The factorial of a number n is exactly divisible by (2^11×11^2) then the least posible value of n is?
a.22 b.25 c. does not exist d.none of this

Hi Mainak,

For n! to be visible by $$2^{11}*11^2$$, n! must contain 2 11s and 11 2s.
The least value of n for which n! contains 2 11s is 22.
In 22!, the number of 2s is equal to $$\left \lfloor \frac{22}{2} \right \rfloor + \left \lfloor \frac{22}{2^2} \right \rfloor + \left \lfloor \frac{22}{2^3} \right \rfloor + \left \lfloor \frac{22}{2^4} \right \rfloor + ... $$ = 11 + 5 + 2 + 1 = 19
Hence, 22 satisfies both the required conditions => The least value of n is 22.

I think the answer will be 22 only because for 2^11 X 11^2 to be divisible by a number,it should have eleven 2's and two 11's atleast.. Now 22 is the least number whose factorial will have two 11's (22/11= 2) and the number of 2's in 22! will be= 22/2 + 22/4 + 22/8 + 22/16 = 11+5+2+1 = 19. Hence 22 is the answer I guess

For n! to be divisible by 2^11 and 11^2, it requires atleast 2 multiples of 11 in n! and atleast 11 multiples of 2..so 22 satisfies both conditions..
ANSWER: 22

22

22