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 | + | 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]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:55, 31 August 2024
Problem
The number of distinct positive integral divisors of excluding and is
Solution
The prime factorization of is , so the prime factorization of is . Therefore, the number of positive divisors of is . However, we have to subtract to account for and , 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 | |
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