Difference between revisions of "1982 AHSME Problems/Problem 15"

Latest revision as of 22:31, 29 June 2025

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 $0<b<1$ .

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 $\boxed{D}$ .

~Alvie567

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 25: / Line 25: @@
 ~Alvie567
+==See Also==
+{{AHSME box|year=1982|num-b=14|num-a=16}}
+{{MAA Notice}}

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Difference between revisions of "1982 AHSME Problems/Problem 15"

Latest revision as of 22:31, 29 June 2025

Contents

Problem

Solution

Solution 2 (RIGID)

See Also