# 1982 AHSME Problems/Problem 15

## Problem

Let $[z]$ denote the greatest integer not exceeding $z$. Let $x$ and $y$ satisfy the simultaneous equations

\begin{align*} y&=2[x]+3 \\ y&=3[x-2]+5. \end{align*}

If $x$ is not an integer, then $x+y$ is

$\text {(A) } \text{ an integer} \qquad \text {(B) } \text{ between 4 and 5} \qquad \text{(C) }\text{ between -4 and 4}\qquad\\ \text{(D) }\text{ between 15 and 16}\qquad \text{(E) } 16.5$

## Solution

We simply ignore the floor of $x$. Then, we have $y$ = $2x + 3$ = $3(x-2)+5$. Solving for $3x - 1 = 2x + 3$, we get $x = 4$. For the floor of $x$, we have $x$ is between $4$ and $5$. Plugging in $8$ + $3$ = $11$ for $y$, we have $y = 11$. We have $11 + 4.x$ = $\boxed {(D)}$

~Arcticturn

## Solution 2 (RIGID)

Since $x$ is not an integer, we let $x=a+b$, where $0.

So $2[x]+3=2a+3$. $3[x-2]+5=3a-1$.

$2a+3=3a-1$. $a=4$. So we know that $x$ is between 4 and 5$.$y=11$. So$x+y$is between$15$and$16$. Select$\boxed{D}\$.

~hastapasta

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