Let m and n be positive integers such that m is divisible by 18 and n is divisible by 15. … — GRE General Test

Let $m$ and $n$ be positive integers such that $m$ is divisible by $18$ and $n$ is divisible by $15$ . Which of the following integers MUST be a divisor of the product $m \cdot n$ ? Select all such values.

$54$ Answer
$90$ Answer
$135$ Answer
D
$36$
E
$60$

Answer

The integers

54

90

, and

135

must be divisors of the product

m \cdot n

Since

m

is a multiple of

18

(

2^1 \times 3^2

) and

n

is a multiple of

15

(

3^1 \times 5^1

), their product

m \cdot n

must be a multiple of

18 \times 15 = 270

. The prime factorization of

270

2^1 \times 3^3 \times 5^1

. Any integer that divides

270

is guaranteed to divide

m \cdot n

for all valid values of

m

and

n

. The numbers

54

(

2^1 \times 3^3

90

(

2^1 \times 3^2 \times 5^1

), and

135

(

3^3 \times 5^1

) are all divisors of

270

Step-by-Step Solution

Express

m

and

n

in terms of their minimal prime factorizations.

m = 18a = 2^1 \times 3^2 \times a

and

n = 15b = 3^1 \times 5^1 \times b

for positive integers

a

and

b

Divisibility conditions specify the minimum prime factors that

m

and

n

must contain.

Find the minimal guaranteed prime factorization of the product

m \cdot n

m \cdot n = (2^1 \times 3^2 \times a) \times (3^1 \times 5^1 \times b) = 2^1 \times 3^3 \times 5^1 \times (ab) = 270 \times ab

Multiplying

m

and

n

combines their guaranteed prime factor powers.

Determine which choices divide

270 = 2^1 \times 3^3 \times 5^1

without requiring additional factors of

a

b

54 = 2^1 \times 3^3

90 = 2^1 \times 3^2 \times 5^1

, and

135 = 3^3 \times 5^1

all divide

270

. Numbers requiring

2^2

(

36

and

60

) do not necessarily divide

270

A number MUST divide

m \cdot n

if its prime factor powers do not exceed the minimum guaranteed powers in

m \cdot n

Key Concept

Divisibility of Integer Products via Prime Factorization

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