2007 AMC 10B Problems/Problem 19
Problem 19
The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by 
and the second number is divided by 
The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=2; pair O=(0,0); pair A=(0,2), A1=(0,-2); draw(A--A1); pair B=(sqrt(3),1), B1=(-sqrt(3),-1); draw(B--B1); pair C=(sqrt(3),-1), C1=(-sqrt(3),1); draw(C--C1); path circleO=Circle(O,r); draw(circleO); pair[] ps={O}; dot(ps); label("$6$",(-0.6,1)); label("$1$",(0.6,1)); label("$2$",(0.6,-1)); label("$9$",(-0.6,-1)); label("$7$",(1.2,0)); label("$3$",(-1.2,0)); label("$pointer$",(-4,0)); draw((-5.5,0.5)--(-5.5,-0.5)--(-3,-0.5)--(-2.5,0)--(-3,0.5)--cycle); fill((4,0)--(4,1)--(5,1)--(5,0)--cycle,gray); fill((6,2)--(6,1)--(5,1)--(5,2)--cycle,gray); fill((6,0)--(6,-1)--(5,-1)--(5,0)--cycle,gray); fill((6,0)--(6,1)--(7,1)--(7,0)--cycle,gray); fill((4,-1)--(5,-1)--(5,-2)--(4,-2)--cycle,gray); fill((6,-1)--(7,-1)--(7,-2)--(6,-2)--cycle,gray); draw((4,2)--(7,2)--(7,-2)--(4,-2)--cycle); draw((4,1)--(7,1)); draw((4,0)--(7,0)); draw((4,-1)--(7,-1)); draw((5,2)--(5,-2)); draw((6,2)--(6,-2)); label("$1$",midpoint((4,-1)--(4,-2)),W); label("$2$",midpoint((4,0)--(4,-1)),W); label("$3$",midpoint((4,1)--(4,0)),W); label("$4$",midpoint((4,2)--(4,1)),W); label("$1$",midpoint((4,-2)--(5,-2)),S); label("$2$",midpoint((5,-2)--(6,-2)),S); label("$3$",midpoint((7,-2)--(6,-2)),S); [/asy]](http://latex.artofproblemsolving.com/d/1/6/d16f5082e7ba12a9e32d1bd5aad7b6620c1985ae.png)
Solution
When dividing each number on the wheel by the remainders are 
and 
Each column on the checkerboard is equally likely to be chosen.
When dividing each number on the wheel by 
the remainders are and 
The probability that a shaded square in the 
st or 
rd row of the st or rd column is
![\[\frac{2}{3} \times \frac{3}{6} = \frac{1}{3}\]](http://latex.artofproblemsolving.com/5/0/8/5085076484cb2b4c737dae329bacc077edc4922e.png)
The probability that a shaded square in the 
nd or 
th row of the nd column is
![\[\frac{1}{3} \times \frac{3}{6} = \frac{1}{6}\]](http://latex.artofproblemsolving.com/7/6/d/76da33bad7bfda26ad29f4bbdf0278c7bde0d552.png)
Add those two together
![\[\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \boxed{\textbf{(C)} \frac{1}{2}}\]](http://latex.artofproblemsolving.com/9/4/1/94112e565703159cbc8443b1d20984d0a6e60d34.png)
