what is the last two digits of 1!+2!+3!+...........+99!+100! ???

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- Nov 12th 2010, 01:12 PMearthboylast two digits
what is the last two digits of 1!+2!+3!+...........+99!+100! ???

- Nov 12th 2010, 01:19 PMWilmer
....0313

- Nov 12th 2010, 01:27 PMPlato
Here is a huge hint: If $\displaystyle K\ge 10$ then $\displaystyle K!$ ends in at least two zeros.

- Nov 13th 2010, 09:57 AMrtblue
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.