What must be true about the integer n in relation to the divisibility by 2500 for n^3?

To determine the conditions under which ( n^3 ) is divisible by 2500, it is essential to analyze the prime factorization of 2500. The prime factorization of 2500 is ( 2^2 \times 5^4 ). For ( n^3 ) to be divisible by 2500, it must contain at least these prime factors with the required powers.

Since ( n^3 ) expresses the prime factors of ( n ), we can identify how the prime factorization of ( n ) must be structured. Specifically, if ( n ) is expressed in its prime factorization as ( n = 2^a \times 5^b \times \text{(other primes)} ), then:

• ( n^3 = 2^{3a} \times 5^{3b} \times \text{(other primes)}^3 )

To ensure ( n^3 ) is divisible by ( 2^2 ), it necessitates that ( 3a \geq 2 ). This gives ( a \geq \frac{2}{3} ). Since ( a ) must be an integer,