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05 Nov 2016, 14:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
58% (01:16) correct
42%
(01:32)
wrong
based on 507
sessions
History
Date
Time
Result
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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
A)6
B)12
C)18
D)19
E)66
06 Nov 2016, 01:59
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stonecold
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 etc
Now, Power of prime 11 in 200! = [200/11] + [200/11^2] + [200/11^3] = 18+1+0 = 19
Answer: Option E
General Discussion
05 Nov 2016, 15:28
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33^n= 3^n * 11^n
Highest prime factor will be the limiting factor.
200/11 + 200/121 = 18+1 = 19
D
Sent from my iPhone using GMAT Club Forum mobile app
Highest prime factor will be the limiting factor.
200/11 + 200/121 = 18+1 = 19
D
Sent from my iPhone using GMAT Club Forum mobile app
