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

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==Additional Note 2==
 
==Additional Note 2==
 
This problem is similar to 2007 AMC10A Problem 16. View it here: https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16
 
This problem is similar to 2007 AMC10A Problem 16. View it here: https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16
~coolmath_2018
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2020|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:22, 22 January 2021

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 $2(2 + 4) = 12$ 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 1

Education, The Study of Everything

https://youtu.be/D34FxUr9TvI


https://youtu.be/RKlG6oZq9so

~IceMatrix

Video Solution 2

https://youtu.be/3bRjcrkd5mQ?t=1

~ pi_is_3.14

Additional Note 1

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


Additional Note 2

This problem is similar to 2007 AMC10A Problem 16. View it here: https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16

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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