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March 4th, 2016, 09:17 PM   #1
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How to find the sum of the multiples

Like the title says what is the general formula to find the sum of the multiples of a number?
For example how to find the sum of the multiples of 2 under 529.
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March 4th, 2016, 11:19 PM   #2
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Originally Posted by popoder View Post
Like the title says what is the general formula to find the sum of the multiples of a number?
For example how to find the sum of the multiples of 2 under 529.
$\displaystyle 2+4+6+8+ ... + 528 =2(1+2+3+4+ ... +264)= \cancel{2}\dfrac{264\cdot265}{\cancel{2}}=264\cdot 265 =69 960$
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March 5th, 2016, 02:18 AM   #3
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$\displaystyle 2+4+6+8+ ... + 528 =2(1+2+3+4+ ... +264)= \cancel{2}\dfrac{264\cdot265}{\cancel{2}}=264\cdot 265 =69 960$
Thanks for the reply. But one more thing, if i want to make the sum of multiples of two numbers how does it work?
Like find the sum of the multiples of 17 or 19 under 587
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March 5th, 2016, 11:47 AM   #4
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You find the sum of the multiples of 17, then the sum of the multiples of 19, and, finally, the sum of the multiples of these two numbers.
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March 5th, 2016, 09:03 PM   #5
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You find the sum of the multiples of 17, then the sum of the multiples of 19, and, finally,
the sum of the multiples of these two numbers.
If you were to add up those three sums, that would be an overcount.

The multiples of 17 include the multiples of these two numbers, and the
multiples of 19 also include the multiples of these two numbers.

The number 323 occurs both in the list of multiples of 17 and the list of multiples of 19,
and there is no larger common multiple of those two prime numbers under 587.

So, you can add the sum of the multiples of 17 to sum of the multiples of 19 and then
subtract 323 to not have the overcount.
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Last edited by Math Message Board tutor; March 5th, 2016 at 09:24 PM.
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March 6th, 2016, 05:20 AM   #6
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The "sum of the multiples" of a number, n, is n+ 2n+ 3n+ ... + An where An is less than or equal to the upper limit. Factoring out n gives n(1+ 2+ 3+ ...+ A) so that the sum is just n times the sum of 1+ 2+...+ A which is easily proved to be A(A+ 1)/2.

To find the sum of "all multiples of 2 under 259" first note that the largest multiple of 2 under 259 is 258= 2(129) so A= 129. 1+ 2+ ...+ 129= 129(130)/2. The sum of all such multiples of 2 is 2(129)(130)/2= 129(130).

To find "the sum of the multiples of 17 or 19 under 587", first note that 17 divides into 587 34 times with remainder. A= 34 so the sum of all multiples of 34 under 587 is 17(1+ 2+ 3+ ...+ 34)= 17(34)(35)/2.

Similarly for the sum of the multiples of 19. It is not clear whether you intend this as two separate problems (the sum of all multiples of 17 and then the sum of all multiples of 19) or a single problem (the sum of all numbers that are multiples of either 17 or 19). If it is the latter, you can almost get the answer by doing them as two separate problem, the sum of all multiples of 17 and the sum of all multiples of 19, and adding the two, as . The reason it is "almost" is that, as Math Message Board Tutor said, you would be counting numbers that are multiples of both 17 and 19 twice. Since 17 and 19 have no common factor (in fact, they are prime) you can fix that by calculating the sum of all numbers that are multiples of (17)(19)= 323 and subtracting it from the previous answer.
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