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Math 7 · Mathematics
Numbers · Natural numbers

Prime Factorization

Before you start, make sure you know: Prime factorization (Math 6)
Before you start, make sure you know: Divisibility rules (Math 6)

A bakery packs $24$ cupcakes in boxes: $2 \times 12$, $3 \times 8$, $4 \times 6$, and more. Lots of ways! But if you break $24$ all the way down into primes, there is only one answer. What is it, and why is it so useful?

Prime or composite?

Prime numbers are the building blocks of every other number, so you need to spot them fast.

The first primes are $2, 3, 5, 7, 11, 13, 17, 19, \ldots$ Here they are on a chart of the numbers from $1$ to $30$.

The numbers 1 to 30 in a grid, with the primes 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 shaded. 12345678910 11121314151617181920 21222324252627282930 Shaded = prime. The number 1 is neither. 2 is the only even prime.

To test a number $n$, divide it by the primes $2, 3, 5, 7, \ldots$ and stop when the prime squared is bigger than $n$. Why can you stop? If $n = a \times b$ with $a \le b$, then $a \times a \le n$. So a factor pair always has a small member you would have already tried.

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Classify the following natural numbers into prime or composite. Remember that prime numbers are those that only have two divisors: $1$ and themselves.…

Classify the following natural numbers into prime or composite. Remember that prime numbers are those that only have two divisors: $1$ and themselves.

$14, 17, 21, 23, 27, 29, 33, 37$

Solve
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They want to design a special barcode based on a number $K$. This number is formed by multiplying the first $n$ prime numbers consecutively (for example, if $n=3$, the…

They want to design a special barcode based on a number $K$. This number is formed by multiplying the first $n$ prime numbers consecutively (for example, if $n=3$, the number would be $2 \cdot 3 \cdot 5 = 30$).

If we want the number $K$ to:

  1. Be greater than $1000$ but less than $5000$.

  2. Have exactly $6$ distinct prime factors in its prime factorization.

What is the value of $K$? Show the list of prime numbers you have used and check that the result is within the requested range.

p₁ p₂ p₃ p₄ p₅ p₆ K 1000 < K < 5000
Solve

The Fundamental Theorem of Arithmetic

Now back to $24$. Break it into smaller factors until every factor is prime.

The prime factorization of 24 written as a repeated product and then with exponents: 2 cubed times 3. 24 = 2 × 2 × 2 × 3 = 2³ × 3 No matter how you split it, you always end up with the same prime factors.

When a prime repeats, write it as a power: $2 \times 2 \times 2 \times 3 = 2^3 \times 3$. By convention, list the primes from smallest to largest.

You can also go backward. To find the number, evaluate the powers first, then multiply. And if a problem says a number has only the prime factors $2$ and $3$, it must look like $2^a \times 3^b$.

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State which number the following prime factorizations correspond to: a) $2^{2} \cdot 3 \cdot 5$ b) $2^{3} \cdot 7$ c) $3^{2} \cdot 11$

State which number the following prime factorizations correspond to:

  1. a) $2^{2} \cdot 3 \cdot 5$
  1. b) $2^{3} \cdot 7$
  1. c) $3^{2} \cdot 11$
Solve
Practice★☆☆Not started
What is the largest two-digit number that has only $2$ and $3$ as prime factors in its factorization? Help example: the number $6 = 2 \cdot 3$ meets the condition, but…

What is the largest two-digit number that has only $2$ and $3$ as prime factors in its factorization?

Help example: the number $6 = 2 \cdot 3$ meets the condition, but it is not the largest two-digit number.

Solve

The factor tree

A visual way to factor: split the number into any two factors, then keep splitting each branch that isn't prime. Circle a branch when it reaches a prime, because it stops there.

A factor tree for 24: it splits into 6 and 4, then into the primes 2, 3, 2 and 2. 24 6 4 2 3 2 2 24 = 2 × 3 × 2 × 2 = 2³ × 3

Every number in the tree equals the product of its two children. That gives you two quick checks. If a number is missing, divide the parent by the child you know. And when you multiply all the end primes, you must get the starting number back.

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Complete the following factor tree for the number $24$. You must find two numbers that multiplied give the level above, until all the numbers at the ends are prime.…

Complete the following factor tree for the number $24$. You must find two numbers that multiplied give the level above, until all the numbers at the ends are prime.

× × × 24 6 ? 2 3 ? ?

Finally, write the factorization of $24$ as a product of its prime factors.

Solve
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Copy this prime factorization diagram into your notebook and fill in the empty gaps (marked with the letter $x$) with the corresponding numbers:

Copy this prime factorization diagram into your notebook and fill in the empty gaps (marked with the letter $x$) with the corresponding numbers:

180 x 45 15 x 2 2 x 3
Solve

The division ladder

For bigger numbers, a ladder is faster. Write the number, draw a bar to its right, and divide by the smallest prime that works. Write the quotient below and repeat until you reach $1$. (Some problems call this the vertical line method.)

A division ladder for 60: divide by 2, 2, 3 and 5 until the quotient is 1. 602 302 153 55 1 60 = 2² × 3 × 5

The numbers on the right of the bar are the prime factors. Use the divisibility rules to choose each divisor: even numbers → $2$; digit sum a multiple of $3$ → $3$; ends in $0$ or $5$ → $5$.

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Fill in the gaps of the following prime factorization of a number using the vertical line method: Write the final result of the factorization using powers.

Fill in the gaps of the following prime factorization of a number using the vertical line method:

60 ? 15 5 1 2 2 ? 5

Write the final result of the factorization using powers.

Solve
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Find the prime factorization of the following numbers. Express the final result as a product of powers: a) $36$ b) $45$ c) $80$

Find the prime factorization of the following numbers. Express the final result as a product of powers:

  1. a) $36$
  1. b) $45$
  1. c) $80$
Solve

Reading divisibility from a factorization

Once a number is factored, you can answer questions about its multiples and divisors without computing its value.

Checking whether N = 2 cubed times 3 times 5 squared is divisible by 12 and by 9 using only its prime factors. N = 2³ × 3 × 5² Divisible by 12 = 2² × 3? 2² fits inside 2³, and 3 is there → yes Divisible by 9 = 3²? N has only one 3 → no
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Look at the prime factorization of the following numbers: $A = 2^{2} \cdot 3 \cdot 7$ $B = 2 \cdot 3^{2} \cdot 5$ Without performing the multiplications, answer with…

Look at the prime factorization of the following numbers:

$A = 2^{2} \cdot 3 \cdot 7$

$B = 2 \cdot 3^{2} \cdot 5$

Without performing the multiplications, answer with reasoning:

  1. a) Is the number $A$ a multiple of $7$?
  2. b) Is the number $B$ divisible by $10$? (Hint: observe if it has the necessary factors to form $10$).
  3. c) Is the number $3$ a divisor of $A$?
Solve
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From the prime factorization of the number $150$, answer the following questions reasoning your answer: a) Is $150$ divisible by $10$? b) Is $150$ divisible by $4$? c)…

From the prime factorization of the number $150$, answer the following questions reasoning your answer:

  1. a) Is $150$ divisible by $10$?
  2. b) Is $150$ divisible by $4$?
  3. c) Is $150$ divisible by $15$?
Solve
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Find the number whose prime factorization is $2^3 \cdot 3^2 \cdot 11$. Once found, explain whether that number is a multiple of $6$ and $22$ by looking only at its prime…

Find the number whose prime factorization is $2^3 \cdot 3^2 \cdot 11$. Once found, explain whether that number is a multiple of $6$ and $22$ by looking only at its prime factors (without performing divisions).

Solve

Finding all the factors

Every factor of $N$ is made from the primes of $N$, using each prime between $0$ and its full exponent times. Multiply every combination and you get all the factors.

Here is $20 = 2^2 \times 5$. The rows are the choices for the $2$ ($2^0, 2^1, 2^2$) and the columns are the choices for the $5$ ($5^0, 5^1$).

A grid of all products of powers of 2 (2 to the 0, 1, 2) and powers of 5 (5 to the 0 and 1), giving the factors 1, 5, 2, 10, 4 and 20 of the number 20. Factors of 20 = 2² × 5 5⁰5¹ 2⁰2¹2² 15 210 420

Factors come in pairs that multiply to $N$: $1 \times 18$, $2 \times 9$, $3 \times 6$. That is how you split $N$ items into equal groups ("$3$ groups of $6$"). The same idea works for a box with volume $V = \ell \times w \times h$: the prime factors of $V$ are shared out among the three edges. For example, a box of volume $30 \text{ in.}^3$ can have edges $2$, $3$, $5$, or $1$, $5$, $6$.

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A Physical Education teacher wants to organize their $28$ students into equal groups of more than $2$ people but fewer than $10$, so that no student is left over. Use…

A Physical Education teacher wants to organize their $28$ students into equal groups of more than $2$ people but fewer than $10$, so that no student is left over.

Use the prime factorization of $28$ to find all the possible ways of making the groups.

Solve
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A professional Instagram photographer wants to organize their photo feed. They have $210$ images and want to post them in equal blocks. a) Find the prime factorization…

A professional Instagram photographer wants to organize their photo feed. They have $210$ images and want to post them in equal blocks.

  1. a) Find the prime factorization of $210$.
  2. b) Write down four different ways of grouping the photos (for example: "3 blocks of 70 photos") using combinations of their prime factors.
Solve
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A logistics company uses small cubic boxes to store parts. The total volume of a cargo container is $151200$ $cm^{3}$. The container must be completely filled with…

A logistics company uses small cubic boxes to store parts. The total volume of a cargo container is $151200$ $cm^{3}$. The container must be completely filled with rectangular packages measuring $a$ in wide, $b$ in long and $c$ in high, where $a$, $b$ and $c$ are distinct prime numbers.

If we know that the container can hold exactly a number of packages that is a multiple of $10$:

  1. Find the prime factorization of $151200$.

  2. Identify the possible dimensions of the packages if the volume of each package must be greater than $100$ $cm^{3}$.

V = 151200 in³ a b c V > 100 in³
Solve

Counting the divisors

You don't need to list the factors just to count them. In the grid for $20$, there were $3$ rows and $2$ columns, so $3 \times 2 = 6$ factors.

Counting the divisors of 60 = 2 squared times 3 times 5: there are 3 choices for the exponent of 2, 2 for 3 and 2 for 5, so 3 times 2 times 2 equals 12 divisors. N = 2² × 3 × 5 = 60 2²3¹5¹ exponent is0, 1, or 2 exponent is0 or 1 exponent is0 or 1 3 choices2 choices2 choices 3 × 2 × 2 = 12 divisors
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A video game designer wants to organize a virtual warehouse of items. To do this, they need to know how many divisors the number of available boxes has, which is $360$.…

A video game designer wants to organize a virtual warehouse of items. To do this, they need to know how many divisors the number of available boxes has, which is $360$.

  1. a) Find the prime factorization of $360$.
  2. b) Using the rule of exponents (add one to each exponent of the factorization and multiply them), calculate the total number of divisors that $360$ has.
Solve
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A 7th grade student claims to have found a number $X$ whose prime factorization is $2^{2} \cdot 3^{4} \cdot 6^{2} \cdot 5$. 1. Explain why this expression is not a valid…

A 7th grade student claims to have found a number $X$ whose prime factorization is $2^{2} \cdot 3^{4} \cdot 6^{2} \cdot 5$.

  1. Explain why this expression is not a valid prime factorization.

  2. Correct the expression so that it is a real prime factorization.

  3. Without calculating the numerical value of $X$, determine how many divisors it has. Show the calculation process step by step.

Solve
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A hacker has blocked access to a server and left the following clue about the numerical password, which is a four-digit number $N$: The prime factorization of $N$ is of…

A hacker has blocked access to a server and left the following clue about the numerical password, which is a four-digit number $N$:

  • The prime factorization of $N$ is of the form $2^{a} \cdot 3^{b} \cdot 5^{c}$.

  • The total number of divisors of $N$ is exactly $30$.

  • The number is a multiple of $100$ but is not a multiple of $8$.

Determine with reasoning the value of the password $N$ and show its complete prime factorization.

sys.auth.decrypt_N() 1. Structure: N = 2ᵃ · 3ᵇ · 5ᶜ 2. Divisors: (a+1) · (b+1) · (c+1) = 30 3. Multiples: N = 100·k a≥2, c≥2 N ≠ 8·p a<3 PASSWORD: [ _ _ _ _ ]
Solve

Greatest common factor

The greatest common factor (GCF) of two numbers is the largest number that divides both. (Some problems write it as M.C.D.) With prime factorizations you can find it quickly.

Table of the prime factorizations of 72 and 120 with the smaller exponent of each common prime in the GCF row, giving 24. 235 72 120 GCF 2³3²– 2³3¹5¹ 2³3¹– = 24 smaller exponent; skip any prime missing from a number

Working backward. Sometimes the GCF is given and an exponent is unknown. Suppose $A = 2^4 \times 5^x$ and $B = 2^y \times 5^3 \times 7$ have $\text{GCF} = 2^2 \times 5^3$.

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Find the prime factorization of the numbers $720$ and $1260$ using the vertical line method. Once the expressions in power form are obtained, state which prime factors…

Find the prime factorization of the numbers $720$ and $1260$ using the vertical line method. Once the expressions in power form are obtained, state which prime factors they have in common and what is the highest exponent with which each one appears in both factorizations.

Solve
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Consider two natural numbers, $A$ and $B$, whose prime factorizations are as follows: $A = 2^{3} \cdot 3^{x} \cdot 5^{2}$ $B = 2^{y} \cdot 3^{2} \cdot 7^{1}$ Knowing…

Consider two natural numbers, $A$ and $B$, whose prime factorizations are as follows:

$A = 2^{3} \cdot 3^{x} \cdot 5^{2}$

$B = 2^{y} \cdot 3^{2} \cdot 7^{1}$

Knowing that the greatest common factor of both is $M.C.D.(A, B) = 36$, find the value of the unknowns $x$ and $y$. Once found, calculate the value of the least common multiple $m.c.m.(A, B)$ also expressed in its prime factorization form.

Solve

Least common multiple

The least common multiple (LCM) is the smallest number that is a multiple of each number. (Some problems write it as m.c.m.) It tells you when repeating cycles line up again.

Two number lines marking the multiples of 12 and the multiples of 18 up to 72. They first meet at 36 and again at 72. multiples of 12 0122436486072 018365472 LCM = 36 multiples of 18

Check: for any two numbers, $\text{GCF} \times \text{LCM}$ equals the product of the numbers. For $72$ and $120$: $24 \times 360 = 8{,}640 = 72 \times 120$. ✓ (Here $\text{LCM}(72, 120) = 2^3 \times 3^2 \times 5 = 360$.)

Working backward. If an LCM is given, each exponent in it is the largest exponent among the numbers. So an unknown exponent can't be bigger than the one in the LCM, and at least one number must reach it exactly.

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In a satellite network, Satellite A completes an orbit every $1260$ minutes, Satellite B every $1584$ minutes and Satellite C every $n$ minutes. It is known that the…

In a satellite network, Satellite A completes an orbit every $1260$ minutes, Satellite B every $1584$ minutes and Satellite C every $n$ minutes.

It is known that the last time the three coincided in the sky over Chicago was exactly the time indicated by their $m.c.m$ ago.

If the prime factorization of $n$ is $2^{3} \cdot 3^{2} \cdot 7$, calculate:

  1. The prime factorization of the times of satellites A and B.

  2. The total time (in minutes) that must pass for the three satellites to coincide again. Leave the result expressed as a product of powers of prime factors.

Coincidence vector (LCM) Sat A: 1260 min Sat B: 1584 min Sat C: 2³·3²·7 min CHICAGO
Solve
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We have three natural numbers decomposed into prime factors: $n_1 = 2^{3} \cdot p \cdot 7$ $n_2 = 2^{q} \cdot 3^{2} \cdot 5$ $n_3 = 3 \cdot 5^{2} \cdot 11$ If we know…

We have three natural numbers decomposed into prime factors:

$n_1 = 2^{3} \cdot p \cdot 7$

$n_2 = 2^{q} \cdot 3^{2} \cdot 5$

$n_3 = 3 \cdot 5^{2} \cdot 11$

If we know that:

  • $M.C.D.(n_1, n_2) = 8$

  • $m.c.m.(n_2, n_3) = 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 11$

Find the values of the unknown prime numbers or exponents $p$ and $q$. Justify your answer using the properties of prime factorization for calculating the $M.C.D.$ and the $m.c.m$.

Solve

Perfect squares and perfect cubes

A perfect square is a number like $36 = 6^2$. In prime factors, squaring a number doubles every exponent. So a number is a perfect square exactly when all exponents are even. For a perfect cube, all exponents are multiples of $3$.

Taking the root goes the other way: divide every exponent by $2$ (square root) or by $3$ (cube root). For example, $\sqrt{2^4 \times 3^2} = 2^2 \times 3 = 12$.

Making 72 = 2 cubed times 3 squared a perfect square: the exponent 3 is odd, so multiply by 2 to get 144 = 2 to the fourth times 3 squared, which is 12 squared. 72 = 2³ × 3² 2³: exponent 3 is odd → need one more 2 3²: exponent 2 is even → fine 72 × 2 = 144 = 2⁴ × 3² All exponents even → 144 = 12²

For a number that must be a perfect square and a multiple of several numbers, find the LCM first, then fix the odd exponents. For example, for $12 = 2^2 \times 3$ and $20 = 2^2 \times 5$: $\text{LCM} = 2^2 \times 3 \times 5$. The exponents of $3$ and $5$ are odd, so multiply by $3 \times 5$ to get $2^2 \times 3^2 \times 5^2 = 900 = 30^2$.

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We say that a number is a perfect square if all the exponents of its prime factorization are even numbers. a) Find the prime factorization of the number $180$. Is it a…

We say that a number is a perfect square if all the exponents of its prime factorization are even numbers.

  1. a) Find the prime factorization of the number $180$. Is it a perfect square?
  2. b) What is the smallest number by which you would have to multiply $180$ so that the result is indeed a perfect square? Justify your answer by analyzing the exponents.
Solve
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The volume of a cube-shaped box is exactly $1728$ in$^3$. a) Find the prime factorization of the number $1728$. b) Knowing that the volume of a cube is calculated as…

The volume of a cube-shaped box is exactly $1728$ in$^3$.

  1. a) Find the prime factorization of the number $1728$.
  2. b) Knowing that the volume of a cube is calculated as $V = l^3$ (where $l$ is the edge), use the previous factorization to group the factors into three identical groups and find the measurement of the edge.
l l l V = 1728 in³ = l³
Solve
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Find out which is the smallest number that simultaneously meets these three conditions: It is a perfect square (all the exponents of its prime factorization are even).…

Find out which is the smallest number that simultaneously meets these three conditions:

  • It is a perfect square (all the exponents of its prime factorization are even).

  • It is a multiple of $42$.

  • It is a multiple of $60$.

To solve it, use the prime factorization of $42$ and $60$ and reason which factors and exponents you must add to obtain the minimum perfect square.

Solve

Back to the start: the cupcakes. $24 = 2^3 \times 3$ is the only prime factorization, and from it you can read everything: it has $(3+1)(1+1) = 8$ factors ($1, 2, 3, 4, 6, 8, 12, 24$), so there are exactly $8$ box sizes that work.

Summary

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