1988 AHSME Problems/Problem 12

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Problem

Each integer $1$ through $9$ is written on a separate slip of paper and all nine slips are put into a hat. Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random. Which digit is most likely to be the units digit of the sum of Jack's integer and Jill's integer?

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{each digit is equally likely}$

Solution

We can draw a sample space diagram, and find that of the $9^2 = 81$ possibilities, $9$ of them give a sum of $0$, and each other sum mod $10$ (from $1$ to $9$) is given by $8$ of the possibilities (and indeed we can check that $9 + 8 \times 9 = 81$). Thus $0$ is the most likely, so the answer is $\boxed{\text{A}}$.

Solution (alternative explanation)

The largest sum that can be formed from two slips of these papers is 18, and if we list the composition of each of the numbers greater than 1 and less than 19, we can find that there are 9 combinations in total which make a sum of 10 ($1+9$, $2+8$, etc). Notice that the order in which either number appears doesn't matter since they all count toward the number of possibilities.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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