15*7 = 105
142*7 = 994
So the question is equivalent to finding the sum: 105 + 112 + 119 + ... + 994
= 7*(15 + 16 + 17 + ... + 142)
= 7*(142*143/2 - 14*15/2)
= 7*(10153 - 105)
= 7*10048
= 70336
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This is the sum S = 105 + 110 + ... + 995
This is a sequence whose nth term is a+r*n where a = 100, r (the common difference) = 5 and n = 1,2,3, ... , 179, so that N = max(n) = 179.
Then S = N *[a + (N+1)*r/2] = 179*[100+180*5/2] = 179*(100+450) = 179*550 = 98450.
Another way to solve this is to work in one-fifths and then multiply the answer by 5 at the end. So,
(1) add all the whole numbers from 1 to 199 (similar to multiples of 5 from 5 to 995);
(2) subtract from this the sum of all whole numbers from 1 to 20 (multiples in subtract 5 to 100) so you are left with the sum of all whole numbers from 21 to 199 (105 to 995);
(3) multiply this answer by 5.
There are 67 multiples of 6 and 50 multiples of 8 in that range. Their total, 117, will include numbers that are both.
There are 6 multiples of 15 in 100. They are: 15, 30, 45, 60, 75, and 90
There are no multiples of 500 in 100.
100, 200, 300, 400, 500, 600, 700, 800, 900, 1000 and keep adding 100 until you get to infinity.
77 is the only common multiple of 7 and 77 between 1 and 100.