Middle School Math Quiz: Factors And Multiples

5 questions · exam conditions

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Factors And MultiplesQuestion 1 of 5

A number $N$ has the property that when divided by 12, the remainder is 8, and when divided by 18, the remainder is 14. What is the remainder when $N$ is divided by the least common multiple of 12 and 18?

8

14

20

32

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Middle School Math Quiz

Middle School Math Quiz: Factors And Multiples

Practice Factors And Multiples 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 Factors And Multiples, 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

$8$
$14$
$20$
$32$ (correct answer)

Explanation: We have

N \equiv 8 \pmod{12}

and

N \equiv 14 \pmod{18}

. Notice that

N = 12q + 8

and

N = 18r + 14

for some integers

q

and

r

. Also,

N + 4 \equiv 0 \pmod{12}

and

N + 4 \equiv 0 \pmod{18}

, so

N + 4

is a multiple of both 12 and 18. The LCM of 12 and 18 is

\text{lcm}(12,18) = \frac{12 \times 18}{\gcd(12,18)} = \frac{216}{6} = 36

. So

N + 4 \equiv 0 \pmod{36}

, which means

N \equiv -4 \equiv 32 \pmod{36}

Question 2

How many positive integers less than 100 are multiples of 6 but not multiples of 9?

$10$
$11$ (correct answer)
$12$
$16$

Explanation: Multiples of 6 less than 100:

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96

. That's

\lfloor 99/6 \rfloor = 16

multiples. Among these, we need to exclude those that are also multiples of 9. A number is a multiple of both 6 and 9 if and only if it's a multiple of

\text{lcm}(6,9) = 18

. Multiples of 18 less than 100:

18, 36, 54, 72, 90

. That's

\lfloor 99/18 \rfloor = 5

multiples. Therefore, there are

16 - 5 = 11

positive integers less than 100 that are multiples of 6 but not multiples of 9.

Question 3

The product of two positive integers is 360. If their greatest common factor is 6, what is the minimum possible value of their least common multiple?

$60$ (correct answer)
$90$
$120$
$360$

Explanation: Let the two numbers be

a

and

b

. We know

ab = 360

and

\gcd(a,b) = 6

. Since

\gcd(a,b) \times \text{lcm}(a,b) = ab

, we have

6 \times \text{lcm}(a,b) = 360

, so

\text{lcm}(a,b) = 60

. The least common multiple is always the same value (60) regardless of which specific pair

(a,b)

we choose, as long as

ab = 360

and

\gcd(a,b) = 6

Question 4

The least common multiple of two numbers is 180, and their greatest common factor is 12. If one of the numbers is 36, what is the other number?

$45$
$60$ (correct answer)
$72$
$90$

Explanation: For any two numbers

a

and

b

, we have

\gcd(a,b) \times \text{lcm}(a,b) = a \times b

. Given that

\gcd = 12

\text{lcm} = 180

, and one number is 36, we have:

12 \times 180 = 36 \times b

, so

2160 = 36b

, which gives

b = 60

. We can verify:

\gcd(36, 60) = 12

and

\text{lcm}(36, 60) = 180

Question 5

If $x$ and $y$ are positive integers such that $x$ divides $y^2$ and $y$ divides $x^3$ , and $\gcd(x,y) = 1$ , what can be concluded about $x$ and $y$ ?

Both $x$ and $y$ must be perfect squares
Both $x$ and $y$ must be perfect cubes
Both $x$ and $y$ must equal 1 (correct answer)
$x$ must be a perfect cube and $y$ must be a perfect square

Explanation: Since

\gcd(x,y) = 1

x

and

y

share no common prime factors. If

x

divides

y^2

, then every prime power in the factorization of

x

must appear in

y^2

with at least the same exponent. But since

\gcd(x,y) = 1

x

and

y

have no common prime factors, so

x

can only divide

y^2

x = 1

. Similarly, since

y

divides

x^3

and

\gcd(x,y) = 1

, we must have

y = 1

. Therefore, both

x = 1

and

y = 1