If n is a positive integer such that n^2 is divisible by 72, what is the smallest possible… — GRE General Test

If $n$ is a positive integer such that $n^2$ is divisible by 72, what is the smallest possible value of $n$ ?

The prime factorization of 72 is

2^3 \times 3^2

. For

n^2

to be a multiple of 72, the prime factorization of

n^2

must contain at least three factors of 2 and two factors of 3. Since

n^2

is a perfect square, the exponents of all prime factors in

n^2

must be even. Therefore, the exponent of 2 in

n^2

must be at least 4, making the minimum value of

n^2

equal to

2^4 \times 3^2 = 144

. Taking the square root yields

n = 12

, which is the smallest positive integer.

Find the prime factorization of 72.

72 = 2^3 \times 3^2

Decomposing 72 into prime factors reveals the minimum prime factor requirements for

n^2

Determine the prime factorization required for

n^2

For

n^2

to be divisible by

2^3 \times 3^2

, its prime factorization must contain at least three factors of 2 and two factors of 3. Because

n^2

is a perfect square, all exponents in its prime factorization must be even numbers. Thus,

n^2

must contain at least

2^4 \times 3^2 = 144

Exponents of prime factors in any perfect square must be even integers.

Take the square root to find the minimum value of

n

n = \sqrt{2^4 \times 3^2} = 2^2 \times 3^1 = 12

The smallest positive integer

n

must equal the square root of the minimal perfect square multiple of 72.

Prime Factorization and Divisibility of Perfect Squares

Question