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The exponent of $11$ in the prime factorization of $ 300!$ is

  1. $27$
  2. $28$
  3. $29$
  4. $30$

My attempt:

According to Exponent of $p$ in the prime factorization of $n!$

$\left\lfloor\dfrac{300}{11}\right\rfloor+\left\lfloor\dfrac{300}{11^2}\right\rfloor=27+2=29$

Can you explain in alternative/formal way? Please.

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  • $\begingroup$ Do you mean to imply that the given method is not formal? $\endgroup$ – Hagen von Eitzen Apr 16 '16 at 12:35
  • $\begingroup$ @Hagen Thanks. No, I'm agree with given explanation, but I'm looking to understand in other way, if there. $\endgroup$ – ً ً Apr 16 '16 at 12:41
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See the terms having factor $11$ are $11,22,33,..121,...220,..297$ so excluding $121,242$ we have $25$ numbers which give only one $11$ and $121,242$ gives two $11$ so total exponent is $11^{25}.11^4=11^{29}$

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  • $\begingroup$ Yes, $121=11\times 11$ and $242=11\times 11\times 2$ will be include in exactly two times $11$, but other $25$ number included $11$ exactly once. $\endgroup$ – ً ً Apr 16 '16 at 12:56
  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – ً ً Apr 16 '16 at 12:57
  • $\begingroup$ You are welcome . $\endgroup$ – Archis Welankar Apr 16 '16 at 13:02
  • $\begingroup$ The general formula for the largest exponent of the prime $p$ in $n!$ is $\sum_{j=1}^{\infty}[np^{-j}]$, where $ [x]$ denotes the largest integer not exceeding $x$. Note that only finitely many terms in the summation are non-zero. We can replace the supscript "$ j=\infty$" with "$ j=n$". $\endgroup$ – DanielWainfleet Apr 16 '16 at 14:32

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