Find four consecutive even integers such that 3 times the 2nd plus twice the 4th equals 78.?
104227
Too tired from working on other one.  Stresses my brain.
822074
..ll<3,2>ll  = sqrt( <3,2> . <3,2> )\n= sqrt( 4(3)(3) +7(2)(2) ) \n=sqrt( 36 +28) = sqrt ( 64)\n= 8
523701
If you add the numbers together then they equal an even number.   1 1  1+1=2
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
611894
8 to 3  (56 to 21 and divide by 7)
820779
Consecutive integers are n, n + 1, n + 2, n + 3\n\nThe sum of the second and fourth equals the third increased by 17.\n\n(n + 1) + (n + 3) = (n + 2) + 17
1136505
(7,0) \nXdiff = X1 - X2\nYdiff = Y1 - Y2\nThe first (X1 or Y1) should be the larger number of the 2\nso Xdiff = 4 - -3 = 7\nand Ydiff = 0 - 0 = 0
1110742
12 simply place the 1 where the x is\n\n(1+3)(1+2)\n4*3\n12
1099830
27\n\n(+7 -4 +8 -3 +9 -2 +10)
770779
Call the integers x, x+1, x+2\n\nx+1+x+2 = x/2 + 24\n\n2x + 3 = x/2 + 24\n2x = x/2 + 21\n4x = x + 42\n3x = 42\nx = 14\n\nThe three integers are 14, 15, 16
208376
6+8+10=24 ........................................................................
47484
Y=X+1 (consecutive numbers)\n4Y=3X+23\n4(X+1)=3X+23\n4X+4=3X+23\n\nTherefore:\n\nX=19\nY=20
853291
4 is the answer. Here's how:\n4 appears after a break of two numbers, so the number that follows 40 must be 4
734849
x+78 [greater than or equal to] 120\nx [greater than or equal to] 120-78\nx [greater than or equal to] 42
897152
let x = smaller number\n2x + 8 = larger number\n2(2x + 8) + 3x = 65\n4x + 16 + 3x = 65\n7x + 16 - 16 = 65 - 16\n7x = 49\n7x/7 = 49/7\nx =7
403326
1+2+3+4+5+6+7+8+9/3+8*2-4\n=36+3+16-4\n=51
650616
These three sequntal clock wise numbers \n2+3+4 --> 9+4 --> 1 (adding the last no.(4)\nBut not works for all