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Primes and Composites: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The factor total

A whole number has exactly two distinct positive factors. Their sum is $174$ . What is the number?
1. Hint 1
  A whole number with exactly two distinct positive factors is prime.
2. Hint 2
  Its two factors must be $1$ and the number itself.
Answer

$173$ .

Full solution

The factors must be $1$ and the number itself, so subtract the first factor from their sum.
$174-1=173$

As a check, $173$ is odd, its digit sum is $11$ , and it does not end in $0$ or $5$ , so $2$ , $3$ , and $5$ do not divide it.
Division by $7$ , $11$ , and $13$ leaves a remainder.
$\begin{aligned}173&=7\times24+5\\173&=11\times15+8\\173&=13\times13+4\end{aligned}$
The next prime candidate is $17$ , and $17\times17=289$ passes $173$ , so no further trial is needed.
Its factors are exactly $1$ and $173$ .

Answer

$173$ .

Key idea

The two factors of a prime are one and the prime itself.
Problem 1: Mark complete
Problem 2 Two even numbers

Both $2$ and $142$ are even. Classify each as prime or composite.
1. Hint 1
  Being even means $2$ is a factor, so ask what that factor adds to each number's list of factors.
2. Hint 2
  Compare the factor $2$ with $1$ and with the number itself: for which of the two numbers is it a new factor?
3. Hint 3
  A prime has exactly two distinct factors, and a composite has more than two.
Answer

$2$ is prime; $142$ is composite.

Full solution

The only whole numbers that divide $2$ are $1$ and $2$ .
That is exactly two distinct factors, so $2$ is prime.

The number $142$ is even, so $2$ divides it.
$142=2\times71$
So $2$ is a factor of $142$ besides $1$ and $142$ itself.

Then $142$ has at least the three distinct factors $1$ , $2$ , and $142$ .
That is more than two, so $142$ is composite.

For $2$ , the factor $2$ is the number itself, so being even adds no new factor.
For every even number greater than $2$ , the factor $2$ is an extra one, which is why $2$ is the only even prime.

Answer

$2$ is prime; $142$ is composite.

Key idea

Every even number greater than $2$ has $2$ as a third factor, so $2$ is the only even prime.
Problem 2: Mark complete
Problem 3 Four numbers sorted

Classify each of $1$ , $59$ , $95$ , and $141$ as prime, composite, or neither.
1. Hint 1
  Compare each number's count of distinct factors with the definitions of prime and composite.
2. Hint 2
  For each number above $1$ , try the tests for $2$ , $3$ , and $5$ first, then divide by $7$ where the stopping rule still requires it.
Answer

$1$ : neither; $59$ : prime; $95$ : composite; $141$ : composite.

Full solution

The number $1$ has only one distinct factor, itself, so it is neither prime nor composite.

The number $59$ is odd, its digit sum is $14$ , and it does not end in $0$ or $5$ , so $2$ , $3$ , and $5$ do not divide it.
Division by $7$ leaves a remainder.
$59=7\times8+3$
The next prime candidate is $11$ , and $11\times11=121$ passes $59$ , so $59$ is prime.

The number $95$ ends in $5$ , so $5$ is a factor besides $1$ and $95$ .
$95=5\times19$
So $95$ is composite.

The digit sum of $141$ is $6$ , a multiple of $3$ , so $3$ is a factor besides $1$ and $141$ .
$141=3\times47$
So $141$ is composite.

Answer

$1$ : neither; $59$ : prime; $95$ : composite; $141$ : composite.

Key idea

Counting distinct factors sorts a number: one factor means neither, two means prime, and more than two means composite.
Problem 3: Mark complete
Problem 4 The inner factors

List every positive factor of $165$ other than $1$ and $165$ , and justify that your list is complete.
1. Hint 1
  Factors come in pairs, so each smaller partner you find also gives you its larger partner.
2. Hint 2
  Test each whole number from $2$ upward, and stop once the number times itself passes $165$ .
3. Hint 3
  The divisibility rules settle most candidates: $165$ is odd, ends in $5$ , and has digit sum $12$ .
Answer

$3$ , $5$ , $11$ , $15$ , $33$ , and $55$ .

Full solution

Factors come in pairs whose product is $165$ , and the smaller partner of a pair, times itself, is at most $165$ .
Since $13\times13=169$ passes $165$ , every pair has a smaller partner from $1$ to $12$ .

The number $165$ is odd, so none of $2$ , $4$ , $6$ , $8$ , $10$ , and $12$ divides it.
Its digit sum is $12$ , a multiple of $3$ but not of $9$ , so $3$ divides it and $9$ does not.
It ends in $5$ , so $5$ divides it.

The candidates $7$ and $11$ need division.
$\begin{aligned}165&=7\times23+4\\165&=11\times15\end{aligned}$
So $7$ leaves a remainder and $11$ divides exactly.

Dividing by $3$ and by $5$ gives the partners $55$ and $33$ .
The factor pairs are $1\times165$ , $3\times55$ , $5\times33$ , and $11\times15$ , so the factors other than $1$ and $165$ are $3$ , $5$ , $11$ , $15$ , $33$ , and $55$ .

Answer

$3$ , $5$ , $11$ , $15$ , $33$ , and $55$ .

Key idea

Testing every possible smaller partner up to the stopping point finds every factor pair.
Problem 4: Mark complete
Problem 5 The sorting cards

A sorter has a pile of cards numbered $2$ through $60$ . It sets aside the smallest card in the pile, removes every larger multiple of that number from the pile, and repeats until the pile is empty.

List the cards it sets aside, and list the cards it removes at the step where it sets aside $7$ .
1. Hint 1
  This is the Sieve of Eratosthenes, so each card set aside has no smaller card dividing it.
2. Hint 2
  Work through $2$ , $3$ , and $5$ first, and keep track of which cards are already gone before the step for $7$ .
3. Hint 3
  Once the card being set aside, times itself, passes $60$ , every card still in the pile is prime.
Answer

Set aside: $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , $19$ , $23$ , $29$ , $31$ , $37$ , $41$ , $43$ , $47$ , $53$ , $59$ . Removed at the step for $7$ : $49$ only.

Full solution

Setting aside $2$ removes every larger even card.
Setting aside $3$ next removes the odd multiples of $3$ still in the pile: $9$ , $15$ , $21$ , $27$ , $33$ , $39$ , $45$ , $51$ , and $57$ .

Setting aside $5$ then removes $25$ , $35$ , and $55$ , the multiples of $5$ that are still in the pile.

At the step for $7$ , its larger multiples up to $60$ are $14$ , $21$ , $28$ , $35$ , $42$ , $49$ , and $56$ .
Every one except $49$ has a factor of $2$ , $3$ , or $5$ and is already gone.
$49=7\times7$
So $49$ is the only card removed at this step.

The next card in the pile is $11$ , and $11\times11=121$ passes $60$ .
By the stopping rule every card still in the pile is prime: a composite card up to $60$ has a factor pair whose smaller partner is from $2$ to $7$ , since $8\times8=64$ passes $60$ .
If that partner is $2$ , $3$ , $5$ , or $7$ , it divides the card directly; if it is $4$ or $6$ , the card is even.
So every composite card up to $60$ is a multiple of $2$ , $3$ , $5$ , or $7$ , and was removed at one of the first four steps.
Each remaining card is set aside in turn, and no more cards are removed.

The cards set aside are $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , $19$ , $23$ , $29$ , $31$ , $37$ , $41$ , $43$ , $47$ , $53$ , and $59$ .

Answer

Set aside: $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , $19$ , $23$ , $29$ , $31$ , $37$ , $41$ , $43$ , $47$ , $53$ , $59$ . Removed at the step for $7$ : $49$ only.

Key idea

A sieve removes each composite card at the step for the smallest prime that divides it.
Problem 5: Mark complete
Problem 6 The display request

A display needs two prime numbers whose product is $176$ . Can the request be filled? Explain.
1. Hint 1
  If two primes multiply to $176$ , together they form one of its factor pairs.
2. Hint 2
  Consider separately the case where one prime is even and the case where both primes are odd.
3. Hint 3
  For two odd primes, the smaller one times itself is at most $176$ , so only a few odd candidates need testing.
Answer

No.

Full solution

Suppose two primes multiply to $176$ .
Then they form a factor pair of $176$ .

If one of them is $2$ , the other is forced.
$176\div2=88$
But $88$ is even and greater than $2$ , so it has $2$ as an extra factor and is composite.

Otherwise neither prime is $2$ , so both are odd, because $2$ is the only even prime: every even number greater than $2$ has $2$ as a factor besides $1$ and itself.
The smaller prime, times itself, is at most $176$ , and $14\times14=196$ passes $176$ , so it is one of $3$ , $5$ , $7$ , $11$ , and $13$ .

The digit sum of $176$ is $14$ and it ends in $6$ , so $3$ and $5$ do not divide it.
Division by $7$ and by $13$ leaves a remainder.
$\begin{aligned}176&=7\times25+1\\176&=13\times13+7\end{aligned}$
That leaves $11$ .
$176=11\times16$
Its partner $16$ is even and greater than $2$ , so it is composite.

No pair of primes has product $176$ , so the request cannot be filled.

Answer

No.

Key idea

Checking the factor pairs of a number settles whether it is a product of two primes.
Problem 6: Mark complete
Problem 7 The covered labels

Two covered labels each show a prime number strictly between $310$ and $320$ . What is the greatest possible distance between the numbers on the number line? Justify that no other allowed labels give a greater distance.
1. Hint 1
  The greatest distance comes from the smallest and the largest eligible labels, so find those first.
2. Hint 2
  After the tests for $2$ , $3$ , and $5$ , the candidates left are $311$ , $313$ , $317$ , and $319$ .
3. Hint 3
  Since $19\times19=361$ passes every candidate, trial division by $7$ , $11$ , $13$ , and $17$ settles each one.
Answer

$6$ units.

Full solution

Every even number here is composite, and $315$ ends in $5$ .
The digit sums of $311$ , $313$ , $317$ , and $319$ are $5$ , $7$ , $11$ , and $13$ , so $3$ divides none of them.

The greatest distance uses the smallest and the largest eligible labels, so test from each end.
Since $17\times17=289$ and $19\times19=361$ , the prime candidates to try are $7$ , $11$ , $13$ , and $17$ .

The largest candidate, $319$ , fails at $11$ .
$319=11\times29$
So it is composite, although none of the tests for $2$ , $3$ , and $5$ caught it.

Each trial leaves a remainder for $311$ .
$\begin{aligned}311&=7\times44+3\\311&=11\times28+3\\311&=13\times23+12\\311&=17\times18+5\end{aligned}$
So $311$ is prime, and it is the smallest eligible label.

Each trial leaves a remainder for $317$ as well.
$\begin{aligned}317&=7\times45+2\\317&=11\times28+9\\317&=13\times24+5\\317&=17\times18+11\end{aligned}$
So $317$ is prime, and it is the largest eligible label.

The widest separation is between these two.
$317-311=6$
Any other pair of eligible labels lies between $311$ and $317$ , so it is closer together.

Answer

$6$ units.

Key idea

The greatest distance between two primes in a range joins the smallest and the largest primes in it.
Problem 7: Mark complete
Problem 8 The factor claim

A student says that every positive factor of a composite number must also be composite. Is the student right? Use the factors of $62$ to explain.
1. Hint 1
  Test the claim on the actual factors of $62$ , one factor at a time.
2. Hint 2
  Write $62$ as a product of two factors and classify each factor you find.
3. Hint 3
  A single factor of $62$ that is not composite is enough to settle the claim.
Answer

No; for example, $2$ is a factor of $62$ and is prime. The factors $1$ and $31$ are also not composite.

Full solution

Split $62$ into a factor pair.
$62=2\times31$
So $62$ has the factor $2$ besides $1$ and $62$ , which makes $62$ composite.
Yet $2$ is prime, not composite.

One factor that is not composite is enough to disprove a claim about every factor.
The factor $1$ is another counterexample, since $1$ is neither prime nor composite, and so is the prime $31$ .

Answer

No; for example, $2$ is a factor of $62$ and is prime. The factors $1$ and $31$ are also not composite.

Key idea

Being composite describes a number, not each of its factors: $1$ divides every composite number and is never composite.
Problem 8: Mark complete
Problem 9 The written list

Alex writes $1$ , $139$ , $1$ , $139$ as the factor list of $139$ and calls it composite. Is that classification valid? Explain.
1. Hint 1
  Repeated entries do not create new factors.
2. Hint 2
  Test whether $139$ has any factor besides the two different numbers Alex wrote.
Answer

No; $139$ is prime.

Full solution

The list contains only two distinct numbers, $1$ and $139$ .
Repeating them does not increase the factor count.

The number $139$ is odd, its digit sum is $13$ , and it does not end in $0$ or $5$ .
These observations rule out $2$ , $3$ , and $5$ .
Division by $7$ and by $11$ leaves a remainder.
$\begin{aligned}139&=7\times19+6\\139&=11\times12+7\end{aligned}$
The next prime candidate is $13$ , and $13\times13=169$ passes $139$ .
There are no additional factors, so $139$ is prime.

Answer

No; $139$ is prime.

Key idea

Classification depends on different factors, not the number of times a factor is written.
Problem 9: Mark complete
Problem 10 The five-divisor claim

A student claims that every whole number from $2$ to $168$ that is not divisible by any of $2$ , $3$ , $5$ , $7$ , and $11$ is prime. Is the claim correct? Explain.
1. Hint 1
  Think about what a composite number's factor pairs look like, and how large the smaller partner can be.
2. Hint 2
  The smaller partner of a factor pair, times itself, is at most the number, so compare $13\times13$ with $168$ .
3. Hint 3
  Check that every whole number from $2$ to $12$ is divisible by at least one of $2$ , $3$ , $5$ , $7$ , and $11$ .
Answer

Yes, the claim is correct.

Full solution

Take a number from $2$ to $168$ that none of $2$ , $3$ , $5$ , $7$ , and $11$ divides, and suppose it were composite.
Then it has a factor pair whose smaller partner is greater than $1$ .

That smaller partner, times itself, is at most the number, so at most $168$ .
$13\times13=169$
So the smaller partner is a whole number from $2$ to $12$ .

If the partner is $2$ , $3$ , $5$ , $7$ , or $11$ , it divides the number directly.
If it is $4$ , $6$ , $8$ , $10$ , or $12$ , it is even, so the number is even and $2$ divides it.
If it is $9$ , the number is a multiple of $9$ and so of $3$ .

Every case contradicts the choice of the number, so it cannot be composite, and it is prime.

The upper limit matters: $169$ is divisible by none of the five, yet it is composite.
$169=13\times13$

Answer

Yes, the claim is correct.

Key idea

A composite number below $169$ has a factor pair whose smaller partner is from $2$ to $12$ , so one of $2$ , $3$ , $5$ , $7$ , and $11$ divides it.
Problem 10: Mark complete

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