1988 AHSME Problems/Problem 12

Revision as of 07:53, 24 June 2022 by Plainoldnumbertheory (talk | contribs) (Solution)

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 (needs to be checked)

The largest sum that can be formed from two slips of these papers is 18

See also

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