Question

The largest value of $n$, so that $10^n$ divides $51!$ is

• A. $47$
• B. $12$
• C. $35$
• D. $59$

Answer 1

Answer: (B)

Fact :
$E_{p}\left(n!\right) = \left[\frac{n}{P}\right] +\left[\frac{n}{P^{2}}\right] +\left[\frac{n}{P^{3}}\right]+... +\left[\frac{n}{P^{s}}\right]$ such that
$P^s < n < P^{s + 1}$, where $[x]$ represents $G.I.F$.
Now, prime factors of $10 = 2 \times 5$
$\because E_2(51!)$
$=\left[\frac{51}{2}\right]+\left[\frac{51}{2^{2}}\right]+\left[\frac{51}{2^{3}}\right] + \left[\frac{51}{2^{4}}\right]+\left[\frac{51}{2^{5}}\right]+\left[\frac{51}{2^{6}}\right]$
$= 25 + 12 + 6 + 3 + 1 + 0 = 47 $
Again, $E_{5} \left(51!\right) = \left[\frac{51}{5}\right]+\left[\frac{51}{5^{2}}\right]+\left[\frac{51}{5^{3}}\right] $
$= 10 + 2 = 12 $
$E_{10} (51!)=$ number of zeroes in $51!$
$\therefore E_{10} (51!) =$ Minimum of $(47, 12) = 12$
$\therefore 10^n = 10^{12}$ so $n =12$.
Short Cut Method :
$E_2(51!) = E_2(2 \cdot 4 \cdot 6 \cdot 8 \cdot .... 50)$
$=E_{2}\left(2^{25}\left(1\cdot2\cdot3\cdot4\cdot5\cdot....25\right)\right) $
$ = 25 + E_{2}\left(1\cdot2\cdot3\cdot....25\right) $
$= 25 + E_{2}\left(2 \cdot 4 \cdot 6 \cdot 8\cdot...24\right) $
$ = 25+E_{2}\left(2^{12}\left(1\cdot2\cdot3\cdot4...12\right)\right) $
$= 25+12 + E_{2}\left(1\cdot2\cdot3\cdot4....12\right) $
$37+E_{2}\left(2\cdot4\cdot...12\right) $
$= 37 + E_{2}\left(2^{6}\left(1\cdot2\cdot...6\right)\right) $
$ = 37+ 6 +E_{2}\left(2\cdot4\cdot6\right) $
$ = 37 + 6 +3 E_{2}\left(1\cdot2\cdot3\right)$
$= 37+6+3+1+E_{2}\left(1\cdot3\right) = 47$
Similarly, $E_5(51!) = 12$
For $E_{10}(51!)$, we need at least one $2$ and one $5$
$\therefore $ Max. exponent of $10 =$ Minimum of $E_2 (51!)$ and
$E_5(51!)$ i.e. Minimum of $(47, 12)$
$\therefore n = 12$