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 :
$E_2(51!)=E_2(2\cdot 4\cdot 6\cdot 8\cdot ....50)$
$=E_2\left(2^{25}\left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ....25\right)\right)$
$=25+E_2\left(1\cdot 2\cdot 3\cdot ....25\right)$
$=25+E_2\left(2\cdot 4\cdot 6\cdot 8\cdot ...24\right)$
$=25+E_2\left(2^{12}\left(1\cdot 2\cdot 3\cdot \mathrm{4...12}\right)\right)$
$=25+12+E_2\left(1\cdot 2\cdot 3\cdot \mathrm{4....12}\right)$
$37+E_2\left(2\cdot 4\cdot ...12\right)$
$=37+E_2\left(2^6\left(1\cdot 2\cdot ...6\right)\right)$
$=37+6+E_2\left(2\cdot 4\cdot 6\right)$
$=37+6+3E_2\left(1\cdot 2\cdot 3\right)$
$=37+6+3+1+E_2\left(1\cdot 3\right)=47$
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 $(47,12)$
$\therefore n=12$