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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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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).

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.

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

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

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

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30 Aug 2017, 21:04

2.- If 83! has 16 zeroes at the end, how many zeroes will 90! have at the end?

A. 16 B. 17 C. 18 D. 19 E. 20
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]

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30 Aug 2017, 21:08

gmatchile wrote:

2.- If 83! has 16 zeroes at the end, how many zeroes will 90! have at the end?

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

83! = 16 zeroes at the end 90! = 83! 84*85*86*87*88*89*90=(16 ceros)*(2*42)*(5*17)*(2*43)*(3*29)*(2*44)*89*(9*10)= =(10^16)*(2*5)*42*17*2*43*3*29*2*44*89*9*10=(10^16)*10*10*42*17*43*3*29*2*44*89*9=(10^18)*…… 90! Increment two zeroes at the end. 16 + 2 = 18 zeroes at the end. c)
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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

Since we are given that 73! has 16 zeros at the end, determining the number of zeros 80! will have at the end is the same as determining the number of extra zeros that will be produced from the following product:

74 x 75 x 76 x 77 x 78 x 79 x 80

To determine the number of zeros, we need to determine the number of pairs of fives and twos that are created from the above product. Note that we need five-and-two pairs because 5 x 2 = 10, and each 10 creates an additional trailing zero.

Since we know there are going to be fewer fives than twos, let’s determine the number of fives.

80 = 2^4 x 5

75 = 5^2 x 3

We see that since there are 3 fives, there will be 3 five-two pairs, and thus 3 trailing zeros. Therefore, there will be a total of 19 trailing zeros.

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

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02 Sep 2017, 10:23

To figure out how many trailing zeros are in 80!, one will need to be aware that the only way to get a zero at the end of a number is if that number has a 10 as a factor, and by having 10 as a factor, that number will also have 2 and 5 as factors. Effectively, all you have to do is figure out how many factors of 5 are there in that number.

Since the question is asking for 1*2*3*...*80, we know that there are 16 multiples of 5 (for example: one for 5, another one for 10, another one for 15, and so on). We are not done yet because we have to account for any factors with extra 5's. In this case we have 25, 50, and 75.

25: 5*5 -> an extra 5 50: 5*5*2 -> an extra 5 75: 5*5*3 -> an extra 5

All together we have 16 multiples of 5 plus three extra 5's for each multiple of 25's. Answer is D=19.