GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 08 Dec 2019, 13:27 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # What is the remainder when 25! is divided by 10^7

Author Message
TAGS:

### Hide Tags

NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1025
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

16 00:00

Difficulty:   95% (hard)

Question Stats: 11% (01:12) correct 89% (02:01) wrong based on 123 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
Math Expert V
Joined: 02 Aug 2009
Posts: 8290
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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
_________________
Intern  B
Joined: 09 Sep 2017
Posts: 5
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

2
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?

NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1025
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

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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9855
Location: Pune, India
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 22 Sep 2016
Posts: 2
What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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  G
Joined: 14 Jun 2018
Posts: 211
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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 V
Joined: 02 Aug 2009
Posts: 8290
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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
_________________
Manager  G
Joined: 06 Sep 2018
Posts: 171
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

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?
Math Expert V
Joined: 02 Aug 2009
Posts: 8290
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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..
_________________
Manager  P
Joined: 31 Jul 2017
Posts: 197
Location: Tajikistan
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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[/quote]

Dear chetan2u,
in highlighted part above, do you mean there are more 2s? I cannot see how there are more 5s
Manager  G
Joined: 25 Jul 2018
Posts: 62
Location: Uzbekistan
Concentration: Finance, Organizational Behavior
GRE 1: Q168 V167 GPA: 3.85
WE: Project Management (Investment Banking)
Re: What is the remainder when 25! is divided by 10^7  [#permalink]

### Show Tags

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

Hi chetan2u,

Will we have such a long calculation requiring question during the actual exam? I would have been able to solve it, If I had done such manual calculation writing down all multipliers and figuring out similarities and patterns, but I was thinking that GMAT questions should not be solved in this way and there had to be some logical approach. I dont see any logic being tested here except for fast calculating capability. Am I missing smth? Re: What is the remainder when 25! is divided by 10^7   [#permalink] 29 May 2019, 02:59
Display posts from previous: Sort by

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