what is the last two digits of 1!+2!+3!+...........+99!+100! ???
While Plato has "spilled the beans" I will give you the reason behind his explanation:
$\displaystyle 5*2=10$
very simple math, but very essential in understanding the answer.
When calculating factorials 10, or greater, consider this, for 10!:
$\displaystyle 1*2*3*4*5*6*7*8*9*10$
When something is multiplied by ten, it automatically ends in a zero: 345*10 = 3450.
When something is multiplied by two ten's, it will end in two zero's: 345*10*10 = 34500.
$\displaystyle 1*2*3*4*5*6*7*8*9*10$ in this, we see two tens: one coming
from 2*5, and one ten by itself. Therefore, the last two digits must both be zero.
If you wanted to calculate the total number of zeros that are at the end of 100!, you would find
how many tens could be produced.