find the sum of all three digit numbers which when divided by 11 leave remainder 5
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The sum of all 3 digit numbers which when divided by 11 leaves remainder 5 is equal to 45059.
The numbers that will leave a remainder of 5 on dividing by 11 must be of the form 11n + 5, where n is a natural number.
for the number to be a 3 digit number n must be greater than 9 and lest than 90.
So we get an AP with 104 as the first term, common difference equal to 11 and 995 as the last term.
Sum=104+115+126+...............+995(total 82 terms)
Sum of AP = no. of terms ×(first term +last term)/2
Evaluating the above expression , we get our sum to be 45059
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