Difference between revisions of "2020 AMC 10A Problems/Problem 18"
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<math>\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192</math> | <math>\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192</math> | ||
− | == Solution == | + | == Solution 1 == |
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>. | ||
Revision as of 23:26, 31 January 2020
Contents
[hide]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 1
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
- I believe this is incorrect. When finding the evens, is incorrect because you are assuming the first number is always even/odd.
is the correct answer.
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 |
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