Hello , i got a pretty difficult task given by my Proffesor. I need to find how many 0's (in the row) at the end of number 2014! are there. If i wasn't clear enough , there's an example: number 10! = 3628800 - 2 zeros at the end in the row , 100! got 24 zeros. I doubt there is good "mathelhead" out here but its worth a shot. Have a nice day.
Here's the proceedure for finding the answer
So we have
2014 / 5 = 402.8 = 402
2014 / 52 = 80.56 = 80
2014 / 53 = 16.112 = 16
2014 / 54 = 3.2224 = 3
2014 / 55 = .644 and we can stop
So...summing 402 + 80 + 16 + 3 we have 501 trailing zeroes
You can read about why this proceedure works here : http://www.purplemath.com/modules/factzero.htm
My way of looking at this is:
every number that ends in a 2 multiplied by a number ending in a 5 adds one more zero to the number of zeroes;
also multiplying by a number that ends in a zero adds another zero.
This means that 10! has two zeroes because it has 2 x 5 x 10; one zero for the 2 x 5 and another zero for the 10.
But this means: 1 through 10 adds 2 zeroes; so does 11 through 20; 21 through 30; up through 91 through 100
for a total of 20 zeros
But I forgot about 25, since it contains 2 5's as factors, it will add 2 zeros not just one (there are a lot of extra 2's in the other numbers); so here is an extra zero.
Then, there is also, 50, 75, and 100; here are three more zeros.
OK, that's how you can get 24 zeros in 100!.
Can you use this to finish the analysis for 2014! ?
CPill's answer is far more complete than mine, with exactly the correct procedure. But, in the problem, he transposed 2014 into 2104, so check the last two numbers.