Question

1+2^2+3^3+4^4+.....+10^10
when is is divided by 5 what is the remainder!!
show working,please have been trying that i could not,get it

Answer 1

thats the real question

Answer 2

Sum(1) = 1 = 1mod5
sum(2) = 1 + 2^2 = 5 = 0 mod5
sum(3) = 5 + 3^3 = 32 = 2 mod5
sum(4) = 32 + 4^4 = 288 = 3 mod5

To divide by 5 all we need to do is look at the last digit.
If the digit is form 0 to 4, then that digit is the remainder when dividing by 5
If the digit is from 5 to 9, the remainder is (that digit - 5)
e.g. 245678 leaves a remainder of 3

on my calculator continuing this until
Sum(8) = 873612 + 8^8 = 17650828
on the next step, many calculators will kick into scientific notation and lose the unit digits.
however, 9^9 ends in ...9 , so adding this to
Sum(9) ends in ..... 7
and we know that 10^10 ends in 0
so
Sum(0) ends in ....0

so the remainder when divided by 5 is 0

btw, the notation
288 = 3 mod5
means that the remainder is 3 when 288 is divided by 5