Find the last two digits of the following sum:

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

Atroshus
Sep 5, 2017

#2**+2 **

sum_(n=1)^20 (5 n)! = 93326215454274131171424014083629874498929842578166214433940153388687946517415655599163396465910182277066190137342795918966177385078001619556987788**165518636920. This is because of 5! which is =120. The rest of the terms all end in at least 2 zeros.**

Guest Sep 5, 2017

#3**+2 **

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 :)

Melody
Sep 5, 2017