Suppose n is an integer such that the sum of digits on n is 2, and $$10^{10} < n < 10^{11}$$. The number of different values of n is
Solution
The sum of digits should be 2. The possibilities are 1000000001,1000000010,10000000100,..these 10 cases . Also additional 1 case where 20000000000. Hence option A .
For 1000000000, number of cases with sum of digits equal to 2 should be 9, as there are nine zeroes that can be replaced with a 1. So answershould be 10
Hi, 100...1, 1000..10, ........here the number of cases will be 10 because $$10^{10}$$ will have 10 zeroes and '1' can take place of each zero in subsequent cases.