Difference between revisions of "2020 AMC 10A Problems/Problem 18"
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== Solution == | == Solution == | ||
In order for <math>a\cdot d-b\cdot c</math> to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are <math>2 \cdot 4 = 8</math> ways to pick numbers to obtain a even product. There are <math>2 \cdot 2 = 4</math> ways to obtain an odd product. Therefore, the total amount of ways to make <math>a\cdot d-b\cdot c</math> odd is <math>2 \cdot (8 \cdot 4) = \boxed{\text{B}, 64}</math>. | In order for <math>a\cdot d-b\cdot c</math> to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are <math>2 \cdot 4 = 8</math> ways to pick numbers to obtain a even product. There are <math>2 \cdot 2 = 4</math> ways to obtain an odd product. Therefore, the total amount of ways to make <math>a\cdot d-b\cdot c</math> odd is <math>2 \cdot (8 \cdot 4) = \boxed{\text{B}, 64}</math>. | ||
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| + | -Midnight | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:52, 31 January 2020
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.)
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
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. 
