You did in broad lines.

Still, even Paul Kellet is a little bit messy in his mathematics.

At one point he states :"we need to figure the highest powers of 2 and 5 dividing 100! and take the lesser of the two exponents." But he then omits determining the highest power of 2 dividing 100! and omits showing that its power is larger then the power of 5 dividing 100!.

Also it still needs to be proven that the numbers that can not be factorized into 2's and'or 5's can never produce another number that can be factorized along either number after multiplication. Paul Kellet omits that part of the proof as well. Basically he describes a method for calculating the minimum number of trailing zero's, but does not proof that this number is also the maximum number of trailing zero's.



The idea behind his proof is actually very elegant however, although it is beyond novice level. His explanation is certainly beyond novice level mathematics. It can be made full and be worded much simpler, so that is complete and even novices can understand it.

Basically he is saying that the total number of trailing zero's in the number 100! is equal to the number of times you can devide that number by the number 10 without having to write down decimals. As the number 10 can be factorized into a product of the numbers 2 and 5 as 2*5 = 10, you can know the number of "even" (= no resulting decimals) divisions by 10 by factorizing all numbers between 1 to 100 along the numbers 2 and 5. You then group as many of the factored out 2's and 5's together into pair (2*5) as you can (forming 10's) and count these combo's => the number of combo's produces the minimum number of trailing 0's.

a small example for the other readers out here ;

the result of 55*54*53*52*51*50 has 3 trailing zero's. Proof :

55 factors in 11*5
54 factors in 27*2
53 can not be factored in multiple of 2's or 5's
52 factors in 26*2
51 can not be factored in multiple of 2's or 5's
50 factors in 5*5*2

All the leftover parts that can not be further factorized along either the numbers 2 or 5 are members of the "number space" that satisfies teh condition "the number is an integer with its last digit being 1,2,3,4,6,7,8 or 9'. This space is closed under multiplication, meaning that no multiplication using numbers from this number space can result in a number that does not belong to this number space as well. As a consequence, the result (multiplication of leftovers) can never produce any trailing zero's as that digit is not part of the set 1,2,3,4,6,7,8 and 9. This is turn proofs that the method described above not only gives the minimum amount of trailing zero's but also the maximum amount of trailing zero's

Combining these results we have found 3 times a factor 5 and three times a factor 2 allowing us to form 3 combo's of 2*5=10 ergo the resulting multuplication has 3 trailing zero's


P.S. I used the identifier "number space" here for clearity, but I think the correct name for such a mathematic body is a "ring"

Wouter