Question

The decimal representation of $2005!$ ends with $m$ zeroes then $m=$

• A. $500$
• B. $501$
• C. $502$
• D. $499$

Answer 1

The correct option is A $500$

We know that a number gets a zero at the end of it if the number has 10 as a factor.

So I need to find out how many times 10 is a factor in the expansion of 23!.

But since 5×2 = 10, I need to account for all the products of 5 and 2.

Looking at the factors in the above expansion, there are many more numbers that are multiples of 2 (2, 4, 6, 8, 10, 12, 14,...) than are multiples of 5 (5, 10, 15,...).

That is, if I take all the numbers with 5 as a factor, I'll have way more than enough even numbers to pair with them to get factors of 10 (and another trailing zero on my factorial).

No. of multiples of 5 between 1 and 2005!= 401

No. of multiples of 25= 80

no. of multiples of 125=16

no. of multiples of 625=3

Hence total multiples of 5 and hence 10 are 500.

There are 500 zeroes in the end of 2005!.