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# What is the remainder when 25! is divided by 10^7

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Director
Joined: 18 Jul 2018
Posts: 502
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
What is the remainder when 25! is divided by 10^7  [#permalink]

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03 Oct 2018, 18:30
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85% (hard)

Question Stats:

11% (00:45) correct 89% (01:41) wrong based on 74 sessions

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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

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Posts: 7108
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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03 Oct 2018, 19:19
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
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Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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04 Oct 2018, 03:52
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?

Director
Joined: 18 Jul 2018
Posts: 502
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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04 Oct 2018, 04:00
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?

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.
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Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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04 Oct 2018, 04:14
1
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?

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 non-zero digit is 4 (as shown by chetan2u above), the remainder will be 4*10^6.
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What is the remainder when 25! is divided by 10^7  [#permalink]

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05 Oct 2018, 01:21
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 device

Hi 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: 221
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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05 Oct 2018, 06:01
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 device

Hi 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/everything-a ... 85592.html
Math Expert
Joined: 02 Aug 2009
Posts: 7108
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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05 Oct 2018, 07:27
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 device

Hi 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
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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05 Oct 2018, 08:12
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?
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Posts: 7108
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

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05 Oct 2018, 08:44
1
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/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: What is the remainder when 25! is divided by 10^7 &nbs [#permalink] 05 Oct 2018, 08:44
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# What is the remainder when 25! is divided by 10^7

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