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.)
Solutions
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 is .
-Midnight
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 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
~lpieleanu (reformatting and minor edits)
Solution 3 (Complementary Counting)
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: , which is . The number of ways to get an odd product can be counted like so: , which is , or . So, for one product to be odd the other to be even: (order matters). ~ Anonymous and Arctic_Bunny
Solution 4 (Solution 3 but more in-depth)
We use complementary counting: If the difference is even, then we can subtract those cases. There are a total of cases.
For an even difference, we have (even)-(even) or (odd-odd).
From Solution 3:
"There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. even products:(number)*(number)-(odd)*(odd): . odd products: (odd)*(odd): ."
With this, we easily calculate .
~kevinmathz
Solution 5-Casework
As in solution 1, we must have (even)-(odd) or (odd)-(even). We see that there are two cases, if is even and is odd and if is odd and is even. Because of symmetry, we can multiply by two for when is odd and is even. Let denote an even number and let denote an odd number.
If is even and is odd, there are three cases:
For each of these cases, there are ways to choose from the set as there are 2 even's and 2 odd's; because there are three cases, we multiply this by 3. Also, because of there are 2 cases ( is even and is odd and if is odd and is even), we multiply this by 2. This gives us:
Solution 6
For parity reasons, if is to be odd, we must have odd and even or even and odd. By symmetry, these cases are identical, so we consider the first one and multiply by two at the end. For to be odd, we must have both and odd, and there are ways to do so. To count the cases where is odd, we use PIE. there are ways for to be odd and ways for to be odd, and there are ways for both to be odd. Thus, there are ways for to be even. Multiplying out, there are ways to have odd for a total of .
~ cxsmi
Video Solutions
Education, The Study of Everything
~IceMatrix
https://youtu.be/3bRjcrkd5mQ?t=1
~ pi_is_3.14
Additional Notes
Additional Note 1
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
Additional Note 2
This problem is similar to 2007 AMC10A Problem 16. View it here: https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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