Question

Find the sum of all four digit numbers which leave remainder 2 when divided by 5


IT IS 4 DIGIT NOR 3 OK​

Answer 1

Answer:

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

In this A.P

a (first term)= 102

d (Common difference)= 5

|(last term ) = 997 %3|

I= an = a + (n - 1) d

997= 102 + (n – 1) x 5

5 (n – 1) = 997 – 102 = 895

n-1= 885/5

(n – 1) = 179

n = 179 +1

n = 180

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

Sn = n/2 [a+I]

Sn= 180/2 [ 102+ 997]

Sn= 90 x 1099

Sn= 98910

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

Answer 2

Answer:

hope it helps you.....