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

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== Problem ==
 
== Problem ==
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Let <math>(a,b,c,d)</math> be an ordered quadruple of not necessarily distinct integers, each one of them in the set <math>{0,1,2,3}.</math> For how many such quadruples is it true that <math>a\cdot d-b\cdot c</math> is odd? (For example, <math>(0,3,1,1)</math> is one such quadruple, because <math>0\cdot 1-3\cdot 1 = -3</math> is odd.)
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Let <math>(a,b,c,d)</math> be an ordered quadruple of not necessarily distinct integers, each one of them in the set <math>\{0,1,2,3\}.</math> For how many such quadruples is it true that <math>a\cdot d-b\cdot c</math> is odd? (For example, <math>(0,3,1,1)</math> is one such quadruple, because <math>0\cdot 1-3\cdot 1 = -3</math> is odd.)
  
 
<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>
  
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== Solution ==
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==Solutions==
−
In order for <math>a\cdot d-b\cdot c</math> to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). Note that by
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===Solution 1 (Parity)===
 +
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(2 + 4) = 12</math> ways to pick numbers to obtain an 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 (12 \cdot 4) = \boxed{\bold{(C)}\ 96}</math>.
 +
 
 +
~Midnight
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 +
===Solution 2 (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 <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 \cdot 4 \cdot 12= \boxed{ \text{(C) } 96}</math> total values of <math>ad-bc.</math>
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 +
~lpieleanu (Reformatting and Minor Edits)
 +
 
 +
===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: <math>\text{P(any number)} \cdot \text{P(any number)}-\text{P(odd)}\cdot\text{P(odd)}</math>, which is <math>4 \cdot 4 - 2 \cdot 2=12</math>.
 +
The number of ways to get an odd product can be counted like so: <math>\text{P(odd)}\cdot\text{P(odd)}</math>, which is <math>2 \cdot 2</math>, or <math>4</math>.
 +
So, for one product to be odd the other to be even: <math>2 \cdot 4 \cdot 12= \boxed{ \text{(C) } 96}</math> (order matters).
 +
 
 +
~Anonymous and Arctic_Bunny
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 +
===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 <math>4^4=256</math> 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.
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even products:(number)*(number)-(odd)*(odd): <math>4 \cdot 4 - 2 \cdot 2=12</math>.
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odd products: (odd)*(odd): <math>2 \cdot 2 =4</math>."
 +
 
 +
With this, we easily calculate <math>256-12^2-4^2=\textbf{(C)96}</math>.
 +
 
 +
~kevinmathz
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 +
===Solution 5 (Casework)===
 +
As in solution 1, we must have (even)-(odd) or (odd)-(even). We see that there are two cases, if <math>ad</math> is even and <math>bc</math> is odd and if <math>ad</math> is odd and <math>bc</math> is even. Because of symmetry, we can multiply by two for when <math>ad</math> is odd and <math>bc</math> is even. Let <math>e</math> denote an even number and let <math>o</math> denote an odd number.
 +
 
 +
If <math>ad</math> is even and <math>bc</math> is odd, there are three cases:
 +
 
 +
 
 +
<cmath>(e,o,o,e)</cmath>
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<cmath>(e,o,o,o)</cmath>
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<cmath>(o,o,o,e)</cmath>
 +
 
 +
 
 +
For each of these cases, there are <math>2^4</math> ways to choose from the set <math>(0,1,2,3)</math> as there are 2 even's and 2 odd's; because there are three cases, we multiply this by 3. Also, because of there are 2 cases (<math>ad</math> is even and <math>bc</math> is odd and if <math>ad</math> is odd and <math>bc</math> is even), we multiply this by 2. This gives us:
 +
 
 +
<math>2^4 \cdot 3 \cdot 2= \textbf{(C)96}</math>
 +
 
 +
===Solution 6===
 +
For parity reasons, if <math>ad - bc</math> is to be odd, we must have <math>ad</math> odd and <math>bc</math> even or <math>ad</math> even and <math>bc</math> odd. By symmetry, these cases are identical, so we consider the first one and multiply by two at the end. For <math>ad</math> to be odd, we must have both <math>a</math> and <math>d</math> odd, and there are <math>2 \cdot 2</math> ways to do so. To count the cases where <math>bc</math> is odd, we use PIE. there are <math>2 \cdot 4 = 8</math> ways for <math>b</math> to be odd and <math>4 \cdot 2 = 8</math> ways for <math>c</math> to be odd, and there are <math>2 \cdot 2 = 4</math> ways for both to be odd. Thus, there are <math>8 + 8 - 4 = 12</math> ways for <math>bc</math> to be even. Multiplying out, there are <math>2 \cdot 4 \cdot 12</math> ways to have <math>ad - bc</math> odd for a total of <math>\boxed{\textbf{(C)}~96}</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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 +
==Video Solutions==
 +
 
 +
Education, The Study of Everything
 +
 
 +
https://youtu.be/D34FxUr9TvI
 +
 
  
−
==Video Solution==
 
 
https://youtu.be/RKlG6oZq9so
 
https://youtu.be/RKlG6oZq9so
  
 
~IceMatrix
 
~IceMatrix
 +
 +
 +
https://youtu.be/3bRjcrkd5mQ?t=1
 +
 +
~ pi_is_3.14
 +
 +
==Additional Notes==
 +
===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 <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]]
 +
 +
===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==
 
==See Also==

Latest revision as of 21:35, 4 November 2024

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

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 (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 $2 \cdot 4 \cdot 12= \boxed{ \text{(C) } 96}$ total values of $ad-bc.$

~lpieleanu (Reformatting and Minor Edits)

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: $2 \cdot 4 \cdot 12= \boxed{ \text{(C) } 96}$ (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

Solution 5 (Casework)

As in solution 1, we must have (even)-(odd) or (odd)-(even). We see that there are two cases, if $ad$ is even and $bc$ is odd and if $ad$ is odd and $bc$ is even. Because of symmetry, we can multiply by two for when $ad$ is odd and $bc$ is even. Let $e$ denote an even number and let $o$ denote an odd number.

If $ad$ is even and $bc$ is odd, there are three cases:


\[(e,o,o,e)\] \[(e,o,o,o)\] \[(o,o,o,e)\]


For each of these cases, there are $2^4$ ways to choose from the set $(0,1,2,3)$ as there are 2 even's and 2 odd's; because there are three cases, we multiply this by 3. Also, because of there are 2 cases ($ad$ is even and $bc$ is odd and if $ad$ is odd and $bc$ is even), we multiply this by 2. This gives us:

$2^4 \cdot 3 \cdot 2= \textbf{(C)96}$

Solution 6

For parity reasons, if $ad - bc$ is to be odd, we must have $ad$ odd and $bc$ even or $ad$ even and $bc$ odd. By symmetry, these cases are identical, so we consider the first one and multiply by two at the end. For $ad$ to be odd, we must have both $a$ and $d$ odd, and there are $2 \cdot 2$ ways to do so. To count the cases where $bc$ is odd, we use PIE. there are $2 \cdot 4 = 8$ ways for $b$ to be odd and $4 \cdot 2 = 8$ ways for $c$ to be odd, and there are $2 \cdot 2 = 4$ ways for both to be odd. Thus, there are $8 + 8 - 4 = 12$ ways for $bc$ to be even. Multiplying out, there are $2 \cdot 4 \cdot 12$ ways to have $ad - bc$ odd for a total of $\boxed{\textbf{(C)}~96}$.

~ cxsmi

Video Solutions

Education, The Study of Everything

https://youtu.be/D34FxUr9TvI


https://youtu.be/RKlG6oZq9so

~IceMatrix


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

~ pi_is_3.14

Additional Notes

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 (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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