Question

M is a set of six consecutive even integers. When the least three integers of set $M$ are summed, the result is $x$. When the greatest three integers of set $M$ are summed, the result is $y$. Mark the true \[\]equation.
A.$y=x-18$\[\]
B.$y=x+18$\[\]
C.$y=2x$\[\]
D.$y=2x+4$\[\]

Answer 1

Hint: We denote the six even consecutive integers from the set $M$ in ascending order as$n,n+2,n+4,n+6,n+8,n+10$. We add the first three integers and equate to $x$ as given in the question and express $n$ in terms of $x$. We add the latter three integers and equate to $y$as given in the question. We put $n$ in terms of $x$ and simplify to obtain the equation. \[\]

Complete step-by-step solution
We are given in the question that $M$ is a set of six consecutive even integers. We know that even integers are multiples of 2 and two consecutive even integers will have a difference of 2. If we denote an arbitrary even number as $n$, then we can represent the next 5 consecutive even integers in ascending order can be written as $n+2,n+4,n+6,n+6,n+10$. So the set $M$ can be written in listed form as,
\[M=\left\{ n,n+2,n+4,n+6,n+8,n+10 \right\}\]
We are also given the question that the least three integers of set $M$ are summed, the result is $x$. The least three integers are$n,n+2,n+4$. So we have,
\[\begin{align}
  & n+n+2+n+4=x \\
 & \Rightarrow 3n+6=x \\
 & \Rightarrow n=\dfrac{x-6}{3}.......\left( 1 \right) \\
\end{align}\]
We are further given the question that when the greatest three integers of set $M$ are summed, the result is $y$.The greatest three integers in the set $M$ are$n+4,n+6,n+10$. So we have,
\[\begin{align}
  & n+4+n+6+n+10=y \\
 & \Rightarrow 3n+20=y \\
\end{align}\]
We put $n=\dfrac{x-6}{3}$ obtained from (1) in the above step to have,
\[\begin{align}
  & \Rightarrow 3\dfrac{x-6}{3}+24=y \\
 & \Rightarrow x-6+24=y \\
 & \Rightarrow y=x+18 \\
\end{align}\]
The above equation is the required equation and hence the correct option is B.

Note: The obtained equation is a linear equation in two variables whose standard form is $ax+by=c$ where $a,b,c$ are real numbers and $a\ne 0,b\ne 0$. There are infinite solutions for one linear equation and we can get integral solutions only when the greatest common divisor of $a,b$ exactly divides $c$. That linear equation is called Diophantine equation.