Question

Find the sun of all 3 digit nos. which leaves the same remainder 2 when
divided by 5

Answer 1

Answer :

Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910

Explanation :

Given that :

Three digit numbers which leave the remainder 2 when divided by 5 are 102, 107, 112,117......, 997.

102, 107, 112,......, 997 is an A.P

To Find :

Find the sum of all 3 digit nums.

Solution :

As given that So,

$\textbf{\underline{\underline{Consider as - }}}$

In this A.P  

=> (First term) a = 102

=> (Common difference) d = 5

=> (last term ) I = 997.

Now,

$\tt \implies l= an = a + (n - 1) d\\ \\\implies 997= 102 + (n - 1) \times 5\\ \\\implies 5 (n - 1) = 997 - 102 = 895\\ \\\implies n-1= \dfrac{885}{5}\\ \\\implies (n - 1) = 179\\ \\\implies n = 179 +1\\ \\\implies n = 180$

Sum of all 3 digit numbers which leaves remainder 2 when divided by 5 :

Sum of all three digit numbers which leaves remainder 2

When divided by 5 is 98910 :

$\sf \implies S_n = \dfrac{n}{2}, (a+l)\\ \\\implies S_n= \dfrac{180}{2},(102+ 997)\\ \\\implies S_n= 90 \times 1099\\ \\\implies S_n= 98910$

∴ Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910