$y$ and $n$ are positive integers.
$1!\times2!\times3!\times...\times26! = y\times13^n$
$n$ is equals ?
($n$ is above)
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Do you mean $y*13^n$ with positive integers $y$ and $n$? – n55 Jul 07 '15 at 15:33
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yes exactly ... – Kaan Şan Jul 07 '15 at 15:34
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Write down the prime factorisation of the whole number, and then see how many $13$'s you've got. (Note that it's semi-factorised already, so it shouldn't be too hard.) – Arthur Jul 07 '15 at 15:37
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There are multiple answers of course, with $1\leq n\leq 15$. Are you looking for a specific one? Maybe the one with the largest $n$? – JMoravitz Jul 07 '15 at 15:39
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yes the largest n – Kaan Şan Jul 07 '15 at 15:40
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answers are 11 12 13 14 15 – Kaan Şan Jul 07 '15 at 15:41
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See http://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n. – lhf Jul 07 '15 at 15:44
2 Answers
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In $x!$, where $ 13 \le x \le 25$ there is only one 13 in each number: this gives $25-13+1 = 13$ of the nmber Thirteen. Note that when $x=26$, the number has 2 "Thirteens" so the total answer is $n = 15$
Perhaps a more interesting or instructive question would be to count the number of zeroes at the end of the your number.
pMarkov
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I'm assuming you're looking for the largest $n$.
Hint: Since $13$ is prime, $1!$, $2!$, ..., $12!$ do not contain a factor of $13$. Each of $13!$, $14!$, ..., $25!$ have exactly one factor of $13$, and $26!$ has two factors of $13$ (one coming from $13$ and the other from $26=2*13$)... Add those together and that should be your answer.
n55
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