Question

X! is completely divisible by ${11}^{51}$ but not by ${11}^{52}$. What is the sum of digits of largest such number X?___

Answer 1

X! is completely divisible by ${11}^{51}$. So the value of X should be less than $11\times 51=561$
Highest power of 11 in $561!$ $\left[\frac{561}{11}\right]+\left[\frac{561}{{11}^2}\right]+\dots \dots =51+4=55$
If we subtract $11\times 3=33$ from 561 we get 561 - 33 = 528. Highets power of 11 in 528! is
$\left[\frac{528}{11}\right]+\left[\frac{527}{{11}^2}\right]=47+4=51$
So the required number is 527
Sum of the digits = 5 + 2 + 7 = 14