1974 AHSME Problems/Problem 26

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Problem

The number of distinct positive integral divisors of $(30)^4$ excluding $1$ and $(30)^4$ is

$\mathrm{(A)\ } 100 \qquad \mathrm{(B) \ }125 \qquad \mathrm{(C) \  } 123 \qquad \mathrm{(D) \  } 30 \qquad \mathrm{(E) \  }\text{none of these}$

Solution

The prime factorization of $30$ is $2\cdot3\cdot5$, so the prime factorization of $30^4$ is $2^4\cdot3^4\cdot5^4$. Therefore, the number of positive divisors of $30^4$ is $(4+1)(4+1)(4+1)=125$. However, we have to subtract $2$ to account for $1$ and $30^4$, so our final answer is $125-2=123, \boxed{\text{C}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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