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

Revision as of 11:57, 2 February 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

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 $ad-bc.$ $(C)$

Solution 3

There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. even products:(number)*(number)-(odd)*(odd): $4 \cdot 4 - 2 \cdot 2=12$ . odd products: (odd)*(odd): $2 \cdot 2 =4$ . One product is odd the other is even: $2 \cdot 4 \cdot 12=\boxed{(C)96}$ (order matters)

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 12: / Line 12: @@
 ===Solution 2===
 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>
+==Solution 3==
+There are 4 ways to choose any number independently and 2 ways to choose any odd number independently.
+even products:(number)*(number)-(odd)*(odd): <math>4 \cdot 4 - 2 \cdot 2=12</math>.
+odd products: (odd)*(odd): <math>2 \cdot 2 =4</math>.
+One product is odd the other is even: <math>2 \cdot 4 \cdot 12=\boxed{(C)96}</math>(order matters)
 ==Video Solution==

2020 AMC 10A (Problems • Answer Key • Resources)
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