Difference between revisions of "1974 AHSME Problems/Problem 26"
(→Solution) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==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>}<math> | |
| + | </math>Solution By RNVAA$ | ||
==See Also== | ==See Also== | ||
Revision as of 06:51, 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 
}$$ (Error compiling LaTeX. Unknown error_msg)Solution 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
