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

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==Solution==
 
==Solution==
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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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==
 
{{AHSME box|year=1974|num-b=25|num-a=27}}
 
{{AHSME box|year=1974|num-b=25|num-a=27}}
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

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 (ProblemsAnswer KeyResources)
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
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