Difference between revisions of "2020 AMC 10A Problems/Problem 18"
Pi is 3.14 (talk | contribs) (→Video Solution) |
Sweetmango77 (talk | contribs) m (→Additional Note 2) |
||
Line 57: | Line 57: | ||
==Additional Note 2== | ==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 | 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== | ==See Also== | ||
{{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:22, 22 January 2021
Contents
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 1-4
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
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
Video Solution 1
Education, The Study of Everything
~IceMatrix
Video Solution 2
https://youtu.be/3bRjcrkd5mQ?t=1
~ pi_is_3.14
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 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.