Problem:
Let (a,b,c,d) be an ordered quadruple of not necessarily distinct integers, each one of them in the set {0,1,2,3}. For how many such quadruples is it true that a⋅d−b⋅c is odd? (For example, (0,3,1,1) is one such quadruple, because 0⋅1−3⋅1=−3 is odd.)
Answer Choices:
A. 48
B. 64
C. 96
D. 128
E. 192
Solution:
Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set ad to be odd and bc to be even, then multiply by 2.
If ad is odd, both a and d must be odd, therefore there are
2⋅2=4 possibilities for ad
Consider bc. Let us say that b is even. Then there are
2⋅4=8 possibilities for bc
However, b can be odd, in which case we have
2⋅2=4 more possibilities for bc
Thus there are 12 ways for us to choose bc and 4 ways for us to choose ad. Therefore, also considering symmetry, we have
2⋅4⋅12=(C) 96 total values of ad−bc
OR
There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. To get an even products, we count:
P( any number )⋅P( any number )−P( odd )⋅P( odd )=4⋅4−2⋅2=12
The number of ways to get an odd product can be counted like so:
P( odd )⋅P( odd )=2⋅2=4
So, for one product to be odd the other to be even:
2⋅4⋅12=(C) 96 (order matters)
The problems on this page are the property of the MAA's American Mathematics Competitions