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phuongdungnguyen
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If d is a positive integer and f is the product of the first 30 positi 18 Jan 2024, 04:37
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Bunuel
Trailing zeros:
Trailing zeros are a sequence of 0s 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)>n

It'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.

For more on this concept check Everything about Factorials on the GMAT: https://gmatclub.com/forum/everything-a ... 85592.html
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Hi, thank you for your post. But I don’t understand in the second part of the quotation 32/25. I mean 25=5.5 so why we just count 1 time for the second part instead of 2?
You posted this topic for a long time ago, but I really hope that you can see my comment.
Thank you so much

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Bunuel
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If d is a positive integer and f is the product of the first 30 positi 18 Jan 2024, 13:17
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phuongdungnguyen
Bunuel
Trailing zeros:
Trailing zeros are a sequence of 0s 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)>n

It'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.

For more on this concept check Everything about Factorials on the GMAT: https://gmatclub.com/forum/everything-a ... 85592.html

Hi, thank you for your post. But I don’t understand in the second part of the quotation 32/25. I mean 25=5.5 so why we just count 1 time for the second part instead of 2?
You posted this topic for a long time ago, but I really hope that you can see my comment.
Thank you so much

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I believe, this is already addressed in this topic before but here it goes again: we take only the quotient of the division above, that is for example, 32/25 = 1.
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If d is a positive integer and f is the product of the first 30 positi 01 Feb 2025, 09:14
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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