Difference between revisions of "2020 AMC 10A Problems/Problem 18"

Revision as of 23:49, 31 January 2020

Problem

Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set ${0,1,2,3}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)

$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

Solution 1

In order for $a\cdot d-b\cdot c$ to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are $2 \cdot 4 + 2 \cdot 2 = 12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2 = 4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a\cdot d-b\cdot c$ odd is $2 \cdot (12 \cdot 4) = \boxed{\bold{(C)}\ 96}$ .

-Midnight

Solution 2

Consider parity. We need exactly one term to be odd, one term to be even. Because of symmetry, we can set $ad$ to be odd and $bc$ to be even, then multiply by $2.$ If $ad$ is odd, both $a$ and $d$ must be odd, therefore there are $2\cdot2=4$ possibilities for $ad.$ Consider $bc.$ Let us say that $b$ is even. Then there are $2\cdot4=8$ possibilities for $bc.$ However, $b$ can be odd, in which case we have $2\cdot2=4$ more possibilities for $bc.$ Thus there are $12$ ways for us to choose $bc$ and $4$ ways for us to choose $ad.$ Therefore, also considering symmetry, we have $2*4*12=96$ total values of $ad-bc.$ $(C)$

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

Additional Note

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 $2$ odd integers, it can quickly be deduced that there are $2 \cdot 2 = 4$ possibilities for an odd product. Since the product must be either odd or even, and there are $4 \cdot 4 = 16$ ways to choose factors for the product, there are $16 - 4 = 12$ possibilities for an even product. ~emerald_block

@@ Line 16: / Line 16: @@
 ~IceMatrix
+==Additional Note==
+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 <math>2</math> odd integers, it can quickly be deduced that there are <math>2 \cdot 2 = 4</math> possibilities for an odd product. Since the product must be either odd or even, and there are <math>4 \cdot 4 = 16</math> ways to choose factors for the product, there are <math>16 - 4 = 12</math> possibilities for an even product. ~[[User:emerald_block|emerald_block]]
 ==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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