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03 Oct 2018, 19:30
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C
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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
a) 2
b) 2*\(10^6\)
c) 4*\(10^6\)
d) 6*\(10^6\)
e) None of these
Updated on: 27 Sep 2020, 08:25
Originally posted by CrackverbalGMAT on 27 Sep 2020, 07:16.
Last edited by CrackverbalGMAT on 27 Sep 2020, 08:25, edited 1 time in total.
Last edited by CrackverbalGMAT on 27 Sep 2020, 08:25, edited 1 time in total.
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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
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
03 Oct 2018, 20:19
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Afc0892
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
