AndreG wrote:
How many zeros does 100! end with?
• 20
• 24
• 25
• 30
• 32
expl.
Find how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by 25 = 5^2 . So, the answer is 24.
The correct answer is B.
I have absolutely no Idea what they are telling me...
Can someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?
Thanks!
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of
n!, the factorial of a non-negative integer n, can be determined with this formula:
\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>nIt's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32,
5^2=25 is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:According to above 100! has
\frac{100}{5}+\frac{100}{25}=20+4=24 trailing zeros.
Answer: B.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
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78% (01:18) correct
