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

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
_________________

What I can share here is an equation to solve it.

We know that 25! will have 6 trailing zeroes, and when we divide by \(10^7\), these 6 zeros cancel out. These we keep aside, and we now have a number which ends with non zero digit and when this is divided by 10, the remainder will be this number.

Now a very simple method to find the last non zero digit of any factorial, is to split the factorial as 5a + b which is basically the Dividend = Divisor * Quotient + Remainder.

Then by conducting the operation \(2^a \space * \space a! \space * \space b!\), we get the last non zero digit.


We can test this. We know that 8! = 40320 and the first non zero digit is 2

Dividing 8 by 5, we get Q = 1 and R = 3, therefore in the 5a + b form, we have (5*1) + 3, therefore a = 1 and b = 3

Putting it in the equation above, \(2^a \space * \space a! \space * \space b! = 2^1 * 1! * 3! = 2 * 1 * 6 = 12\)



Now for 25!, we have (5 * 5) + 0.

Here a = 5 and b = 0 and therefore, \(2^5 * 5! * 0! = 32 * 120 * 1 = 3840 \)

The last non zero digit is 4. But the remainder is simply not 4. We have removed 6 zeroes in our initial calculation and this has to be brought back.

Therefore our remainder is \(4 * 10^6\)


Option C

Arun Kumar

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.

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.
_________________

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

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
_________________

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..
_________________

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

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?

Asked: What is the remainder when 25! is divided by \(10^7\)?

Max power of 5 in 25! = 5 + 1 = 6
25! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24*25

25! = 3*6*7*8*9*11*12*13*14*6*4*17*18*19*2*21*22*23*24*10^6
We have to find the last digit of 3*6*7*8*9*11*12*13*14*6*4*17*18*19*2*21*22*23*24 = 4

The remainder when 25! is divided by \(10^7\) = 4*10^6

IMO C

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________