# 1974 AHSME Problems/Problem 26

## 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}}

soln by RNVAA

## See Also

 1974 AHSME (Problems • Answer Key • Resources) Preceded byProblem 25 Followed byProblem 27 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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