1998 AHSME Problems/Problem 20

Problem

Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that

(a) the numbers are all different,
(b) they sum to $13$, and
(c) they are in increasing order, left to right.

First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?

$\textrm{(A)}\ 2 \qquad \textrm{(B)}\ 3 \qquad \textrm{(C)}\ 4 \qquad \textrm{(D)}\ 5 \qquad \textrm{(E)}\ \text{There is not enough information to determine the number.}$

Solution

Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$, $(1,3,9)$, $(1,4,8)$, $(1,5,7)$, $(2,3,8)$, $(2,4,7)$, $(2,5,6)$, and $(3,4,6)$.

If Casey saw the number $3$, she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$.

At this moment, if the last card contained a $10$, $9$, or a $6$, Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$, she would know that the other two are $2$ and $5$.) This eliminates three more possibilities.

Thus before Stacy took her look, we are left with four possible cases: $(2,4,7)$, $(1,4,8)$, $(1,5,7)$, and $(2,3,8)$. As Stacy could not find out the exact combination, the middle number must be $\boxed{4}$.

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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

Invalid username
Login to AoPS