Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 09 Feb 2013
Posts: 120

If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
Updated on: 16 Feb 2013, 03:44
Question Stats:
63% (00:26) correct 37% (00:26) wrong based on 736 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kudos will encourage many others, like me. Good Questions also deserve few KUDOS.
Originally posted by emmak on 15 Feb 2013, 12:24.
Last edited by Bunuel on 16 Feb 2013, 03:44, edited 1 time in total.
Edited the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 47183

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
16 Feb 2013, 03:51
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Status: Helping People Ace the GMAT
Affiliations: GMAT Answers
Joined: 16 Jan 2013
Posts: 183
Location: United States
Concentration: Finance, Entrepreneurship
GPA: 3.1
WE: Consulting (Consulting)

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
15 Feb 2013, 12:34
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?
_________________
Want to Ace the GMAT with 1 button? Start Here: GMAT Answers is an adaptive learning platform that will help you understand exactly what you need to do to get the score that you want.




Manager
Joined: 13 Dec 2012
Posts: 60

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
06 May 2013, 19:05
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!
_________________
Got questions? gmatbydavid.com



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2679

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
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
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 30 Oct 2016
Posts: 68
Location: United States

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



SC Moderator
Joined: 22 May 2016
Posts: 1835

If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
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.
_________________
In the depths of winter, I finally learned that within me there lay an invincible summer.  Albert Camus, "Return to Tipasa"



Math Expert
Joined: 02 Sep 2009
Posts: 47183

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 27 Jul 2017
Posts: 52

Re: If 73! has 16 zeroes at the end, how many zeroes will 80!
[#permalink]
Show Tags
05 Apr 2018, 08:14
I feel more comfortable solving the above question by using trailing zero concepts. https://gmatclub.com/forum/everythinga ... 85592.htmlThe answer must be D.
_________________
Ujjwal Sharing is Gaining!




Re: If 73! has 16 zeroes at the end, how many zeroes will 80! &nbs
[#permalink]
05 Apr 2018, 08:14






