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

Question

Two digits are selected at random from digits 1 through 9. If sum is even, find the probability that both numbers are odd?

Answer 1

330

Solution:
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 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 N natural number sum is n*(n+1)/2
n=100
sum= 101*100/2=5050
if we replace all 6 by 9 then sum is increment by 3.
total sum is 5050+30 =5080

Comments

66 replaced by 99 means increment of 33 not 3+3, u may like to correct the answer