Difference between revisions of "1996 AJHSME Problems/Problem 21"

Revision as of 23:25, 4 July 2013

Problem

How many subsets containing three different numbers can be selected from the set $\{ 89,95,99,132, 166,173 \}$ so that the sum of the three numbers is even?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution

To have an even sum with three numbers, we must add either $E+O+O$ , or $E + E + E$ , where $O$ represents an odd number, and $E$ represents an even number.

Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.

There are $2$ choices for the even number.

There are $4$ choices for the first odd number. There are $3$ choices for the last odd number. But the order of picking these numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$ . Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.

In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum. This is option $\boxed{D}$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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
All AJHSME/AMC 8 Problems and Solutions

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

@@ Line 25: / Line 25: @@
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "1996 AJHSME Problems/Problem 21"

Revision as of 23:25, 4 July 2013

Problem

Solution

See Also