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

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~Arcticturn
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== Solution 2 (RIGID) ==
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Since <math>x</math> is not an integer, we let <math>x=a+b</math>, where <math>0<b<1</math>.
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So <math>2[x]+3=2a+3</math>. <math>3[x-2]+5=3a-1</math>.
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<math>2a+3=3a-1</math>. <math>a=4</math>. So we know that <math>x</math> is between 4 and 5. <math>y=11</math>. So <math>x+y</math> is between <math>15</math> and <math>16</math>. Select <math>\boxed{D}</math>.
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~Alvie567
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==See Also==
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{{AHSME box|year=1982|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 23:31, 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

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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