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Difference between revisions of "1974 AHSME Problems/Problem 26"

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

Revision as of 01:26, 4 May 2024

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 (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

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