Let <em>S</em> be the sum of the integers 25-210:
<em>S</em> = 25 + 26 + 27 + … + 208 + 209 + 210
Let <em>S₃</em>, <em>S₄</em>, and <em>S₁₂</em> denote the sums of the integers in <em>S</em> that are multiples of 3, 4, or 12, respectively. We'll also count how many terms each sum involves; it'll be useful later.
<em>S₃</em> = 27 + 30 + 33 + … + 204 + 207 + 210 … … … (<u>62</u> terms)
<em>S₃</em> = 3 (9 + 10 + 11 + … + 68 + 69 + 70)
<em>S₄</em> = 28 + 32 + 36 + … + 200 + 204 + 208 … … … (<u>46</u> terms)
<em>S₄</em> = 4 (7 + 8 + 9 + … + 50 + 51 + 52)
<em>S₁₂</em> = 36 + 48 + 60 + … + 180 + 192 + 204 … … … (<u>15</u> terms)
<em>S₁₂</em> = 12 (3 + 4 + 5 + … + 15 + 16 + 17)
Let's look at <em>S₃ </em>:
<em>S₃</em> = 3 (9 + 10 + 11 + … + 68 + 69 + 70)
By reversing the order of the sum, we get
<em>S₃</em>* = 3 (70 + 69 + 68 + … + 11 + 10 + 9)
Of course <em>S₃</em> = <em>S₃</em>*, I'm just calling it something else temporarily. Notice that every term in the same position of either sum adds up the same number.
9 + 70 = 79
10 + 69 = 79
11 + 68 = 79
and so on. Then
<em>S₃</em> + <em>S₃</em>* = 3 (79 + 79 + 79 + … + 79 + 79 + 79)
or
2<em>S₃</em> = 3 × <u>62</u> × 79 ==> <em>S₃</em> = 7,347
We can compute the other two sums in the same way.
<em>S₄</em> = 4 (7 + 8 + 9 + … + 50 + 51 + 52)
<em>S₄</em>* = 4 (52 + 51 + 50 + … + 9 + 8 + 7)
==> 2<em>S₄</em> = 4 × <u>46</u> × 59 ==> <em>S₄</em> = 5,428
<em>S₁₂</em> = 12 (3 + 4 + 5 + … + 15 + 16 + 17)
<em>S₁₂</em>* = 12 (17 + 16 +15 + … + 5 + 4 + 3)
==> 2<em>S₁₂</em> = 12 × <u>15</u> × 20 ==> <em>S₁₂</em> = 1,800
Then the sum you want is
<em>S₃</em> + <em>S₄</em> - <em>S₁₂</em> = 10,975
We subtract <em>S₁₂</em> because each of its terms is counted twice (once in <em>S₃</em> and again in <em>S₄</em>).
