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