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

Revision as of 18:48, 21 October 2021

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

@@ Line 13: / Line 13: @@
 ==Solution==
-We simply ignore the floor of <math>x</math>. Then, we have <math>y</math> = <math>2x + 3</math> = <math>3(x-2)+5</math>. Solving for <math>3x - 1 = 2x + 3</math>, we get <math>x = 4</math>. For the floor of <math>x</math>, we have <math>x</math> is between <math>4</math> and <math>5</math>. Plugging in <math>8</math> + <math>3</math> = <math>11</math> for <math>y</math>, we have <math>y = 11</math>. We have <math>11 + 4.x</math> = $\boxed {(D)}
+We simply ignore the floor of <math>x</math>. Then, we have <math>y</math> = <math>2x + 3</math> = <math>3(x-2)+5</math>. Solving for <math>3x - 1 = 2x + 3</math>, we get <math>x = 4</math>. For the floor of <math>x</math>, we have <math>x</math> is between <math>4</math> and <math>5</math>. Plugging in <math>8</math> + <math>3</math> = <math>11</math> for <math>y</math>, we have <math>y = 11</math>. We have <math>11 + 4.x</math> = <math>\boxed {(D)}</math>
 ~Arcticturn

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

Revision as of 18:48, 21 October 2021

Problem

Solution