1982 AHSME Problems/Problem 15: Difference between revisions

Revision as of 17:40, 2 May 2022

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}$.

~hastapasta

@@ Line 16: / Line 16: @@
 ~Arcticturn
+== Solution 2 (RIGID) ==
+Since <math>x</math> is not an integer, we let <math>x=a+b</math>, where <math>0<b<1</math>.
+So <math>2[x]+3=2a+3</math>. <math>3[x-2]+5=3a-1</math>.
+<math>2a+3=3a-1</math>. <math>a=4</math>. So we know that <math>x</math> is between 4 and 5<math>. </math>y=11<math>. So </math>x+y<math> is between </math>15<math> and </math>16<math>. Select </math>\boxed{D}$.
+~hastapasta

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1982 AHSME Problems/Problem 15: Difference between revisions

Revision as of 17:40, 2 May 2022

Problem

Solution

Solution 2 (RIGID)