2020 AMC 10A Problems/Problem 18

Revision as of 18:47, 9 May 2020 by Kokonut pi (talk | contribs) (Solutions 1-4)

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.)

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

Solutions 1-4

Solution 1 (Parity)

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 (Basically Solution 1 but more in depth)

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 (Complementary Counting)

There are 4 ways to choose any number independently and 2 ways to choose any odd number independently. To get an even products, we count: $\text{P(any number)} \cdot \text{P(any number)}-\text{P(odd)}\cdot\text{P(odd)}$, which is $4 \cdot 4 - 2 \cdot 2=12$. The number of ways to get an odd product can be counted like so: $\text{P(odd)}\cdot\text{P(odd)}$, which is $2 \cdot 2$, or $4$. So, for one product to be odd the other to be even: $2 \cdot 4 \cdot 12=\boxed{(C)96}$(order matters). ~ Anonymous and Arctic_Bunny

Solution 4 (Solution 3 but more in depth)

We use complementary counting: If the difference is even, then we can subtract those cases. There are a total of $4^4=256$ cases.

For an even difference, we have (even)-(even) or (odd-odd).

From 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): $2 \cdot 2 =4$."

With this, we easily calculate $256-12^2-4^2=\textbf{(C)96}$.

~kevinmathz

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

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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