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Prime Factorization: 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 two pieces

Write the value of $15\times22$ in prime-power form.
1. Hint 1
  Break down each of the two factors into primes.
2. Hint 2
  Collect the prime factors from both pieces into one product, listing primes from smallest to largest.
Answer

$2\times3\times5\times11$ .

Full solution

Break each factor into its prime pieces.
$15=3\times5$ $22=2\times11$
Both branches now contain primes only.
Putting the leaves together gives
$15\times22=2\times3\times5\times11$

No prime repeats, so no exponent above $1$ is needed.
Both forms multiply to $330$ .

Answer

$2\times3\times5\times11$ .

Key idea

The prime factors of a product come from completing the breakdown of each factor.
Problem 1: Mark complete
Problem 2 The division record

A record shows the numbers $182$ , $91$ , $13$ , $1$ , in that order. Each new number was obtained by dividing the previous one by the smallest prime that divides it exactly. What divisors were used, in order?
1. Hint 1
  Each divisor is the missing factor connecting two neighboring numbers in the record.
2. Hint 2
  Divide the earlier number by the later number at each step.
Answer

$2$ , $7$ , $13$ .

Full solution

Reconstruct each exact division.
$182\div91=2$ $91\div13=7$ $13\div1=13$
These are all primes.

They are also the smallest primes that fit at their respective steps: $182$ is even; $91$ is odd, has digit sum $10$ , and does not end in $0$ or $5$ ; and $13$ is itself prime.

Answer

$2$ , $7$ , $13$ .

Key idea

An exact division record preserves the prime factor removed at every step.
Problem 2: Mark complete
Problem 3 The covered leaves

A completed factor tree starts at $525$ . It has four leaves: two of them show $3$ and $7$ , and the other two are covered. What are the covered leaves?
1. Hint 1
  The product of all the leaves of a completed tree equals the number at its top.
2. Hint 2
  Divide $525$ by the product of the two visible leaves to find what the covered leaves multiply to.
3. Hint 3
  Each covered leaf is prime, so split that product into two primes.
Answer

$5$ and $5$ .

Full solution

The two visible leaves have product
$3\times7=21$
The two covered leaves supply the rest of the product.
$525\div21=25$

So the covered leaves are two primes whose product is $25$ .
The only way to write $25$ as a product of two primes is $5\times5$ , so both covered leaves are $5$ .

Check by multiplying all four leaves.
$3\times5\times5\times7=525$

Answer

$5$ and $5$ .

Key idea

Dividing the top of a tree by its visible leaves shows what the covered leaves must multiply to.
Problem 3: Mark complete
Problem 4 The shelf inventory

A cabinet has $7$ shelves with $20$ boxes on each shelf and $14$ clips in each box. Find the total number of clips and write that total in prime-power form.
1. Hint 1
  The total is the product of the three counts.
2. Hint 2
  Break down $20$ and $14$ before collecting repeated primes.
Answer

$1{,}960$ clips; $2^3\times5\times7^2$ .

Full solution

First count the boxes, then the clips.
$7\times20=140$ $140\times14=1{,}960$
Thus the cabinet holds $1{,}960$ clips.

For the prime factors, the $7$ is already prime, $20=2^2\times5$ , and
$14=2\times7.$
Together these supply three $2$ s, one $5$ , and two $7$ s.
$1{,}960=2^3\times5\times7^2$

Check by multiplying the prime-power form back out.
$8\times5\times49=1{,}960$

Answer

$1{,}960$ clips; $2^3\times5\times7^2$ .

Key idea

Counts multiplied in a setting can also be broken into primes and collected into one factorization.
Problem 4: Mark complete
Problem 5 The regrouped primes

A number has prime-power form $2^3\times3\times11$ . Regroup its prime factors to write the number as a product of two whole numbers that are each more than $10$ and less than $30$ . Give every such pair.
1. Hint 1
  Each member of a pair uses some of the prime copies, and its partner uses all the rest.
2. Hint 2
  Sort the possible members by whether or not they contain the prime $11$ .
3. Hint 3
  A member without the $11$ is built from $2$ s and at most one $3$ , so it is a factor of $2^3\times3$ .
Answer

$11\times24$ and $12\times22$ .

Full solution

Multiplying out gives the number.
$2^3\times3\times11=264$
In a pair, one member uses some of these prime copies and its partner uses all the rest, so both members are factors of $264$ .

First take a member that contains the $11$ .
It is $11$ itself or $2\times11=22$ , because any other choice adds a $3$ or a second $2$ to the $11$ and so is at least $3\times11=33$ , which is over $30$ .
The partner of $11$ is made of all the other primes.
$2^3\times3=24$
The partner of $22$ is made of what is left after one $2$ and the $11$ .
$2^2\times3=12$
This gives the pairs $11\times24$ and $12\times22$ , and every member is in range.

Now take a member without the $11$ .
It is built from $2$ s and at most one $3$ , so it is one of the factors $1$ , $2$ , $3$ , $4$ , $6$ , $8$ , $12$ , and $24$ of
$2^3\times3=24.$
Only $12$ and $24$ are in range, and their partners, $22$ and $11$ , give the two pairs already found.

Answer

$11\times24$ and $12\times22$ .

Key idea

Splitting the prime copies of a factorization into two groups gives a factor pair of the number.
Problem 5: Mark complete
Problem 6 The branching record

At the first split of a factor tree, one branch ends at $5$ . The other branch splits into two numbers that are both $9$ . Find the number at the top, and write its finished prime factorization in prime-power form.
1. Hint 1
  Rebuild the number from the two branches, then finish splitting any composite numbers.
2. Hint 2
  Each $9$ contributes two copies of $3$ .
Answer

$405$ ; $3^4\times5$ .

Full solution

The branch that splits into two $9$ s has value
$9\times9=81$
Combining it with the other branch gives the original number.
$5\times81=405$

Each $9$ splits into $3\times3$ , so the two $9$ s contribute four copies of $3$ in all.
The remaining leaf is the prime $5$ .
$405=3^4\times5$
The power $3^4$ equals $81$ , checking the original branch.

Answer

$405$ ; $3^4\times5$ .

Key idea

Each repeated composite branch must contribute all its prime copies to the final count.
Problem 6: Mark complete
Problem 7 The stopping point

Start with $780$ and repeatedly divide by the smallest prime that fits. Stop as soon as the current quotient is prime. Give all the quotients reached up to that stopping point and write $780$ in prime-power form.
1. Hint 1
  Use the divisibility rules on the current quotient after each division.
2. Hint 2
  Keep removing $2$ while the quotient is even, then test $3$ and $5$ in order.
Answer

Quotients: $390,195,65,13$ ; $780=2^2\times3\times5\times13$ .

Full solution

The first two divisions remove $2$ while the current number is even.
$780\div2=390$ $390\div2=195$

The quotient $195$ is odd and has digit sum $15$ , so its smallest prime divisor is $3$ .
$195\div3=65$
The quotient $65$ is not divisible by $2$ or $3$ , but ends in $5$ .
$65\div5=13$

The current quotient is now the prime $13$ , so the requested stopping point is reached.
Include that last prime with the divisors already used.
$780=2^2\times3\times5\times13$
Multiplying the prime factors gives $780$ .

Answer

Quotients: $390,195,65,13$ ; $780=2^2\times3\times5\times13$ .

Key idea

If repeated division stops at a prime quotient, that quotient still belongs in the factorization.
Problem 7: Mark complete
Problem 8 The factor requirement

A student says the prime factorization of a composite number must contain at least two different primes. Does $1{,}331$ support or disprove the claim? Explain.
1. Hint 1
  A prime can occur several times without introducing a different prime.
2. Hint 2
  After ruling out $2$ , $3$ , and $5$ , work upward through the next primes in turn.
Answer

$1{,}331$ disproves the claim; $1{,}331=11^3$ .

Full solution

The number $1{,}331$ is odd, its digit sum is $8$ , and it does not end in $0$ or $5$ , so $2$ , $3$ , and $5$ do not divide it.
The next prime, $7$ , does not divide it either, since $7\times190=1{,}330$ leaves a remainder of $1$ .

The prime $11$ does divide it.
$1{,}331\div11=121$
Dividing again by $11$ leaves the prime $11$ .
$121\div11=11$
Thus
$1{,}331=11\times11\times11$ $1{,}331=11^3$

So $1{,}331$ is composite, yet every factor in its prime factorization is the same prime.
It disproves the claim: a prime factorization needs prime factors, not two different primes.

Answer

$1{,}331$ disproves the claim; $1{,}331=11^3$ .

Key idea

A composite number's prime factorization can consist of copies of a single prime.
Problem 8: Mark complete
Problem 9 The two methods

A student says that a completed factor tree and repeated division by the smallest prime always give the same number of copies of each prime, for every whole number greater than $1$ . Is this correct? Explain.
1. Hint 1
  Ask what the primes produced by each method multiply to.
2. Hint 2
  Both methods end in a list of primes, so compare two prime lists that have the same product.
3. Hint 3
  Recall what the Fundamental Theorem of Arithmetic says about writing a number as a product of primes.
Answer

Yes.

Full solution

Every split in a factor tree keeps the product the same, and a completed tree has only primes as leaves.
So its leaves are primes that multiply to the original number.

Repeated division by the smallest prime that fits uses only prime divisors and stops when the quotient reaches $1$ .
So its divisors are also primes that multiply to the original number.

By the Fundamental Theorem of Arithmetic, every whole number greater than $1$ has exactly one prime factorization apart from the order of the factors.
The two lists are both prime factorizations of the same number, so they contain the same primes with the same number of copies, and the student is correct.

Answer

Yes.

Key idea

Two methods that each end in a product of primes for the same number agree on every prime and its count.
Problem 9: Mark complete
Problem 10 The proposed leaf

Could a correctly completed factor tree for $27\times35$ have a leaf labeled $13$ ? Justify your answer without first multiplying $27$ by $35$ .
1. Hint 1
  Find the complete prime lists of the two displayed factors.
2. Hint 2
  Every correctly completed tree for the same number contains the same primes with the same counts.
Answer

No.

Full solution

Complete the breakdown of each displayed factor.
$27=3^3$ $35=5\times7$
Together they give this prime-power form.
$27\times35=3^3\times5\times7$

This completed prime list has three $3$ s, one $5$ , and one $7$ , with no $13$ .
Uniqueness of prime factorization forces any other correctly completed tree for the same number to have precisely that list, so a leaf labeled $13$ is impossible.

Answer

No.

Key idea

One complete prime factorization fixes which prime leaves any other correct tree can contain.
Problem 10: Mark complete

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