If x is a positive integer such that 120 is a factor of x^2, what is the least possible va… — GRE General Test

If $x$ is a positive integer such that $120$ is a factor of $x^2$ , what is the least possible value of $x$ ?

The prime factorization of 120 is

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

. For

x^2

to be divisible by 120,

x^2

must contain at least three 2s, one 3, and one 5. Because

x^2

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

x^2

2^4 \times 3^2 \times 5^2

, which means

x

must be at least

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

Find the prime factorization of 120.

120 = 2^3 \times 3^1 \times 5^1

Decomposing 120 into prime factors reveals the minimal prime factor exponents required for

x^2

Determine the prime factors required for

x^2

to be divisible by 120.

The prime factorization of

x^2

must contain even exponents for all its prime factors and must be divisible by

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

. Therefore,

x^2

must contain at least

2^4 \times 3^2 \times 5^2

Because

x^2

is a perfect square, every prime factor in its prime factorization must have an even exponent.

Calculate the minimum value of

x

x = \sqrt{2^4 \times 3^2 \times 5^2} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60

Taking the square root of

2^4 \times 3^2 \times 5^2

gives the least positive integer

x

satisfying the condition.

Prime Factorization and Perfect Squares

Question