Difference between revisions of "2020 AMC 10A Problems/Problem 18"
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== Problem == | == Problem == | ||
− | 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.) | + | 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> | ||
− | == | + | ==Solutions== |
− | ===Solution 1=== | + | ===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 | + | 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 | |
− | ===Solution 2=== | + | ===Solution 2 (Solution 1 but more in-depth)=== |
− | |||
− | ===Solution 3=== | + | 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> |
+ | |||
+ | ~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. | 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>. | 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>. | 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{(C)96}</math>(order matters). | + | 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). |
− | |||
− | ===Solution 4=== | + | ~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 <math>4^4=256</math> cases. | 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. | ||
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~kevinmathz | ~kevinmathz | ||
− | ==Video | + | ===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> | ||
+ | <cmath>(e,o,o,o)</cmath> | ||
+ | <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] | ||
+ | |||
+ | ==Video Solutions== | ||
+ | |||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/D34FxUr9TvI | ||
+ | |||
+ | |||
https://youtu.be/RKlG6oZq9so | https://youtu.be/RKlG6oZq9so | ||
~IceMatrix | ~IceMatrix | ||
− | ==Additional Note== | + | |
+ | 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]] | 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 14:09, 13 July 2024
Problem
Let be an ordered quadruple of not necessarily distinct integers, each one of them in the set
For how many such quadruples is it true that
is odd? (For example,
is one such quadruple, because
is odd.)
Solutions
Solution 1 (Parity)
In order for to be odd, consider parity. We must have (even)-(odd) or (odd)-(even). There are
ways to pick numbers to obtain an even product. There are
ways to obtain an odd product. Therefore, the total amount of ways to make
odd is
.
~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 to be odd and
to be even, then multiply by
If
is odd, both
and
must be odd, therefore there are
possibilities for
Consider
Let us say that
is even. Then there are
possibilities for
However,
can be odd, in which case we have
more possibilities for
Thus there are
ways for us to choose
and
ways for us to choose
Therefore, also considering symmetry, we have
total values of
~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: , which is
.
The number of ways to get an odd product can be counted like so:
, which is
, or
.
So, for one product to be odd the other to be even:
(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 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): .
odd products: (odd)*(odd):
."
With this, we easily calculate .
~kevinmathz
Solution 5 (Casework)
As in solution 1, we must have (even)-(odd) or (odd)-(even). We see that there are two cases, if is even and
is odd and if
is odd and
is even. Because of symmetry, we can multiply by two for when
is odd and
is even. Let
denote an even number and let
denote an odd number.
If is even and
is odd, there are three cases:
For each of these cases, there are ways to choose from the set
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 (
is even and
is odd and if
is odd and
is even), we multiply this by 2. This gives us:
Solution 6
For parity reasons, if is to be odd, we must have
odd and
even or
even and
odd. By symmetry, these cases are identical, so we consider the first one and multiply by two at the end. For
to be odd, we must have both
and
odd, and there are
ways to do so. To count the cases where
is odd, we use PIE. there are
ways for
to be odd and
ways for
to be odd, and there are
ways for both to be odd. Thus, there are
ways for
to be even. Multiplying out, there are
ways to have
odd for a total of
.
~ cxsmi
Video Solutions
Education, The Study of Everything
~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 odd integers, it can quickly be deduced that there are
possibilities for an odd product. Since the product must be either odd or even, and there are
ways to choose factors for the product, there are
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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