Question

If all the 6 are replaced by 9, then the algebraic sum of all the numbers from 1 to 100(both inclusive) varies by ??

Answer 1

Sum of all numbers from 1 to n is given by n(n+1)/2 i.e. 100*(100+1)/2 = 5050

6 comes in 'units' place 10 times in the numbers 1 to 100 i.e. 6,16...96

6 comes in 'tens' place 10 times in the numbers 1 to 100 i.e. 60,61...69

sum of these 6's is 10(6*10)+10(6*1)=600+60=660

If all 6's are replaced by 9
then the sum of those 9's will be 10(9*10)+10(9*1)=990

now algebraic sum will be 5050-660+990= 5380

Difference: 5380 - 5050 = 330

Answer 2

first take the numbers which have 6 in their unit place - 06,16,26,36,46,56,66,76,86,96.

if all 6 in unit plce is replaced with 9 then in each case the increment would be 3.

so for all ten numbers total increment is 10*3=30.

then take all numbers which have 6 in their tenth place - 60,61,62,63,64,65,66,67,68,69

in each case if we replace 6 with 9 -90,91,92,93,94,95,96,97,98,99

then there is increment of 30 in each case .

so total increment for ten numbers is 10*30==300.

so the total increment including unit and tenth place is 30+300=330