100 Factorial
One for the mathematicians this week, with thanks to Paul Kellet for sending it in.
How many zeros are there at the end of 100! (factorial)?
Answer  

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 Here [x] means the integer part of x, or the greatest integer not exceeding x. (Why do we use this [x] instead of x?) All the terms from some point on will be zero, so this is a finite sum. Now for 2's (or any other prime number), the same analysis holds. The answer for the highest power of 2 dividing 100! is [100/2] + [100/2^2] + [100/2^3] + [100/2^4] + ... = 50 + 25 + 12 + 6 + 3 + 1 + 0 + 0 + ... = 97 The smaller of the two is 24, so the highest power of 10 dividing 100! is 10^24, so 100! ends with 24 zeroes.The same analysis works for any factorial n! and any prime p. The highest power of p dividing n! is: [n/p] + [n/p^2] + [n/p^3] + [n/p^4] + ... 
