Mar 20 09:00 PM EDT  10:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Wednesday, March 20th at 9 PM EDT Mar 20 07:00 AM PDT  09:00 AM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. Mar 23 07:00 AM PDT  09:00 AM PDT Christina scored 760 by having clear (ability) milestones and a trackable plan to achieve the same. Attend this webinar to learn how to build trackable milestones that leverage your strengths to help you get to your target GMAT score. Mar 27 03:00 PM PDT  04:00 PM PDT Join a free live webinar and learn the winning strategy for a 700+ score on GMAT & the perfect application. Save your spot today! Wednesday, March 27th at 3 pm PST
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 18 Jul 2018
Posts: 752
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)

What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
03 Oct 2018, 19:30
Question Stats:
9% (01:00) correct 91% (01:59) wrong based on 83 sessions
HideShow timer Statistics
What is the remainder when 25! is divided by \(10^7\)? a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Press +1 Kudo If my post helps!



Math Expert
Joined: 02 Aug 2009
Posts: 7421

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
03 Oct 2018, 20:19
Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentageincreasedecreasewhatshouldbethedenominator287528.html
GMAT Expert



Intern
Joined: 09 Sep 2017
Posts: 5

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
04 Oct 2018, 04:52
chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C If we break up 10^7 to 2^7 and 5^7 . we get no 2's left in the denominator and only 1 5. the remainder cannot be bigger than the quotient. Therefore shouldnt the answer be A) ? which is 2? please confirm.



Director
Joined: 18 Jul 2018
Posts: 752
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
04 Oct 2018, 05:00
sohrabbanker wrote: chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C If we break up 10^7 to 2^7 and 5^7 . we get no 2's left in the denominator and only 1 5. the remainder cannot be bigger than the quotient. Therefore shouldnt the answer be A) ? which is 2? please confirm. You'll be left with 2, But you have to multiply 2 with the common number which you canceled in both the numerator and denominator to get the final remainder.
_________________
Press +1 Kudo If my post helps!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8997
Location: Pune, India

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
04 Oct 2018, 05:14
sohrabbanker wrote: chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C If we break up 10^7 to 2^7 and 5^7 . we get no 2's left in the denominator and only 1 5. the remainder cannot be bigger than the quotient. Therefore shouldnt the answer be A) ? which is 2? please confirm. Here is the thing about cancelling off common terms from the numerator and denominator  if you want the actual remainder, you need to multiply the cancelled terms back. e.g. what is the remainder when 20 is divided by 6? The answer, as you know is 2. The remainder in this case can take 6 values (0/1/2/3/4/5). But if we do 20/6 = 10/3, the remainder is 1. Of course the remainder in this case can take only 3 values 0/1/2. 1 is not the correct answer. Since we cancelled off 2 initially, to get the actual remainder, we need to multiply by 2 back to get the remainder as 1*2 = 2. So when you cancel off 10^6, you will need to multiply it back to the remainder you obtain upon division by 5. Since the last nonzero digit is 4 (as shown by chetan2u above), the remainder will be 4*10^6.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 22 Sep 2016
Posts: 2

What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
05 Oct 2018, 02:21
chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C Posted from my mobile deviceHi Chetan, If u don't mind, can u explain how u concluded that 25! Contains 6 zeroes. Is it like a formula... Kindly explain... Thanks, Subhan



Manager
Joined: 14 Jun 2018
Posts: 223

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
05 Oct 2018, 07:01
Subhan7 wrote: chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C Posted from my mobile deviceHi Chetan, If u don't mind, can u explain how u concluded that 25! Contains 6 zeroes. Is it like a formula... Kindly explain... Thanks, Subhan https://gmatclub.com/forum/everythinga ... 85592.html



Math Expert
Joined: 02 Aug 2009
Posts: 7421

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
05 Oct 2018, 08:27
Subhan7 wrote: chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25! Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C Posted from my mobile deviceHi Chetan, If u don't mind, can u explain how u concluded that 25! Contains 6 zeroes. Is it like a formula... Kindly explain... Thanks, Subhan Hi... 25! = 1*2*3*4*5*....*24*25.. Now 10s depends on number of its prime factors 2s or 5s. Ofcourse there will be more 5s as compared to 2s Now how many 5s are there in 1*2*3*...24*25 When we divide by 5 we get all multiples of 5 so 25/5=5, these are 1*2*..*5....*10...*15...*20...*25 Now by dividing by 5^2 we get numbers that have 5^2 in it her it is 25/5^2=1 So total 5+1=6
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentageincreasedecreasewhatshouldbethedenominator287528.html
GMAT Expert



Manager
Joined: 06 Sep 2018
Posts: 175
Location: Pakistan
Concentration: Finance, Operations
GPA: 2.87
WE: Engineering (Other)

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
05 Oct 2018, 09:12
chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25!Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C How could we know the highlighted part?
_________________
Hitting Kudos is the best way of appreciation.
Eric Thomas, "When you want to succeed as bad as you want to breathe, then you'll be successful."



Math Expert
Joined: 02 Aug 2009
Posts: 7421

Re: What is the remainder when 25! is divided by 10^7
[#permalink]
Show Tags
05 Oct 2018, 09:44
HasnainAfxal wrote: chetan2u wrote: Afc0892 wrote: What is the remainder when 25! is divided by \(10^7\)?
a) 2 b) 2*\(10^6\) c) 4*\(10^6\) d) 6*\(10^6\) e) None of these 25! Will contain 25/5+25/25=5+1=6 zeroes So when we divide 25! by 10^7, remainder will be the last 7 digits of 25!Last 6 are zeroes so we look for the first non zero digit... Now 6 *5s will take 6*2s so we have 1*2/2*3*4/4*5/5*6*7*8*9*10/10*11*12*13*14*15/5*16*17*18*19*20/20*21*22*23*24*25/25 So 1*3*6*7*8*9*11*12*13*14*3*16*17*18*19*21*22*23*24.. Now 3*7 and 9*9 gives us 1 as units digit, so you can remove them in pair 6*8*12*14*3*16*18*22*23*24 2*3 is 6 and 6*6 is again 6, so you can write 6 instead of all the pair 6*8*14*18*24=6*(8*4)*(8*4)=6*2*2=24=4 We are just looking at units digit, so here we have 4 as units digit.. So 25!=xyz____ab4000000 So remainder is 4000000=4*10^6 C How could we know the highlighted part? Take a smaller example say 10^2 so 100.. If you divide any number say 123, remainder is last two digits 23.. The number could be 34276523, remainder will remain last two digits.. If number is 3467900, remainder is 00, that is the number is divisible by 100 Similarly here when you divide something with 10^7 , remainder will be last 7 digits..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentageincreasedecreasewhatshouldbethedenominator287528.html
GMAT Expert




Re: What is the remainder when 25! is divided by 10^7
[#permalink]
05 Oct 2018, 09:44






