The natural numbers 1 to 25 are multiplied together (1 x 2 x 3x..24 x 25). How many zeros are there in the product of this multiplication? I am going to write P for 1 x 2 x 3 ... 24 x 25. Think of the prime factors of P and gather all the 2's and 5's to write P = 2^{k} x 5^{m} x N where N is not divisible by either 2 or 5. Since neither 2 nor 5 divides N, N does not end in zero. Also, since 10 = 2 x 5, the number of zeros at the end of 2^{k} 5^{m} is the minimum of k and m. 5 is a prime factor of 5, 10, 15, 20 and 25. Thus 5^{5} divides P. But 5 is a factor of 25 twice, 25 = 5 x 5 and thus 5^{6} divides P. Thus m = 6. Similarly 2 is a prime factor of 2, 4, 6, ..., 24 so 2^{12} divides P and thus k is at least 12. Hence the minimum of m and k is 6 and 1 x 2 x 3 ... 24 x 25 ends in six zeros. Cheers,Harley
