Question

How many three digit numbers are there, which leave a remainder 2 on division by 6

Answer 1

Answer:

Least 3 digit number is 100 ,

Since divisor is 9, & remainder =2

So, 9 x 11 + 2 = 101 is the first 3 digit number which when divided by 9, leaves remainder =2

Next no will be 101+9 = 110, & next = 110+9 =119

& the last 3 digit no leaving remainder 2, while dividing by 9 is 992

So we get an AP series

101, 110, 119, 128 ………………992

Here, Tn = a + (n-1) *d = 992 ,

Where a= 1st term, d= common difference, & n is the term

=> 101 + (n-1)*9 = 992

=> 9n - 9 = 891

=> 9n = 900

=> n = 100

So, there are 100 such numbers ………