Question

PLEASE ANSWER; MAY NOT BE HARD

Find the sum of all positive 3-digit numbers whose last digit is 2

Answer 1

Explanation:

102+202+302+402+502+602+702+802+902(4518)

+112+212+312+...+ 812+912(4608)

+122+222+322+...+822+922(4698)

+132+232+332+...+932(4788)

..........................................

+192+292+392+...+992(5328)

4518+4608+4698+...+5328

n=10

`s=(10)/(2)(4518+5328)\\=5(9846)\\=49230`