What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder?

Step-by-step solution

The numbers which divided by 3 , leaving remainder 2 will be of the form 3 x+2 Given, 3 x +2 is 2 -digit number So, x can be from 3 to 32 aligned & Sum of numbers = _ x =3 ³²(3 x +2) & =3(3+4+5+ .+32)+(2+2+ .)....(1) & 3,4,5 .32 is an A.P with a =3, ~d =1, ~T _ n =32 & ~T _ n = a +( n -1) 1 & 32=3+( n -1) 1 & n -1=29 n =30 terms & (1) 3(3+4+5+ .+32)+(2+2+ .30 times ) & =3 ( 30 2 (3+32) )+(2 30) ( S _ n = n 2 ( a + ) ) & = 90 2 (35)+60=(45 35)+60 & =1575+60=1635 aligned