If n is a positive integer such that n^3 is divisible by 2500, what must be the value of n^2 in relation to its factors?

If n is a positive integer such that n^3 is divisible by 2500, what must be the value of n^2 in relation to its factors?

• A. 2^2 * 5^4
• B. 5^6
• C. 2^4 * 5^2
• D. 2^3 * 5^6

Answer: (A)

To determine the appropriate value of \( n^2 \) in relation to its factors when \( n^3 \) is divisible by 2500, we begin by examining the prime factorization of 2500. 2500 can be expressed as: \[ 2500 = 25 \times 100 = 5^2 \times (10^2) = 5^2 \times (2 \times 5)^2 = 5^2 \times 2^2 \times 5^2 = 2^2 \times 5^4 \] Thus, the prime factorization of 2500 is: \[ 2500 = 2^2 \times 5^4 \] Next, since \( n^3 \) is divisible by \( 2500 \), it follows that: \[ n^3 = k \cdot (2^2 \times 5^4) \] for some integer \( k \). For \( n^3 \) to contain the factors \( 2^2 \) and \( 5^4 \), we can deduce how many times \( n \) must contain the factors 2 and 5. Let

To determine the appropriate value of ( n^2 ) in relation to its factors when ( n^3 ) is divisible by 2500, we begin by examining the prime factorization of 2500.

2500 can be expressed as:

[

2500 = 25 \times 100 = 5^2 \times (10^2) = 5^2 \times (2 \times 5)^2 = 5^2 \times 2^2 \times 5^2 = 2^2 \times 5^4

]

Thus, the prime factorization of 2500 is:

[

2500 = 2^2 \times 5^4

]

Next, since ( n^3 ) is divisible by ( 2500 ), it follows that:

[

n^3 = k \cdot (2^2 \times 5^4)

]

for some integer ( k ). For ( n^3 ) to contain the factors ( 2^2 ) and ( 5^4 ), we can deduce how many times ( n ) must contain the factors 2 and 5.

Let