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I got a question when I started factorials

Q. If $a^8$ and $8^a$ is completely divisible by $50!$ Then which one of the following is true about 'highest value of a'?

(A) $10<a<14$
(B) $14<a<16$
(C) $16<a<18$
(D) $18<a<20$

My approach : If we divide $8^a$ by $50!$ then the quotient will not be an integer for any value of a.

But the options in this question doesn't say so. What am I doing wrong?

EDIT: Solution given in the book is as follows:

Since highest power of $2$ in $50!$ is $47$ hence highest power of $8$ in $50!$ is $[47/3]$ = $15$. Here [] sign denotes greatest integer function.
If we assume a = $15$ then $15^8$ or $3^8$ and $5^8$.
So we have to check that whether $3^8$ and $5^8$ are divisible by $50!$ or not, Since highest power of $3$ in $50!$ is $22$ and that of $5$ is $12$, hence $a = 15$ will satisfy the condition.

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manshu
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  • If $a<20$ (as all the answers state), we can also find primes that divide $50!$ but don't divide $a^8$ (for instance $23, 29, 31...$), thus making $a^8 / 50!$ not an integer as well. Are you sure that the question is not: "If $a^8$ and $8^a$ divide $50!$, then find $a$"? – Paolo Franchi Jan 09 '16 at 22:17
  • @PaoloFranchi i double cross-checked...no mistake in typing the question – manshu Jan 09 '16 at 22:19
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    I can only imagine that $a$ is not an integer, but if $a$ is irrational, then there are limited ways in which $a^8$ can be an integer. Moreover, if somehow $a$ satisfies the divisibility criteria, so will $2a$. So it seems to me the problem must somehow be garbled. My suspicion is that $a$ should be an integer and you've reversed the order of division, i.e. that $50!$ should be divisible by both $a^8$ and $8^a$. Please check. – hardmath Jan 09 '16 at 22:21
  • See if this helps:http://math.stackexchange.com/questions/391067/power-of-a-number-in-a-factorial – NoChance Jan 09 '16 at 22:26
  • @PaoloFranchi just added the book solution. – manshu Jan 09 '16 at 22:28
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    @hardmath just added the book solution. – manshu Jan 09 '16 at 22:29
  • @NoChance I did these kind of questions an hour ago. thanks for the reference though. – manshu Jan 09 '16 at 22:30
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    So unless you misread after all (another double cross-check wouldn't hurt), the author goofed. Clearly they meant $a^8$ and $8^a$ divide $50!$. – Daniel Fischer Jan 09 '16 at 22:30
  • @DanielFischer Another triple check this time. No mistake by my side. – manshu Jan 09 '16 at 22:33

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As you note in the comments, you're right. The author made a typo. This is something that people mess up all the time and is well worth practicing: $a|b$, "a divides b" is not the same as $\frac{a}{b}\in\mathbb{Z}$. Last week a math PhD student got massively confused in a grad CS class after making this very mistake. Always be on your guard about it.

Stella Biderman
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