what is the number of zeros at the end of the product of the numbers from 1 to 100?

Answer 1

1*2*3*...*98*99*100 = 100! (100 factorial)

The answer is 24

The trick here is not to calculate 100! on your calculator (which only gives you ten digits of accuracy), but to figure out how high a power of 10 goes into 100! evenly. For every trailing zero, there is a power of 10 that divides 100! evenly. In order to do that, since 10 = 2*5, we need to figure the highest powers of 2 and 5 dividing 100! and take the lesser of the two exponents.

(Why?) Consider what happens when we multiply together 1*2*3*4*5*6*..., starting with the lowest numbers first. Every fifth number, starting with 5, is divisible by 5. That gives you 100/5 = 20 factors of 5 in 100!. But there are more. Every 25th number, starting with 25, has an extra factor of 5 beyond the ones already counted. That gives you 100/25 = 4 more factors of 5 in 100!. To get a third factor of 5 from a single number, it has to be a multiple of 125, and no number <= 100 is, so that is all.

The answer is: [100/5] + [100/5^2] + [100/5^3] + ... = 20 + 4 + 0 + ... = 24

Answer 2

24

Check the divisiblity of every no. starting from 1 to 100 by 5
The sum of no. of times a number is divisible by 5 for all 100 numbers will be the no. of zeros.
It seems complicated but will hardly take more than 2mins to arrive at the answer...
now do it.

Answer 3

21 zero's
2 because of 100
10 because of 2's and 5's
9 bcauz of 10-90