Question

How many three digit numbers leave the remainder 2 when divided by 9?

Answer 1

A.P = 101,110,119,..............,992

An= a+(n-1)d

a=101

d=9

An=992

992 = 101 +(n-1)9

992-101 = (n-1)9

891 = (n-1)9

(n-1) =99

n=100

there are 100 terms

Answer 2

Given: Three digits number leaves a remainder 2 when divided by 9

To find: The count of such numbers

Step-by-step explanation:

Therefore we have Three digits numbers leaves a remainder 2 when divided by 9 are

101, 110 , ........................ , 992

It forms an A.P. where first term is 101 , last term is 992 and common difference d is 9

As we know

$a_n= a+(n-1)d$ where a is first term , d is common difference, n is number of terms and $a_n$ is the nth term of an A.P.

$992= 101+(n-1)\times 9\\\\\Rightarrow 992-101=(n-1)\times 9\\\\\Rightarrow 891=(n-1)\times 9\\\\\Rightarrow 99 =(n-1) \\\\\Rightarrow n= 100$

Hence there are 100 3 digits number which leave 2 as remainder when divided by 9