Difference between revisions of "2020 AMC 10A Problems/Problem 18"
(→Solution) |
(→Solution) |
||
| Line 8: | Line 8: | ||
-Midnight | -Midnight | ||
| + | |||
| + | == Solution == | ||
| + | Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set <math>ad</math> to be odd and <math>bc</math> to be even, then multiply by <math>2.</math> If <math>ad</math> is odd, both <math>a</math> and <math>d</math> must be odd, therefore there are <math>2\cdot2=4</math> possibilities for <math>ad.</math> Consider <math>bc.</math> Let us say that <math>b</math> is even. Then there are <math>2\cdot4=8</math> possibilities for <math>bc.</math> However, <math>b</math> can be odd, in which case we have <math>2\cdot2=4</math> more possibilities for <math>bc.</math> Thus there are <math>12</math> ways for us to choose <math>bc</math> and <math>4</math> ways for us to choose <math>ad.</math> Therefore, also considering symmetry, we have <math>2*4*12=96</math> total values of <math>ad-bc.</math> <math>(C)</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:53, 31 January 2020
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
In order for to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are 
ways to pick numbers to obtain a even product. There are 
ways to obtain an odd product. Therefore, the total amount of ways to make odd is 
.
-Midnight
Solution
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
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. 
