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Bunuel
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Bunuel
If the sum of four consecutive even integers is equal to the sum of eight consecutive integers, what is the difference between the sums of the non-identical elements in the two sets?

A. 0
B. 2
C. 4
D. 6
E. 8

Let's assume

  • Set A consists of four consecutive even numbers

    \(A =\) {\(x-2, \quad x, \quad x+2, \quad x+4\)}

  • Set B consists of eight consecutive numbers

    \(B =\) {\( y-3,\quad y-2, \quad y-1, \quad y, \quad y+1,\quad y+2, \quad y+3, \quad y+4\)}

Given "...the sum of four consecutive even integers is equal to the sum of eight consecutive integers..."

\(x - 2 + x + x +2 + x + 4 = y - 3 + y - 2 + y -1 + y + y + 1 + y +2 + y + 3 + y + 4\)

\(4x + 4 = 8y + 4\)

\(4(x + 1) = 4(2y + 1)\)

Dividing by \(4\) on both sides

\(x + 1 = 2y + 1\)

\(x = 2y\)

Let's assume some values to visualize the sets, \(A\) and \(B\), better.

\(x =2 ; y = 1\)

\(A =\) {\(0, \quad 2, \quad 4, \quad 6\)}

\(B =\) {\( -2,\quad -1, \quad 0, \quad 1, \quad 2,\quad 3, \quad 4, \quad 5\)}

  • The sum of non-identical elements of Set \(A = 6\)
  • The sum of non-identical elements of Set \(B = -2 + -1 + 1 + 3 + 5 = 6\)

Difference \(= 6 - 6 = 0\)

Option A
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Bunuel


Right. However, there's a much simpler solution: if the sums of both sets are equal and we exclude the common elements—the sum of which would be the same—then the sums of the remaining unique elements must also be equal, leading to a difference of 0.

:facepalm_man: :facepalm_man: :facepalm_man: :facepalm_man:

Holy Moly !

How did I miss such a simple fact; it seems so obvious in hindsight. :|
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I assumed 1,2,3....till 8 as integers and summed it to 36 and found 6,8,10,12 as those numbers equal 36 too. Hence 22-22=0
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