factorials

The last two zero's?

sum_(n=1)^20 (5 n)! = 93326215454274131171424014083629874498929842578166214433940153388687946517415655599163396465910182277066190137342795918966177385078001619556987788165518636920. This is because of 5! which is =120. The rest of the terms all end in at least 2 zeros.

Find the last two digits of the following sum:

5! + 10! + 15! +... + 100!

20!=2*5*10*other numbers = 100*other numbers      ends in  00

All the factorials bigger than 20! also end in  00  

So add these all up and we get a number that ends in   00

So what is left?

5!+10!+15!  

5! = 2*3*4*5 = 10*12 = 120  the last two digits are 20

10!= 120*6*7*8*9*10 = 6*7*8*9*1200  the last two digits are 00

It follows that the last 2 digits of 15! must also be   00

So the only one that does not end in 00 is   5! which ends in  20

So add them all together and the last two digits must be 20

Just like our guest already said :)