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

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