Find the sum of all three digit numbers which leave the remainder 3 when divided with 5?
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the number will be 33
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smallest 3 digit number which leave the remainder 3 when divided by 5 =103
largest 3 digit number which leave the remainder 3 when divided by 5 =998
3 digit number after 103 which leave the remainder 3 when divided by 5 =108
an ap is formed: 103,108........998
first term=103
common difference=5
let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n
998=103+(n-1)5
180=n
we know that the formula for sum in AP = no of terms*(first term + last term)/2
sum of ap=180*(103+998)/2 = 90*1101
99090
largest 3 digit number which leave the remainder 3 when divided by 5 =998
3 digit number after 103 which leave the remainder 3 when divided by 5 =108
an ap is formed: 103,108........998
first term=103
common difference=5
let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n
998=103+(n-1)5
180=n
we know that the formula for sum in AP = no of terms*(first term + last term)/2
sum of ap=180*(103+998)/2 = 90*1101
99090
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