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}} $

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1974 AHSME Problems/Problem 26

Problem

Solution

See Also