If n is a positive integer such that n is divisible by 12 and n^2 is divisible by 180, wha… — GRE General Test

If $n$ is a positive integer such that $n$ is divisible by $12$ and $n^2$ is divisible by $180$ , what is the least possible value of $n$ ?

The correct answer is 60. The prime factorization of 12 is

2^2 \times 3^1

, requiring

n

to have at least two factors of 2 and one factor of 3. The prime factorization of 180 is

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

. For

n^2

to be divisible by 180,

n

must contribute a factor of 5 (which becomes

5^2

n^2

). Multiplying the minimal prime factors yields

2^2 \times 3^1 \times 5^1 = 60

Analyze the prime factorization required for

n

to be divisible by 12

Since

12 = 2^2 \times 3^1

, any positive integer

n

divisible by 12 must contain at least

2^2

and

3^1

in its prime factorization.

An integer must contain all prime factors of its divisor with at least equal exponents.

Analyze the prime factorization required for

n^2

to be divisible by 180

Since

180 = 2^2 \times 3^2 \times 5^1

n^2

must contain at least

2^2

3^2

, and

5^1

. Because

n^2

doubles all prime exponents of

n

n

must contribute at least

5^1

, which makes the factor of 5 in

n^2

equal to

5^2

The exponent of any prime factor in a perfect square

n^2

must be even.

Combine the minimal prime factor requirements for

n

The minimum prime factorization for

n

2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60

Taking the minimum required power of each prime factor yields the smallest positive integer fulfilling both conditions.

Prime Factorization and Divisibility of Powers

Question