# 2022 AMC 8 Problems/Problem 12

## Problem

The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner $\text{A}$, added to the number on Spinner $\text{B}$. What is the probability that $N$ is a perfect square number? $[asy] //diagram by pog give me 1 billion dollars for this size(6cm); usepackage("mathptmx"); filldraw(arc((0,0), r=4, angle1=0, angle2=90)--(0,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((0,0), r=4, angle1=90, angle2=180)--(0,0)--cycle,lightgray); filldraw(arc((0,0), r=4, angle1=180, angle2=270)--(0,0)--cycle,mediumgray); filldraw(arc((0,0), r=4, angle1=270, angle2=360)--(0,0)--cycle,lightgray*0.5+mediumgray*0.5); label("5", (-1.5,1.7)); label("6", (1.5,1.7)); label("7", (1.5,-1.7)); label("8", (-1.5,-1.7)); label("Spinner A", (0, -5.5)); filldraw(arc((12,0), r=4, angle1=0, angle2=90)--(12,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((12,0), r=4, angle1=90, angle2=180)--(12,0)--cycle,lightgray); filldraw(arc((12,0), r=4, angle1=180, angle2=270)--(12,0)--cycle,mediumgray); filldraw(arc((12,0), r=4, angle1=270, angle2=360)--(12,0)--cycle,lightgray*0.5+mediumgray*0.5); label("1", (10.5,1.7)); label("2", (13.5,1.7)); label("3", (13.5,-1.7)); label("4", (10.5,-1.7)); label("Spinner B", (12, -5.5)); [/asy]$ $\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}$

## Solution 1

First, we calculate that there are a total of $4\cdot4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$ probability that the number formed by the two spinners is a perfect square.

~MathFun1000

## Solution 2

There are $4 \cdot 4 = 16$ total possibilities of $N$. We know $N=10A+B$, which $A$ is a number from spinner $A$, and $B$ is a number from spinner $B$. Also, notice that there are no perfect squares in the $50$s or $70$s, so only $4-2=2$ values of N work, namely $64$ and $81$. Hence, $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$.

~MrThinker

## Solution 3

Just try them!

• If we spin a 5 on the first spinner, there are no solutions.
• If we spin a 6 on the first spinner, there is one solution (64).
• If we spin a 7 on the first spinner, there are no solutions.
• If we spin an 8 on the first spinner, there is one solution (81).

Therefore, there are 2 solutions and $4 \cdot 4 = 16$ total possibilities, so $$\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}$$

~ligonmathkid2

~Math-X

## Video Solution (CREATIVE THINKING!!!)

~Education, the Study of Everything

~Interstigation

~STEMbreezy

~savannahsolver

## Video Solution

~harungurcan

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 