2020 AMC 10 A
Complete problem set with solutions and individual problem pages
Let be an ordered quadruple of not necessarily distinct integers, each one of them in the set . For how many such quadruples is it true that is odd? (For example, is one such quadruple, because is odd.)
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Solution 1 (Parity)： In order for to be odd, consider parity. We must have (even)(odd) or (odd)-(even). There are ways to pick numbers to obtain an even product. There are ways to obtain an odd product. Therefore, the total amount of ways to make odd Solution 2 (Basically Solution 1 but more in depth)： Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set to be odd and to be even, then multiply by 2 . If is odd, both and must be odd, therefore there are possibilities for . Consider . Let us say that is even. Then there are possibilities for . However, can be odd, in which case we have more possibilities for . Thus there are 12 ways for us to choose and 4 ways for us to choose . Therefore, also considering symmetry, we have total values of
