: write the five digits in order from least to greatest. Give the numeral that rule produces here, show why it is not an answer to the question, and state the smallest repair that fixes the rule, and why nothing else in it has to change.
  
  Carry your own answer forward Compare the colleague's rule against the smaller of the two codes you built in part A, whichever code that was.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
The answer

Part A

The largest is $86{,}310$ , where the $8$ contributes $80{,}000$ . The smallest is $10{,}368$ , where the $8$ contributes $8$ .

Part B

It is $10{,}000$ times larger. The $8$ stands four columns further left in the largest code, and each column is worth ten times the one to its right, so four steps multiply by $10$ four times over.

Part C

The rule gives $01368$ , which is not a five-digit code at all, since no code may begin with $0$ . The repair is to lead with the smallest digit that is not $0$ and then continue from least to greatest, $0$ included, which gives $10{,}368$ .

Worked solution

Part A

The highest column is worth the most, so in the largest code it takes the largest digit, the next column the next largest, and so on down:

$86{,}310 = (8 \times 10{,}000) + (6 \times 1{,}000) + (3 \times 100) + (1 \times 10) + (0 \times 1).$

The smallest code runs the same reasoning the other way, with one exception: the leading column may not take the $0$ , so it takes the smallest digit that is allowed there and the $0$ drops to the next column:

$10{,}368 = (1 \times 10{,}000) + (0 \times 1{,}000) + (3 \times 100) + (6 \times 10) + (8 \times 1).$

The $8$ is the same symbol in both codes and contributes $80{,}000$ in one and $8$ in the other.

Part B

Comparing the two contributions directly:

$80{,}000 \div 8 = 10{,}000.$

The same factor can be read off the positions alone. In the largest code the $8$ sits in the ten-thousands column; in the smallest it sits in the ones column. Those are four columns apart, and each step to the left multiplies a contribution by $10$ :

$10 \times 10 \times 10 \times 10 = 10{,}000.$

So the factor depends only on how far the digit moved, and not at all on which digit it was.

Part C

Written from least to greatest the digits give

$0, 1, 3, 6, 8,$

which as a numeral is $01368$ . A leading $0$ contributes nothing and holds a column that no digit needs, so what is actually named there is the four-digit number $1{,}368$ , and the machine forbids the form outright.

The rest of the colleague's rule is sound, and only the leading column needs repair: put the smallest digit that is not $0$ in front, then place the remaining digits from least to greatest with the $0$ among them. That gives $10{,}368$ .

The repair is the smallest one available in the sense that it changes only which digit may lead. Every column after the first is still filled by the colleague's original rule, which is why the two numerals differ only in their first two digits.

In one line

The largest code is $86{,}310$ and the smallest is $10{,}368$ . The $8$ contributes $80{,}000$ in the first and $8$ in the second, a factor of $10{,}000$ , which is the four columns it moved, each worth ten of the one to its right. The colleague's least-to-greatest rule produces $01368$ , which is not a five-digit code, and the repair is to lead with the smallest digit that is not $0$ and then continue from least to greatest.

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

Builds both codes by filling the highest column first and working down, and respects the rule against a leading zero. . Worth 2 points.

Gives what the $8$ contributes in each code as an amount, not as the name of a column. . Worth 2 points.

Part B 3 points

Reports how many times larger one contribution is than the other, as a factor rather than as a difference. . Worth 1 point.

Predicts the same factor from the number of columns the digit has moved, using the ten-times relation between neighbours. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Produces the numeral the colleague's rule actually gives and identifies what disqualifies it. . Worth 2 points.

States a repair that keeps the rest of the rule, and says why the leading column is the only thing that had to change. . Worth 2 points. needs an explanation, not just an answer
7. One factor outside, two different brackets . 12 points. Question 7 of 10.

These two lines look alike. Each has a factor outside a bracket and two numbers inside it, and what stands between those two numbers is the only difference.

$6 \times (7 + 5) \qquad \text{and} \qquad 6 \times (7 \times 5)$
1. Part A.
  Evaluate $6 \times (7 + 5)$ twice: once by settling the bracket first, and once by sending the $6$ to each number inside it. Report both values.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Now try both of those routes on $6 \times (7 \times 5)$ : settle the bracket first, and then send the $6$ to each number inside. Report both values, and state which of them is the value of the line.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Compare the four values you produced in parts A and B, and say which routes agreed and which did not. Then explain what it is about the two brackets that decides whether the factor may be shared across them, and account for any disagreement you found.
  
  Carry your own answer forward Argue from the four values you produced in parts A and B, whichever they were, and say honestly which of them agreed.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

Settling the bracket first gives $6 \times 12 = 72$ . Sending the $6$ inside gives $6 \times 7 + 6 \times 5 = 42 + 30 = 72$ . The two routes agree at $72$ .

Part B

Settling the bracket first gives $6 \times 35 = 210$ , and that is the value of the line. Sending the $6$ to each number inside gives $42 \times 30 = 1{,}260$ , which is not.

Part C

A sum is a count of parts, so six copies of $(7 + 5)$ really are six copies of $7$ together with six copies of $5$ . A product is already a single quantity built by multiplying, so sending the $6$ into it splits nothing and instead inserts a second factor of $6$ : the failed route multiplied the whole line by an extra $6$ .

Worked solution

Part A

Settling the bracket first treats its contents as one quantity:

$6 \times (7 + 5) = 6 \times 12 = 72.$

Sending the $6$ to each number inside leaves two products to add:

$6 \times 7 + 6 \times 5 = 42 + 30 = 72.$

The two routes agree, which is the distributive property doing exactly what it promises.

Part B

Settling the bracket first:

$6 \times (7 \times 5) = 6 \times 35 = 210.$

Sending the $6$ to each number inside, as though the bracket held a sum:

$(6 \times 7) \times (6 \times 5) = 42 \times 30 = 1{,}260.$

The line is worth $210$ . The second route has not mis-multiplied anything: every product in it is correct, and it still gives the wrong number, because $1{,}260 = 6 \times 210$ .

Part C

Look at what each bracket holds.

$7 + 5$ is a count of parts: twelve units, made of a seven and a five. Taking six copies of it takes six copies of each part, which is why part A's two routes had to agree:

$6 \times (7 + 5) = (7 + 5) + (7 + 5) + (7 + 5) + (7 + 5) + (7 + 5) + (7 + 5) = 6 \times 7 + 6 \times 5.$

$7 \times 5$ is not a count of parts. It is already one quantity, $35$ , built by multiplying, and multiplying it by $6$ adds exactly one further factor. Sending the $6$ to both numbers adds two of them:

$(6 \times 7) \times (6 \times 5) = 6 \times 6 \times 7 \times 5 = 6 \times 210.$

So the failed route did not break any rule of arithmetic; it answered a different question, one whose value is six times the line's. Distribution is the rule tying multiplication to ADDITION and SUBTRACTION, and a bracket holding a product offers it nothing to distribute over.

In one line

$6 \times (7 + 5) = 72$ by either route. $6 \times (7 \times 5) = 210$ , while sending the $6$ to both numbers inside gives $1{,}260$ , which is not the value of the line. A bracket holding a sum is a count of parts, so six copies of it are six copies of each part; a bracket holding a product is already one quantity, and sending the factor into it inserts a second $6$ , which is exactly why $1{,}260 = 6 \times 210$ .

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

Carries out both routes in full and reports a value for each. . Worth 2 points.

States that the two routes agree. . Worth 1 point.

Part B 4 points

Carries out both routes in full and reports a value for each. . Worth 2 points.

Identifies which of the two values the line actually has, rather than offering both as answers. . Worth 2 points.

Part C 5 points

Distinguishes the two brackets by what they hold, a count of parts against a single quantity already built by multiplying. . Worth 3 points. needs an explanation, not just an answer

Says what the failed route did to the line, identifying the extra factor rather than calling the result merely wrong. . Worth 2 points. needs an explanation, not just an answer
8. Two rearrangements, tried on two chains . 11 points. Question 8 of 10.

Addition and multiplication carry two freedoms that let a sum or a product be rearranged before it is worked out. The parts below try two of those rearrangements on a subtraction chain and on a division chain.
1. Part A.
  Work out $80 - 25 - 15$ as it is written. Then work out $80 - 15 - 25$ , and then $80 - (25 - 15)$ . Report all three values, and say which of the two rearrangements changed the value of this chain.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Try those same two rearrangements on $720 \div 12 \div 3$ : first exchange the last two numbers, then group them instead. Report all three values, and say which of the two changed the value of this chain.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Look back at what each rearrangement did to the two chains. Decide whether any of it shows that subtraction and division are commutative after all, and say what the exchange does, in each of the two chains, that a swap of the first two numbers would not.
  
  Carry your own answer forward Argue from the six values you computed above, whatever they came to, and say honestly which rearrangements left a chain's value alone.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$80 - 25 - 15 = 40$ , $80 - 15 - 25 = 40$ , and $80 - (25 - 15) = 70$ . Exchanging the last two numbers left the value alone; grouping them changed it.

Part B

$720 \div 12 \div 3 = 20$ , $720 \div 3 \div 12 = 20$ , and $720 \div (12 \div 3) = 180$ . Here too the exchange left the value alone and the grouping changed it.

Part C

No. Commutativity would claim $80 - 25$ equals $25 - 80$ , and taking $80$ from only $25$ runs out before it finishes. The exchange never moves the first number: in the subtraction it reorders two removals from one starting amount, and in the division it reorders two divisions, which together divide by $12 \times 3$ either way.

Worked solution

Part A

A bare chain of subtractions is read left to right, so each subtraction acts on the running total:

$80 - 25 - 15 = 55 - 15 = 40, \qquad 80 - 15 - 25 = 65 - 25 = 40.$

Grouping the last two numbers is a different calculation altogether:

$80 - (25 - 15) = 80 - 10 = 70.$

So exchanging the two numbers being removed left the value at $40$ , while grouping those same two into a difference gave $70$ instead. Only one of the two rearrangements disturbed the chain.

Part B

Left to right again, with every step a whole number:

$720 \div 12 \div 3 = 60 \div 3 = 20, \qquad 720 \div 3 \div 12 = 240 \div 12 = 20.$

Grouping the last two:

$720 \div (12 \div 3) = 720 \div 4 = 180.$

Division has answered the two rearrangements exactly as subtraction did.

Part C

The two moves are not the same claim, and only one of them was tried above.

Commutativity is a claim about the two numbers an operation stands between. At the head of the subtraction chain it would say

$80 - 25 = 25 - 80,$

which is false: $80 - 25$ leaves $55$ , while taking $80$ away from only $25$ runs out before it finishes and cannot leave $55$ . Division is no better, since $720 \div 12$ is $60$ while $12 \div 720$ asks how many $720$ s fit inside $12$ , and not even one does.

The exchange is a different move, and each chain deserves its own account of why it survived.

In the subtraction chain the first number is the starting amount, and the exchange never moves it. The two numbers after it are both taken away from that amount, one after the other, and the order of two removals cannot change what is left at the end.

In the division chain the same holds one level up. Dividing by $12$ and then by $3$ divides by $12 \times 3$ altogether, while dividing by $3$ and then by $12$ divides by $3 \times 12$ , and those two divisors are equal because MULTIPLICATION is commutative. So the exchange survives on the strength of a property that division does not have itself.

The exchanges were therefore safe, the groupings were not, and neither observation makes subtraction or division commutative.

In one line

$80 - 25 - 15 = 80 - 15 - 25 = 40$ , while $80 - (25 - 15) = 70$ ; and $720 \div 12 \div 3 = 720 \div 3 \div 12 = 20$ , while $720 \div (12 \div 3) = 180$ . In each chain the exchange left the value alone and the grouping did not, and neither makes the operation commutative: commutativity would claim $80 - 25 = 25 - 80$ . The exchange leaves the first number untouched, reordering two removals in one chain and two divisions in the other, whose combined divisor $12 \times 3$ may be reordered because multiplication is commutative.

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

Evaluates all three lines correctly, reading the two bare chains left to right. . Worth 2 points.

Presents the three values together and says which of the two rearrangements changed the value. . Worth 1 point.

Part B 3 points

Evaluates all three lines correctly, keeping every step a whole number. . Worth 2 points.

Says which of the two rearrangements changed the value of this chain. . Worth 1 point.

Part C 5 points

Rejects the commutativity reading and shows what that claim would actually assert about the first two numbers of a chain. . Worth 3 points. needs an explanation, not just an answer

Accounts for the surviving exchange in BOTH chains: two removals from one starting amount, and two successive divisions whose combined divisor is a product that may itself be reordered. . Worth 2 points. needs an explanation, not just an answer
9. One product reached from both sides . 13 points. Question 9 of 10.

The rule tying multiplication to addition and subtraction can be travelled in either direction, and each part below travels it once.
1. Part A.
  Two people work out $9 \times 38$ without ever multiplying by $38$ . One writes $9 \times 40 - 9 \times 2$ , the other writes $9 \times 30 + 9 \times 8$ . Check that each pair of parts really does recover $38$ , evaluate both routes, and give the product.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Two products, $9 \times 26$ and $9 \times 14$ , are to be added. Write their sum as a single product and evaluate it.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  One and the same product turned up in both parts above. Name it, say what part it played in each, describe the direction the rule was travelled each time, say which direction you would reach for when one awkward product has to be worked out and which when a sum of two products has to be, and say what the two forms have in common.
  
  Carry your own answer forward Describe the two directions as you actually travelled them above, whichever splits and whichever shared factor you worked with.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 6 points
The answer

Part A

$40 - 2 = 38$ and $30 + 8 = 38$ , so both splits are honest. They give $360 - 18 = 342$ and $270 + 72 = 342$ , so $9 \times 38 = 342$ .
Part B

$9 \times 26 + 9 \times 14 = 9 \times (26 + 14) = 9 \times 40 = 360$ .
- $(26 + 14) \times 9$ is the same single product with the factors written the other way round
Part C

Part A spread a factor across a split, turning one awkward product into two easy ones, with $9 \times 40$ a stepping stone. Part B collected a shared factor into one product, with $9 \times 40$ the destination. Spread when one product must be worked out, collect when a sum of two must be. Either way both forms name one number.
Worked solution

Part A

A split is usable only if its two parts recover the number that was split:

$40 - 2 = 38, \qquad 30 + 8 = 38.$

Both do, so both routes are available. Sending the $9$ across each of them:

$9 \times 40 - 9 \times 2 = 360 - 18 = 342, \qquad 9 \times 30 + 9 \times 8 = 270 + 72 = 342.$

The product is $342$ either way. Note that the sign joining the two products is the sign inside the split, which is why one route subtracts and the other adds.

Part B

Both products are built on the same factor $9$ , so it stands outside a single bracket while the other two numbers go inside:

$9 \times 26 + 9 \times 14 = 9 \times (26 + 14) = 9 \times 40 = 360.$

The inside falls on a round number, so one easy multiplication replaces two hard ones and an addition. The long way agrees: $234 + 126 = 360$ .

Part C

One distributive pattern, read in opposite directions.

Part A read it left to right. One factor was written as two pieces and the other was spread across both, so a single awkward product became two easy ones and $9 \times 40$ was passed through on the way:

$9 \times (40 - 2) = 9 \times 40 - 9 \times 2.$

Part B read it right to left. Two products already sharing a factor were collected into one, and $9 \times 40$ was where the work finished:

$9 \times 26 + 9 \times 14 = 9 \times (26 + 14).$

Which direction to travel is decided by what you are holding. Holding one product that is awkward to multiply, spread it: the split buys a round number and a small correction. Holding a sum of two products that share a factor, collect it: the collection buys one multiplication in place of three operations, and often a round number inside the bracket.

What the two forms have in common is the only thing that matters at the end. Each side of the equation names the same number, so choosing a direction changes the labour and never the answer.

In one line

$9 \times 38 = 342$ by either split, and $9 \times 26 + 9 \times 14 = 9 \times (26 + 14) = 9 \times 40 = 360$ . The first spreads a factor across a split, which is worth doing when one awkward product must be worked out; the second collects a shared factor into a single product, which is worth doing when a sum of two products must be. Both are the same distributive pattern read in opposite directions, so within each part the two forms name the same number and only the labour changes.

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

Checks each split against the number it claims to split before using it. . Worth 2 points.

Evaluates both routes, joining each pair of products with the sign its own split calls for. . Worth 1 point.

Reports a single value for the product. . Worth 1 point.

Part B 3 points

Takes the shared factor outside a single bracket holding both of the other numbers. . Worth 2 points.

Adds inside the bracket first and reports the value of the single product. . Worth 1 point.

Part C 6 points

Names the product that appeared in both parts and describes both directions, one as spreading a factor across a split and the other as collecting a shared factor into one product. . Worth 3 points. needs an explanation, not just an answer

Matches each direction to the kind of calculation it is worth taking, naming what it saves. . Worth 2 points. needs an explanation, not just an answer

Says what both forms have in common, that each names the same number. . Worth 1 point.
10. What a single pair of brackets is worth . 13 points. Question 10 of 10.

As it stands, the line

$20 - 8 \div 4 + 2$

names one number. A single pair of brackets, inserted without moving any number or any sign, can make it name others.
1. Part A.
  Evaluate the line as it stands.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Insert a single pair of brackets to make the line name $16$ , and then, in a different position, a single pair to make it name $5$ . Evaluate each of your two lines to show that it does.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A student says that inserting a pair of brackets always changes what a line names. Produce a placement in this line that leaves its value alone, and state what has to be true of a placement before it can change anything.
  
  Carry your own answer forward Test your placement against the value you found for the bare line in part A, whichever value that was.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

The line names $20$ .

Part B

$20 - (8 \div 4 + 2) = 16$ , and $(20 - 8) \div 4 + 2 = 5$ .

Part C

Bracketing the division, as $20 - (8 \div 4) + 2$ , leaves the value alone: the tiers had already put that division first. A placement can change a line only if it alters the reading the convention would otherwise give, by pulling an operation ahead of one that outranks it or by regrouping within a tier against the left-to-right order.

Worked solution

Part A

Division sits on a higher tier than addition and subtraction, so it is settled first. What remains shares one tier and runs left to right:

$20 - 8 \div 4 + 2 = 20 - 2 + 2 = 18 + 2 = 20.$

The $+2$ is attached to the $2$ beside it, not to a bundle of the last two terms.

Part B

For the first target, gather everything after the subtraction sign so that the whole of it is taken away at once:

$20 - (8 \div 4 + 2) = 20 - (2 + 2) = 20 - 4 = 16.$

For the second, force the subtraction to happen before the division, which the tiers would never do unaided:

$(20 - 8) \div 4 + 2 = 12 \div 4 + 2 = 3 + 2 = 5.$

In both lines every number and every sign stands where it stood; only a pair of brackets has been added.

Part C

Brackets around the division change nothing:

$20 - (8 \div 4) + 2 = 20 - 2 + 2 = 20,$

which is the value part A found for the bare line.

The reason is that the tiers had already put that division first. Those brackets instruct a reader to do what they were going to do anyway, so they are free to write and free to leave out, which is why ordinary formulas are not buried in brackets.

That points at the general test, and it has two halves, both of which this line has already shown. A placement can change the value by pulling an operation ahead of one that OUTRANKS it, as $(20 - 8)$ does by putting a subtraction ahead of a division. It can also change the value without crossing a tier at all, by regrouping operations that SHARE one against the left-to-right reading: $20 - (8 \div 4 + 2)$ took $16$ in part B, and the $+$ and the $-$ there sit on the same tier.

So what a working placement has in common is not a tier crossing but an altered reading: it must change the parse the convention would otherwise supply. The student's claim is too strong in one direction only. A placement that agrees with the convention cannot change anything, while one that alters the parse may or may not: $19 + (26 + 4)$ alters the grouping of a sum and leaves the total exactly where it was.

In one line

As written the line names $20$ . Bracketing it as $20 - (8 \div 4 + 2)$ makes it name $16$ , and as $(20 - 8) \div 4 + 2$ makes it name $5$ . But $20 - (8 \div 4) + 2$ still names $20$ , because those brackets ask for the division first and the tiers had already put it first. A pair of brackets can change a line only if it alters the reading the convention would otherwise give, whether by pulling an operation ahead of one that outranks it or, as in the first of those two, by regrouping within a tier against the left-to-right 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 3 points

Settles the division before the subtraction and the addition, then works left to right. . Worth 2 points.

Reports one number as the value of the line. . Worth 1 point.

Part B 5 points

Finds a placement reaching the first target and shows the working that reaches it. . Worth 2 points.

Finds a placement in a different position reaching the second target, again with the working. . Worth 2 points.

Confirms each line names the target it was built for, with no number and no sign moved, so each is the original line plus one pair of brackets. . Worth 1 point.

Part C 5 points

Produces a specific placement whose value matches the bare line, with the working shown. . Worth 2 points.

States the condition under which a placement can change a value, that it must alter the reading the convention would otherwise give, and covers both a placement crossing a tier and one regrouping within a tier against the left-to-right order. . 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 figure, corrected twice 11 points

Part A 3

Adds the correction in the hundreds column alone, leaving every other column as it was. 2
Names the column that changes and the digit now standing in it, not merely the corrected figure. 1

Part B 4

Writes the expanded form and adds the correction to the hundreds term rather than to the numeral as a whole. 2
Bundles ten of the eleven hundreds into one thousand and reports the figure that results. 1
Names both of the digits that change. 1

Part C 4

Explains the overflow from a column holding only one digit, so ten of its units cannot remain in it. 2
States the ten-times relationship between neighbouring columns and ties it to why the excess lands exactly one column to the left. 2

2. What the rearrangement needed 11 points

Part A 3

Forms pairs that fall on round numbers, using each of the four numbers exactly once. 2
Reports the total the rearranged sum gives. 1

Part B 3

Settles all three brackets correctly without carrying out a multiplication. 2
Names a different property for each bracket, keeping the two identities apart from the zero property. 1

Part C 5

Decides against the classmate and separates a change of grouping from a change of order, saying what each does to the written sum. 3
Names the property the classmate left out and identifies the specific move in part A that needed it. 2

3. One line, one operation at a time 12 points

Part A 4

Settles the power before any of the operations below it, and treats it as repeated multiplication of the base rather than as a product of base and exponent. 1
Runs the multiplication and the division left to right, and leaves the subtraction until both are done. 2
Reports one number as the value of the whole line. 1

Part B 3

Settles the whole numerator and the whole denominator before dividing, rather than dividing one term of the top by one term of the bottom. 2
Reports one number as the value of the whole expression. 1

Part C 5

Names the first expression in the chain that is wrong, rather than only reporting that the final value is wrong. 2
Diagnoses the slip precisely, saying what the power was taken to mean and what an exponent actually counts, and notes that the tier order itself was sound. 2
Gives the corrected value. 1

4. Six kits, and a price the records lost 13 points

Part A 5

Uses the whole order to recover the missing term, rather than guessing a price and testing it. 2
Reads a term of the item-priced form as the number of kits times one item price, and divides by the number of kits to reach a single price. 2
States both amounts in dollars, keeping what all six spools cost apart from what one spool costs. 1

Part B 3

Prices one whole kit by adding the three item prices, then multiplies by the number of kits. 2
Reports the total as an amount of money and sets it against the order. 1

Part C 5

Locates the change in exactly one place in each of the two forms, rather than rebuilding either from scratch. 2
Gives the new total. 1
Says which form shows the cost of the rise more directly, with the reason it does. 2

5. Three terms, and where the collecting stops 13 points

Part A 4

Finds the greatest factor common to all three terms rather than to two of them. 2
Writes the sum as that factor multiplying one bracket holding all three remaining terms. 1
Checks by multiplying back out and recovering all three original terms. 1

Part B 4

Tests candidate factors against all three terms, so that the third term is what settles the answer. 2
Explains that a factor shared by two of the terms is unavailable unless it divides the third as well. 2

Part C 5

Rejects the classmate's conclusion and grounds the rejection in the term that shares nothing with the other two. 3
States the finishing test over all the terms inside the bracket rather than over a pair of them. 2

6. Five digits, and the codes they make 11 points

Part A 4

Builds both codes by filling the highest column first and working down, and respects the rule against a leading zero. 2
Gives what the $8$ contributes in each code as an amount, not as the name of a column. 2

Part B 3

Reports how many times larger one contribution is than the other, as a factor rather than as a difference. 1
Predicts the same factor from the number of columns the digit has moved, using the ten-times relation between neighbours. 2

Part C 4

Produces the numeral the colleague's rule actually gives and identifies what disqualifies it. 2
States a repair that keeps the rest of the rule, and says why the leading column is the only thing that had to change. 2

7. One factor outside, two different brackets 12 points

Part A 3

Carries out both routes in full and reports a value for each. 2
States that the two routes agree. 1

Part B 4

Carries out both routes in full and reports a value for each. 2
Identifies which of the two values the line actually has, rather than offering both as answers. 2

Part C 5

Distinguishes the two brackets by what they hold, a count of parts against a single quantity already built by multiplying. 3
Says what the failed route did to the line, identifying the extra factor rather than calling the result merely wrong. 2

8. Two rearrangements, tried on two chains 11 points

Part A 3

Evaluates all three lines correctly, reading the two bare chains left to right. 2
Presents the three values together and says which of the two rearrangements changed the value. 1

Part B 3

Evaluates all three lines correctly, keeping every step a whole number. 2
Says which of the two rearrangements changed the value of this chain. 1

Part C 5

Rejects the commutativity reading and shows what that claim would actually assert about the first two numbers of a chain. 3
Accounts for the surviving exchange in BOTH chains: two removals from one starting amount, and two successive divisions whose combined divisor is a product that may itself be reordered. 2

9. One product reached from both sides 13 points

Part A 4

Checks each split against the number it claims to split before using it. 2
Evaluates both routes, joining each pair of products with the sign its own split calls for. 1
Reports a single value for the product. 1

Part B 3

Takes the shared factor outside a single bracket holding both of the other numbers. 2
Adds inside the bracket first and reports the value of the single product. 1

Part C 6

Names the product that appeared in both parts and describes both directions, one as spreading a factor across a split and the other as collecting a shared factor into one product. 3
Matches each direction to the kind of calculation it is worth taking, naming what it saves. 2
Says what both forms have in common, that each names the same number. 1

10. What a single pair of brackets is worth 13 points

Part A 3

Settles the division before the subtraction and the addition, then works left to right. 2
Reports one number as the value of the line. 1

Part B 5

Finds a placement reaching the first target and shows the working that reaches it. 2
Finds a placement in a different position reaching the second target, again with the working. 2
Confirms each line names the target it was built for, with no number and no sign moved, so each is the original line plus one pair of brackets. 1

Part C 5

Produces a specific placement whose value matches the bare line, with the working shown. 2
States the condition under which a placement can change a value, that it must alter the reading the convention would otherwise give, and covers both a placement crossing a tier and one regrouping within a tier against the left-to-right order. 3

Multiple choice

Free response