fozzzy wrote:

If we were given a question to find the number of zero's in the product of the first 100 integers then can we use the following

we know that to find the zero's we need 5 and 2 so we can focus on the 5's

100/5 + 100/ 5^2 + 100/5^3 + 100/5^4....

we get 20 + 4 + 0 + 0 ( I know you can stop at 5^3 but just wrote the extra step for a general case)

so there will be 24 zeros? Is this approach correct and can I use it for all such questions

Trailing zeros: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.

For more on trailing zeros check:

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