stonecold wrote:

What is the greatest possible value of integer n if 200! is divisible by 33^n

A)6

B)12

C)18

D)19

E)66

33 = 3*11 i.e. to find the power of 33 in 200! we need to calculate the power of 11 and 3 in 200!

BUT

since power of 11 will be smaller (bigger prime number) than power of 3 (smaller prime number) in 200! so the power of 11 will be judgemental to find the power of 33

CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on

Where,

[n/x] = No. of Integers that are multiple of x from 1 to n

[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step

[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step

And so on.....

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)

i.e. [100/3] = [33.33] = 33

i.e. [100/9] = [11.11] = 11 etcNow, Power of prime 11 in 200! = [200/11] + [200/11^2] + [200/11^3] = 18+1+0 = 19

Answer: Option E

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