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E
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What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24
Source -> Stone Cold's PS and DS Question Collection ( https://gmatclub.com/forum/stonecold-s- ... 17160.html )
A)20
B)21
C)22
D)23
E)24
Source -> Stone Cold's PS and DS Question Collection ( https://gmatclub.com/forum/stonecold-s- ... 17160.html )
22 Nov 2017, 13:26
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stonecold
We need to determine the largest value of n such that 100! is divisible by 15^n. For a number to be divisible by 15, it must be divisible by both 3 and 5. Thus, we need to find the largest value of n such that 100! is divisible by 3^n x 5^n.
Since we know there are fewer 5s in 100! than 3s, we can find the number of 5s and thus be able to determine the number of 5-and-3 pairs.
To determine the number of 5s within 100!, we can use the following shortcut in which we divide 100 by 5, then divide the quotient of 100/5 by 5 and continue this process until we no longer get a nonzero quotient.
100/5 = 20
20/5 = 4
Since 4/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 100!.
Thus, there are 20 + 4 = 24 factors of 5 within 100!
Since there are 24 factors of 5 within 100!, we also know that there are 24 5-and-3 pairs and, thus, the largest value of n is 24.
Answer: E
06 Nov 2016, 00:44
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stonecold
15 = 3*5 i.e. to find the power of 15 in 100! we need to calculate the power of 5 and 3 in 100!
BUT
since power of 5 will be smaller (bigger prime number) than power of 3 (smaller prime number) in 100! so the power of 5 will be judgemental to find the power of 15
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 5 in 100! = [100/5] + [100/5^2] + [100/5^3] = 20+4+0 = 24
Answer: Option E
