2020 AMC 10A Problems/Problem 18
Problem
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.)
Solution 1
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 is .
-Midnight
Solution 2
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 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 ways for us to choose and ways for us to choose Therefore, also considering symmetry, we have total values of
Video Solution
~IceMatrix
Additional Note
When calculating the number of even products and odd products, since the only way to get an odd product is to multiply two odd integers together, and there are odd integers, it can quickly be deduced that there are possibilities for an odd product. Since the product must be either odd or even, and there are ways to choose factors for the product, there are possibilities for an even product. ~emerald_block
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |
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