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 | + | 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>. |
==Video Solution== | ==Video Solution== |
Revision as of 23:52, 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 .
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 |
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