If 73! has 16 zeroes at the end, how many zeroes will 80! : GMAT Problem Solving (PS)
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# If 73! has 16 zeroes at the end, how many zeroes will 80!

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If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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15 Feb 2013, 11:24
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If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end?

A. 16
B. 17
C. 18
D. 19
E. 20
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Feb 2013, 02:44, edited 1 time in total.
Edited the OA.
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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15 Feb 2013, 11:34
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2 x 5 gives you a zero at the end of the number. This means you are only looking for additional 5s (there are plenty of factors of 2 in 73! 75 has 2 5s and 80 has 1. So, there should be 3 more (this means that the answer is actually 19.

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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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15 Feb 2013, 18:15
emmak wrote:
If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end?

A. 16
B. 17
C. 18
D. 19
E. 20

Are these questions from a test? The OA given is (D), not (C)
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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16 Feb 2013, 02:51
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emmak wrote:
If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end?

A. 16
B. 17
C. 18
D. 19
E. 20

We can solve the question using info given in the stem (about 73!) but I find it easier to use direct approach of trailing zeros.

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)>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.

BACK TO THE ORIGINAL QUESTION:

According to above 80! has $$\frac{80}{5}+\frac{80}{25}=16+3=19$$ trailing zeros (take only the quotient into account).

For more on trailing zeros check: everything-about-factorials-on-the-gmat-85592-40.html

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Hope it helps.
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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03 May 2013, 14:29
Nice conceptual thinking by Jim. Faster way to solve.
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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06 May 2013, 18:05
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A tougher question wouldn't have the comment 73! so I chose to handle this one head on, that is, looking for the number of 0's that can be constructed from 80!

Prime Factoring: you need a 2 and a 5 to make a 10 (a "zero"), and there are TONS of 2's so let's skip these and focus on the (rarer) 5s:

80! = 1*2*3*4*5*6...*78*79*80

Since there are 80 consecutive numbers, there are 16 multiples of 5 in there, but if we're prime factoring, we need to remember that SOME multiples of 5 actually contain more than just one 5. Which? 25 comes to mind -- it's got two of them! So all the multiples of 25 actually contain two 5's (ie: 50 and 75)

So, to recap, we have 16 of them, plus 3 more (the additional 5's in 25, 50, and 75), so that makes 19, and since we have more than enough 2's, we know our number will have exactly 19 zeros at the end.

Hope this helps!
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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23 Jan 2015, 20:42
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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24 May 2016, 21:48
Hello from the GMAT Club BumpBot!

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Re: If 73! has 16 zeroes at the end, how many zeroes will 80!   [#permalink] 24 May 2016, 21:48
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