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
100, 200, 300, 400, 500, 600, 700, 800, 900, 1000 and keep adding 100 until you get to infinity.
There are no multiples of 500 in 100.
77 is the only common multiple of 7 and 77 between 1 and 100.
There are 30 multiples of 30 that fall between 100 and 1,000.
100 of them.
110.
297
128!
69342
There are 75 multiples of 12 between 100 and 1000.
Ten of them.
545
Smallest multiple of 7 greater than 100 is 15 Largest multiple of 7 less than 1000 is 142 So number of multiples of 7 = 142 - 15 + 1 = 128
The last "x7" under 100 is 14 x 7, the last under 1000 is 142 x 7, so there are 128 in the range.
There are 30 multiples of 30 between 100 and 999.