Find the sum of all three digit numbers each of which leave remainder 3 when divided by 5....???

Question

serious answers please...!!!!

Answer 1

This is basically the sum of 5x + 3 for 100 < 5x + 3 < 999 and x being an integer

First take 100 < 5x + 3
100 < 5x + 3
97 < 5x
19.4 < x

X is the first integer higher than 19.4 i.e. 20
So the lowest of your series is 5*20 + 3 = 103

Using a similar method, you can work out the highest possible value is 998 (199 * 5 + 3)

So now you have a series of numbers from 103 to 998, with a difference of 5 between each number. (998 - 103)/5 + 1 = 180 numbers in this sequence.

You can now subtract 3 from each number in this sequence, remembering to add back on (180 * 3 = 540) at the end.

So you now have a sum of all multiples of 5 between 100 and 995. Take out the common factor of 5 and you have 5*(the sum of all natural between 20 and 199), which can be calculated as 5*(sum of natural numbers up to 199 minus sum of natural numbers up to 19)

5 * [199(199 + 1)/2 - 19(19 + 1)/2]
= 5 * [199*200/2 - 19*20/2]
= 5 * [19900 - 190]
= 5 * 19710
= 98550

Add back on the 540 we subtracted earlier

99090

Answer 2

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