Let m and n be positive integers such that m^2 n is divisible by 72 and m n^2 is divisible… — GRE General Test

Let $m$ and $n$ be positive integers such that $m^2 n$ is divisible by $72$ and $m n^2$ is divisible by $108$ . What is the minimum possible value of the product $mn$ ?

A
12
B
18
C
24
36Answer
E
72

Answer

The product

(m^2 n)(m n^2) = (mn)^3

must be divisible by

72 \times 108 = 2^5 \times 3^5

. For

(mn)^3

to be a valid cube of an integer, the exponents of 2 and 3 in its prime factorization must be multiples of 3 greater than or equal to 5. The smallest such multiples of 3 are 6, meaning

(mn)^3 \ge 2^6 \times 3^6 = (36)^3

, so

mn \ge 36

. Setting

m = 6

and

n = 6

confirms that

36

satisfies all requirements.

Step-by-Step Solution

Express the given divisibility conditions in terms of prime factorizations.

72 = 2^3 \times 3^2

and

108 = 2^2 \times 3^3

. Therefore,

m^2 n

is a multiple of

2^3 \times 3^2

, and

m n^2

is a multiple of

2^2 \times 3^3

Prime factorization allows us to analyze the minimum exponent requirements for 2 and 3.

Multiply the two expressions to find a lower bound on

(mn)^3

(m^2 n)(m n^2) = m^3 n^3 = (mn)^3

must be divisible by

(2^3 \times 3^2)(2^2 \times 3^3) = 2^5 \times 3^5

Combining the expressions yields a perfect cube

(mn)^3

Determine the minimum prime powers needed for

mn

Since

(mn)^3

is a perfect cube divisible by

2^5 \times 3^5

, the exponents of 2 and 3 in

(mn)^3

must be multiples of 3 that are at least 5. The smallest such exponents are 6 for both 2 and 3. Thus,

(mn)^3 \ge 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3

, so

mn \ge 36

Exponents in a perfect cube factorization must be multiples of 3.

Verify that

mn = 36

is achievable with integer values of

m

and

n

Setting

m = 6

and

n = 6

gives

mn = 36

. Then

m^2 n = 6^3 = 216 = 3 \times 72

(divisible by 72) and

m n^2 = 6^3 = 216 = 2 \times 108

(divisible by 108).

Constructing valid integers

m

and

n

confirms that 36 is achievable.

Key Concept

Properties of Integers and Prime Factorization

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