Find the sum of all three digit numbers which when divided by 11 leave remainder 5
Answers
Answer:
each off the numbers can be constructed as:
z*11+5
where z is a hole number. When listing these numbers we know that the smallest number must be more than 100 (otherwise it would not be three digit) so we need to solve:
zmin*11+5>100 which gives
zmin > (100-5)/11 =8.63, and since imin most be a hole number imin = 9.
likewise zmax can be found to be 90, so our list consists of:
(9*11+5), (10*11+5) ,(11*11+5).... 90*11 + 5
which can be reanged to it looks like this:
((9*11+5)+(90*11+5))+((10*11+5)+(89*11+5))+((11*11+5)+(87*11+5))...=
((9+90)*11+5+5)+((10+89)*11+5+5)+((11+87)*11+5+5)....=
=(99*11+10)+(99*11+10) + (99*11+10)...
=1099 + 1099 + 1099...
now all you have to do is to find out how many elements there are in the list, which should be quite easy.
Step-by-step explanation: