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16 Nov 2021, 08:00
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If K is a positive odd integer and K! has 49 trailing zeros, then how many values of K are possible?
A. 5
B. 4
C. 3
D. 2
E. 1
A. 5
B. 4
C. 3
D. 2
E. 1
16 Nov 2021, 08:24
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I've never seen an official question about "trailing zeros", and if the GMAT did ask a question about them, the question would need to explain what they are -- they're just the zeros at the end of an integer, so a number like 67,000 has three 'trailing zeros'.
For a number to have 49 trailing zeros, it will need to be divisible by 10^49, but not by 10^50. So our number must be divisible by 10^49 = (2*5)^49 = 2^49 * 5^49. Now our number is a factorial, and factorials will have lots more 2's than 5's in them, so really we just want to know how many odd numbers k there are so that k! is divisible precisely by 5^49.^49,
There will be faster ways to do this if one learns things you'd never need on the GMAT, but if I was looking for factorials divisible precisely by 5^49, I'd just do it by inspection. If we take a factorial like 200!, and write it out in full, we'll find we have forty multiples of 5, but we'll also have some multiples of 5^2 = 25 and of 5^3 = 125, so 200! will have roughly the right number of 5's, and we can check exactly how many it has: we have forty multiples of 5, another eight multiples of 5^2, and one multiple of 5^3 in 200!, for a total of forty-nine 5's. Since k is odd, k cannot equal 200, but it can be 201 or 203 (it can't be 205, because then we'd have fifty 5's), for two odd values of k.
For a number to have 49 trailing zeros, it will need to be divisible by 10^49, but not by 10^50. So our number must be divisible by 10^49 = (2*5)^49 = 2^49 * 5^49. Now our number is a factorial, and factorials will have lots more 2's than 5's in them, so really we just want to know how many odd numbers k there are so that k! is divisible precisely by 5^49.^49,
There will be faster ways to do this if one learns things you'd never need on the GMAT, but if I was looking for factorials divisible precisely by 5^49, I'd just do it by inspection. If we take a factorial like 200!, and write it out in full, we'll find we have forty multiples of 5, but we'll also have some multiples of 5^2 = 25 and of 5^3 = 125, so 200! will have roughly the right number of 5's, and we can check exactly how many it has: we have forty multiples of 5, another eight multiples of 5^2, and one multiple of 5^3 in 200!, for a total of forty-nine 5's. Since k is odd, k cannot equal 200, but it can be 201 or 203 (it can't be 205, because then we'd have fifty 5's), for two odd values of k.
30 Sep 2025, 06:38
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Trailing zeros in a factorial are basically the highest power of 10 in any factorial
1! to 4! have 0 trailing zeros
5! to 9! have 1 trailing zero
10! to 14! have 2 trailing zeros and so on
You can see that they're divided into chunks which contain 4 numbers, 2 odd & 2 even
Therefore, if K! has 49 trailing zeros, K can take 4 values and we know that K is odd
Hence, K can only take 2 values.
1! to 4! have 0 trailing zeros
5! to 9! have 1 trailing zero
10! to 14! have 2 trailing zeros and so on
You can see that they're divided into chunks which contain 4 numbers, 2 odd & 2 even
Therefore, if K! has 49 trailing zeros, K can take 4 values and we know that K is odd
Hence, K can only take 2 values.
Bunuel
