Middle School Math Quiz: Prime Factorization

3 questions · exam conditions

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Prime FactorizationQuestion 1 of 3

Two numbers $M$ and $N$ satisfy the equation $M \cdot N = 2^6 \cdot 3^4 \cdot 5^2$ and $\text{GCF}(M, N) = 2^2 \cdot 3^1$ . What is the value of $\text{LCM}(M, N)$ ?

2^4 \cdot 3^3 \cdot 5^2

2^6 \cdot 3^4 \cdot 5^2

2^8 \cdot 3^5 \cdot 5^2

2^4 \cdot 3^2 \cdot 5^1

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Middle School Math Quiz

Middle School Math Quiz: Prime Factorization

Practice Prime Factorization in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Prime Factorization, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two numbers $M$ and $N$ satisfy the equation $M \cdot N = 2^6 \cdot 3^4 \cdot 5^2$ and $\text{GCF}(M, N) = 2^2 \cdot 3^1$ . What is the value of $\text{LCM}(M, N)$ ?

$2^4 \cdot 3^3 \cdot 5^2$ (correct answer)
$2^6 \cdot 3^4 \cdot 5^2$
$2^8 \cdot 3^5 \cdot 5^2$
$2^4 \cdot 3^2 \cdot 5^1$

Explanation: We use the fundamental relationship:

M \cdot N = \text{GCF}(M, N) \cdot \text{LCM}(M, N)

. Given

M \cdot N = 2^6 \cdot 3^4 \cdot 5^2

and

\text{GCF}(M, N) = 2^2 \cdot 3^1

, we can find:

\text{LCM}(M, N) = \frac{M \cdot N}{\text{GCF}(M, N)} = \frac{2^6 \cdot 3^4 \cdot 5^2}{2^2 \cdot 3^1} = 2^{6-2} \cdot 3^{4-1} \cdot 5^2 = 2^4 \cdot 3^3 \cdot 5^2

. Choice A is correct. Choice B incorrectly gives the product

M \cdot N

. Choice C incorrectly adds exponents instead of subtracting. Choice D has incorrect exponents for all primes.

Question 2

A machine produces widgets in batches. The first batch contains $2^2 \cdot 3^1$ widgets, the second batch contains $2^1 \cdot 3^3$ widgets, and the third batch contains $2^3 \cdot 5^2$ widgets. If all widgets from the three batches are to be packed into identical boxes with no widgets left over, what is the largest number of widgets that can be placed in each box?

$2$ (correct answer)
$4$
$6$
$12$

Explanation: To pack all widgets into identical boxes with none left over, the number of widgets per box must be a divisor of each batch size. This means we need the GCF of the three batch sizes. Batch 1:

2^2 \cdot 3^1 = 12

widgets. Batch 2:

2^1 \cdot 3^3 = 54

widgets. Batch 3:

2^3 \cdot 5^2 = 200

widgets. To find GCF(12, 54, 200), we use the minimum exponent for each prime: For prime 2:

\min(2, 1, 3) = 1

. For prime 3:

\min(1, 3, 0) = 0

(since 200 has no factor of 3). For prime 5:

\min(0, 0, 2) = 0

(since 12 and 54 have no factor of 5). Therefore, GCF =

2^1 \cdot 3^0 \cdot 5^0 = 2

. Choice B (4) would not divide 54 evenly since 54 = 2¹ × 3³, and 4 = 2². Choice C (6) would not divide 200 since 200 has no factor of 3. Choice D (12) would not divide 200 evenly since 200 has no factor of 3.

Question 3

Consider all positive integers of the form $2^x \cdot 3^y \cdot 7^z$ where $x$ , $y$ , and $z$ are non-negative integers and $x + y + z = 4$ . How many such integers are there?

$10$
$15$ (correct answer)
$20$
$25$

Explanation: We need to count the number of non-negative integer solutions to

x + y + z = 4

. This is a classic "stars and bars" combinatorics problem. The number of ways to distribute 4 identical objects into 3 distinct bins is

\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15

. Alternatively, we can enumerate: (4,0,0), (3,1,0), (3,0,1), (2,2,0), (2,1,1), (2,0,2), (1,3,0), (1,2,1), (1,1,2), (1,0,3), (0,4,0), (0,3,1), (0,2,2), (0,1,3), (0,0,4). Counting these gives 15 solutions. Choice A (10) incorrectly assumes some constraint on the variables. Choice C (20) might result from incorrectly using

\binom{7}{2}

. Choice D (25) might result from incorrectly calculating

5^2