1

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.

Community
  • 1
hululu
  • 4,913

1 Answers1

2

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 $11^{25}.11^4=11^{29}$

Archis Welankar
  • 16,138
  • 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. – hululu Apr 16 '16 at 12:56
  • Thanks for nice explanation. – hululu Apr 16 '16 at 12:57
  • You are welcome . – Archis Welankar Apr 16 '16 at 13:02
  • 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$". – DanielWainfleet Apr 16 '16 at 14:32