Difference between revisions of "2020 AMC 10A Problems/Problem 18"
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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). ~ Anonymous and Arctic_Bunny | + | So, for one product to be odd the other to be even: <math>2 \cdot 4 \cdot 12=\boxed{(C)96}</math>(order matters). |
| + | ~ Anonymous and Arctic_Bunny | ||
===Solution 4=== | ===Solution 4=== | ||
Revision as of 20:21, 2 February 2020
Contents
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.)
Solution
Solution 1
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
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 
Solution 3
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
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
Video Solution
~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 
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
