Your method may have produced the (correct) number of 24 trailing zero's but it is not mathematically sound. Basically, you followed a pathway that is not "thoroughly logical" and that is most certainly not "closed". The first meaning that you have not taken the full scope of the problem into account. Example ; who is to say that you have covered all eventualities in your "proof", maybe you missed a special number or special combination ? The second means you have failed to proof that there can AT MOST be 24 trailing zero's.


I'll provide a few more details.

You wrote :




In this method you use the number 2 twice; ones in multiplying 2*50 and once in multiplying 2*25. That is obviously incorrect.

You also say : Each number that ends in 5 also creates zero's (so that's 10 zeros), without showing how this is the case. For example, a number that ends in 5 ONLY creates a trailing zero when multiplied by an even number. Your statement still allows it to create trailing zero's when multiplied with any given number.

You forget to show that such a multiplication can produce more then only 1 trailing zero. Example by using the next available even number = 4 we find : 25 * 4 = 100 and that is adding two zero's to the total instead of just 1. So how do you know that there aren't more then 24 trailing zero's ?

Then you leave out the proof that none of the other possible combinations can produce trailing zero's or can produce numbers that in combination with yet another number can produce zero's.

Even with some leniency regarding your phrasing of the proof (using the number 2 twice), you can at most proof that there are At LEAST 24 trailing zero's. Of course that was never the hard part of the proof. The hard part is to proof that there are AT MOST 24 trailing zero's as then you'll have to find a structure in the remaining series of numbers that prevents this series from ever creating more trailing zero's. Only by combining these two parts can the proof be completed.

Wouter