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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, 12:24
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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
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Last edited by Bunuel on 16 Feb 2013, 03:44, edited 1 time in total.
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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, 12:34
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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. What is the source of your question?
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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, 03:51
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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 nonnegative 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 80! has \(\frac{80}{5}+\frac{80}{25}=16+3=19\) trailing zeros (take only the quotient into account). Answer: D. For more on trailing zeros check: everythingaboutfactorialsonthegmat8559240.htmlSimilar questions to practice: if60iswrittenoutasanintegerwithhowmany101752.htmlhowmanyzerosdoes100endwith100599.htmlfindthenumberoftrailingzerosintheexpansionof108249.htmlfindthenumberoftrailingzerosintheproductof108248.htmlifnistheproductofallmultiplesof3between1and101187.htmlifmistheproductofallintegersfrom1to40inclusive108971.htmlifpisanaturalnumberandpendswithytrailingzeros108251.htmlif10252isdivisibleby10nwhatisthegreatest106060.htmlpandqareintegersifpisdivisibleby10qandcannot109038.htmlquestionaboutpprimeintonfactorial108086.htmlifnistheproductofintegersfrom1to20inclusive106289.htmlwhatisthegreatestvalueofmsuchthat4misafactorof105746.htmlifdisapositiveintegerandfistheproductofthefirst126692.htmlif10252isdivisibleby10nwhatisthegreatest106060.htmlHope 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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06 May 2013, 19:05
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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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31 Aug 2017, 10:23
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 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 fiveandtwo 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 fivetwo 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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22 Dec 2017, 14:59
Bunuel wrote: 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 nonnegative 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 80! has \(\frac{80}{5}+\frac{80}{25}=16+3=19\) trailing zeros (take only the quotient into account). Answer: D. For more on trailing zeros check: http://gmatclub.com/forum/everythingab ... 9240.htmlSimilar questions to practice: http://gmatclub.com/forum/if60iswrit ... 01752.htmlhttp://gmatclub.com/forum/howmanyzero ... 00599.htmlhttp://gmatclub.com/forum/findthenumb ... 08249.htmlhttp://gmatclub.com/forum/findthenumb ... 08248.htmlhttp://gmatclub.com/forum/ifnisthep ... 01187.htmlhttp://gmatclub.com/forum/ifmisthep ... 08971.htmlhttp://gmatclub.com/forum/ifpisanat ... 08251.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlhttp://gmatclub.com/forum/pandqarei ... 09038.htmlhttp://gmatclub.com/forum/questionabou ... 08086.htmlhttp://gmatclub.com/forum/ifnisthep ... 06289.htmlhttp://gmatclub.com/forum/whatistheg ... 05746.htmlhttp://gmatclub.com/forum/ifdisapos ... 26692.htmlhttp://gmatclub.com/forum/if10252i ... 06060.htmlHope it helps. Bunel, why does your formula stop working for 90! > I get 21 with your formula but the real answer is 18!



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If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]
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22 Dec 2017, 16:44
arcticTO wrote: 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 Bunel, why does your formula stop working for 90! > I get 21 with your formula but the real answer is 18! arcticTO , the formula does not stop working. Your calculations for 90! are correct. There are 21 trailing zeros in 90! But 90! is not the number in the original prompt. There is some confusion here. In this thread, there are two similar but not identical questions. You are looking at a different question, one whose prompt has incorrect math, and whose answer is incorrect. Here is the topic's question, at the topemmak 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 Three posts above you is this question whose prompt and answer are both mathematically incorrect: gmatchile wrote: 2. If 83! has 16 zeroes at the end WRONG  16 is NOT the number of trailing zeros for 83!, how many zeroes will90! have at the end?
A. 16 B. 17 C. 18 D. 19 E. 20 The numbers have changed. Bunuel , it looks as if the confusing question, located here, whose prompt is inaccurate and not identical to this topic's prompt, should go elsewhere. I'm thinking the virtual shredder might be good.
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80! [#permalink]
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23 Dec 2017, 01:13
genxer123 wrote: arcticTO wrote: 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 Bunel, why does your formula stop working for 90! > I get 21 with your formula but the real answer is 18! arcticTO , the formula does not stop working. Your calculations for 90! are correct. There are 21 trailing zeros in 90! But 90! is not the number in the original prompt. There is some confusion here. In this thread, there are two similar but not identical questions. You are looking at a different question, one whose prompt has incorrect math, and whose answer is incorrect. Here is the topic's question, at the topemmak 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 Three posts above you is this question whose prompt and answer are both mathematically incorrect: gmatchile wrote: 2. If 83! has 16 zeroes at the end WRONG  16 is NOT the number of trailing zeros for 83!, how many zeroes will90! have at the end?
A. 16 B. 17 C. 18 D. 19 E. 20 The numbers have changed. Bunuel , it looks as if the confusing question, located here, whose prompt is inaccurate and not identical to this topic's prompt, should go elsewhere. I'm thinking the virtual shredder might be good. Thank you. Removed confusing posts.
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Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
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