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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>.<math></math>
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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 $ 125-2=123, \boxed{\text{C}}  
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soln by RNVAA
  
 
==See Also==
 
==See Also==

Latest revision as of 06:55, 31 August 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}}


soln by RNVAA

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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