Question

find the sum of all 3 digit numbers which leaves remainder 3 when divided by 5 ​

Answer 1

Answer:

i hope it is 99090. if yes check the solution

Step-by-step explanation:

3 digit nos start from 100 to 999

so when divided by five they need to give reminder 3

So, 100÷5 gives remainder 0

101÷5 gives remainder 1

then we can say that 103÷5 givex remainder 3

103=a

same from the last 999÷5 gives remainder 4

998÷5 gives 3 as remainder

998 will be the last term of AP while 103 is the first with a commom difference(d) of 5

103,108............,998

No. of tems in the AP-

nth term=a+(n-1)d

so, 998=103+(n-1)5

998-103=(n-1)5

895=(n-1)5

895÷5=(n-1)

179+1=n

180=n

998 is the 180th term

the sum of the AP will be the answer

sum=n÷2 (1st term+ last term)

=180÷2 (103+998)

=90 (1101)=99090