Difference between revisions of "1998 AHSME Problems/Problem 20"

Revision as of 18:37, 1 December 2013

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 $10$ 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 two more possibilities.

Thus before Stacy took her look, we are left with four possible cases: $(1,3,9)$ , $(1,4,8)$ , $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\boxed{3}$ .

Thanks

bogus solution?: I think the answer is 4, because once you eliminate $(3,4,6)$ , you are left with $(1,5,7)$ , $(1,4,8)$ , $(1,3,9)$ , $(1,2,10)$ , $(2,5,6)$ , $(2,4,7)$ , and $(2,3,8)$ . You can then eliminate $(1,3,9)$ , $(1,2,10)$ , and $(2,5,6)$ , because those are the ones for which the last term is unique. Finally, since Stacy does not have enough information, then the middle term cannot be unique, so $\boxed{4}$

If I'm wrong please feel free to correct me.

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 Thanks
+*bogus solution?: I think the answer is 4, because once you eliminate <math>(3,4,6)</math>, you are left with <math>(1,5,7)</math>, <math>(1,4,8)</math>, <math>(1,3,9)</math>, <math>(1,2,10)</math>, <math>(2,5,6)</math>, <math>(2,4,7)</math>, and  <math>(2,3,8)</math>. You can then eliminate <math>(1,3,9)</math>, <math>(1,2,10)</math>, and <math>(2,5,6)</math>, because those are the ones for which the last term is unique. Finally, since Stacy does not have enough information, then the middle term cannot be unique, so <math>\boxed{4}</math>
+If I'm wrong please feel free to correct me.
 == See also ==
 {{AHSME box|year=1998|num-b=19|num-a=21}}
 {{MAA Notice}}

1998 AHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1998 AHSME Problems/Problem 20"

Revision as of 18:37, 1 December 2013

Problem

Solution

See also