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Re: Find the number of trailing zeros in the expansion of
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12 Oct 2013, 08:42
madn800 wrote: Buneul, here's my doubt: # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; for calculating trailing zeros up til 24! you did just 20/5=4. but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7) Suppose I want # of trailing zeros in 310! using your concept 310/5+310/5^2=62+12=74 trailing zeroes BUT using the factorial calculator below I am getting 76 trailing zeroes http://www.nitrxgen.net/factorialcalc.phpplease suggest a fool proof method for calculating trailing zeroes of any +ve integer. Please follow the link given in my first post. The # of trailing zeros of 310! is 310/5 + 310/5^2 + 310/5^3 = 62 + 12 + 2 = 76. Questions to practice: ifnisthegreatestpositiveintegerforwhich2nisafact144694.htmlwhatisthelargestpowerof3containedin103525.htmlifnistheproductofallpositiveintegerslessthan103218.htmlifnistheproductofintegersfrom1to20inclusive106289.htmlifnistheproductofallmultiplesof3between1and101187.htmlifpistheproductofintegersfrom1to30inclusive137721.htmlwhatisthegreatestvalueofmsuchthat4misafactorof105746.htmlif6yisafactorof102whatisthegreatestpossible129353.htmlifmistheproductofallintegersfrom1to40inclusive108971.htmlifpisanaturalnumberandpendswithytrailingzeros108251.htmlif73has16zeroesattheendhowmanyzeroeswill147353.htmlfindthenumberoftrailingzerosintheexpansionof108249.htmlhowmanyzerosaretheendof142479.htmlhowmanyzerosdoes100endwith100599.htmlfindthenumberoftrailingzerosintheproductof108248.htmlif60iswrittenoutasanintegerwithhowmanyconsecuti97597.htmlifnisapositiveintegerand10nisafactorofmwhat153375.htmlifdisapositiveintegerandfistheproductofthefirst126692.htmlHope it helps.
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Re: Find the number of trailing zeros in the expansion of
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10 Jan 2014, 07:58
madn800 wrote: Buneul, here's my doubt: # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; for calculating trailing zeros up til 24! you did just 20/5=4. but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7) Suppose I want # of trailing zeros in 310! using your concept 310/5+310/5^2=62+12=74 trailing zeroes BUT using the factorial calculator below I am getting 76 trailing zeroes http://www.nitrxgen.net/factorialcalc.phpplease suggest a fool proof method for calculating trailing zeroes of any +ve integer. The formula for the finding the number of zeroes in N! is N/5 + N/5^2 + N/5^3 + ....+ N/5^r where N < 5^(r+1). So for 310, we need to keep continuing for 125 as well so 310/5 + 310/25 + 310/125 = 62 + 12 + 2 = 76 trailing zeroes.
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Re: Find the number of trailing zeros in the expansion of
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20 Jan 2014, 02:35
Thanks for the reply bunuel.



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Re: Find the number of trailing zeros in the expansion of
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05 May 2014, 04:24
How does one know that the "something" cannot contribute to adding further trailing zeros? If the "something" captures a 2 and a 5 couldn't it add further 0s?



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Re: Find the number of trailing zeros in the expansion of
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Re: trailing zeros question (complicated one)
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06 May 2014, 00:17
Bunuel wrote: feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468 b) 10^469 c) 10^470 d) 10^467 e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? # of trailing zeros in 20!, 21!, 22!, 23!, and 24! will be 4 (20/5=4. For 21!, 22!, 23! and 24!, instead of 20 you'll have 21, 22, ... but the result will be the same) > total of 4*5=20 trailing zeros for these 5 terms. (Note here that this won't always be correct: for example 20 and 50 have one trailing zero each but 20*50=1,000 has three trailing zeros not two. That's because extra 2 in 20 and extra 5 in 50 "produced" one more trailing zero. In our case though, we won't have any extra 5s in any factorial, as all are already used for existing trailing zeros); # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28}*something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\). Total of 468 trailing zeros. Answer: A. Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this. Theory on this topic: everythingaboutfactorialsonthegmat85592.htmlHello Bunuel As you mentioned that any extra 2 or 5 left out from any of the factorial will result in One more Trailing Zero. How can we know that there is no extra 2 or 5 left out ? Also if possible please direct me to such a question where we have to take care of extra 2 or 5 being left out . And can such a concept be tested in GMAT ? Thankssss A LOT !



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Re: trailing zeros question (complicated one)
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06 May 2014, 02:37
niyantg wrote: Bunuel wrote: feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468 b) 10^469 c) 10^470 d) 10^467 e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? # of trailing zeros in 20!, 21!, 22!, 23!, and 24! will be 4 (20/5=4. For 21!, 22!, 23! and 24!, instead of 20 you'll have 21, 22, ... but the result will be the same) > total of 4*5=20 trailing zeros for these 5 terms. (Note here that this won't always be correct: for example 20 and 50 have one trailing zero each but 20*50=1,000 has three trailing zeros not two. That's because extra 2 in 20 and extra 5 in 50 "produced" one more trailing zero. In our case though, we won't have any extra 5s in any factorial, as all are already used for existing trailing zeros); # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28}*something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\). Total of 468 trailing zeros. Answer: A. Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this. Theory on this topic: everythingaboutfactorialsonthegmat85592.htmlHello Bunuel As you mentioned that any extra 2 or 5 left out from any of the factorial will result in One more Trailing Zero. How can we know that there is no extra 2 or 5 left out ? Also if possible please direct me to such a question where we have to take care of extra 2 or 5 being left out . And can such a concept be tested in GMAT ? Thankssss A LOT ! We used all 5's when counting trailing zeros for 20!, 21!, 22!, ..., and 33!. So, there are no more left. When using trailing zero approach shown in my post we use all 5's to get the number of 0's, so I don't know what kind of problems you are talking about. As for similar questions check here: findthenumberoftrailingzerosintheexpansionof10824920.html#p1277367Hope it helps.
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Re: Find the number of trailing zeros in the expansion of
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21 Jul 2014, 00:16
All factorials from 20! > 24! have four 5factors (5,10,15,20). This means that\(4 * 5 = 20\) All factorials from 25! > 29! have six 5factors (5,10,15,20,25) where 25 has two 5 factors. This means that \(6 * 5 = 30\) All factorials from 30! > 34! have seven 5factors (5,10,15,20,25,30) where 25 has two 5 factors. This means that \(7 * 4 = 28\)
\(20+30+28=78\) trailing zeroes
Given the \(3!\) exponent (translating to the power of 6), We multiply this power with the number trailing zeroes
\(78* 6 =468\)
Answer A



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Re: Find the number of trailing zeros in the expansion of
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21 Jul 2014, 04:39
20! to 24! give us 4*5=20 zeros 25! to 29! give us 6*5=30 zeros 30! to 33! give us 7*4=28 zeros
20+30+28=78 zeros in ^3! means 78*6=468, so answer is A



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Re: Find the number of trailing zeros in the expansion of
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26 May 2015, 12:26
Bunuel wrote: feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!. # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28} *something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\). Total of 468 trailing zeros. Answer: A. Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this. Theory on this topic: everythingaboutfactorialsonthegmat85592.htmlWhy is this term "something" in the equation above still there? Why is it needed, I don't understand why you put that in there. Could you explain that?
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Re: Find the number of trailing zeros in the expansion of
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26 May 2015, 18:53
reto wrote: Bunuel wrote: feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!. # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28} *something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\). Total of 468 trailing zeros. Answer: A. Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this. Theory on this topic: everythingaboutfactorialsonthegmat85592.htmlWhy is this term "something" in the equation above still there? Why is it needed, I don't understand why you put that in there. Could you explain that? Each term has many factors other than the 5s and a corresponding number of 2s. For example, 20! = 1*2*3*4*5*6*7...*20 "something" specifies the rest of the factors (3s, 7s, 11s, some 2s etc) after you segregate the 5s and 2s in the form of 10s. We don't care what is leftover since we just need the number of 0s so whatever is leftover is written as "something". It's just a placeholder and you can very well write "x" or "n" or whatever you please there.
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Re: Find the number of trailing zeros in the expansion of
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30 Aug 2016, 15:21
madn800 wrote: Buneul, here's my doubt: # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; for calculating trailing zeros up til 24! you did just 20/5=4. but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7) Suppose I want # of trailing zeros in 310! using your concept 310/5+310/5^2=62+12=74 trailing zeroes BUT using the factorial calculator below I am getting 76 trailing zeroes http://www.nitrxgen.net/factorialcalc.phpplease suggest a fool proof method for calculating trailing zeroes of any +ve integer. Count the number of fives and the number of two's. A ten is formed as follows => 5 multiplied by 2 P.S => Explanation by Bunuel regarding => trailing zeroes must be some multiple of 6
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Re: Find the number of trailing zeros in the expansion of
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19 Oct 2016, 09:19
madn800 wrote: Buneul, here's my doubt: # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; for calculating trailing zeros up til 24! you did just 20/5=4. but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7) Suppose I want # of trailing zeros in 310! using your concept 310/5+310/5^2=62+12=74 trailing zeroes BUT using the factorial calculator below I am getting 76 trailing zeroes http://www.nitrxgen.net/factorialcalc.phpplease suggest a fool proof method for calculating trailing zeroes of any +ve integer. using your concept 310/5+310/5^2=62+12=74 trailing zeroes You must take into account the multiples of 5^3 => 310/5^3 = 2 So the correct function would be 310/5 + 310/5^2 + 310/5^3 =62+12+2=76 trailing zeroes If you get to 625 factorial, you would have to consider 656/5^4 I hope this helps
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Re: Find the number of trailing zeros in the expansion of
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15 Oct 2017, 06:37
feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
A. 468 B. 469 C. 470 D. 467 E. 471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? It is not that difficult.. It will take you a minute to solve. 20! is having 4 5's; multiply the same with 6 as you are give 3! as a power. Same should go for 21,22,23 and 24!. 25! will have 2 more 5's adding up to the solution. So basically. 4*6=24*5= 120 6*6= 36: 36*5=180 7*6=42: 42*4=168 (we are stopping at 33!)
Add them up and your solution arrives.



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Re: Find the number of trailing zeros in the expansion of
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31 Oct 2017, 04:03
Bunuel wrote: Sachin9 wrote: do we get such questions on gmat? Trailing zeros concept is tested on the GMAT (check here: http://gmatclub.com/forum/everythingab ... 85592.html), though this particular question is a bit out of scope. Brunel, this factorial hyperlink is not working.Also some time back I read the formula in terms of 5^k for finding trailing zeros in n!. Couldn't recall. Can you please share? Posted from my mobile device



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Re: Find the number of trailing zeros in the expansion of
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31 Oct 2017, 04:18



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Re: Find the number of trailing zeros in the expansion of
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31 Oct 2017, 04:31
Thanks a lot Brunel, for super quick help Posted from my mobile device



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Re: Find the number of trailing zeros in the expansion of
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10 Oct 2018, 19:50
Shortest solution: We need to find trailing zeroes in (a)^3! or (a)^6 The trailing zeros will be of form 10^6n B, D, E are odd numbers hence out. C is not a multiple of 3 hence out. A is a multiple of 2 and 3 and so 6. Hence correct answer. KUDOS Please




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