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