2020 AMC 10B Problems/Problem 9
- The following problem is from both the 2020 AMC 10B #9 and 2020 AMC 12B #8, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solutions
- 2.1 Solution 1
- 2.2 Solution 2
- 2.3 Solution 3: Solve for x first
- 2.4 Solution 4: Solve for y first
- 2.5 Solution 5: Similar to solution 4
- 2.6 Solution 6: (Casework)
- 2.7 Solution 7: (Casework)
- 2.8 Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)
- 2.9 Video Solution 1 by TheBeautyOfMath
- 2.10 Video Solution by WhyMath
- 2.11 Video Solution by Sohil Rathi
- 3 See Also
Problem
How many ordered pairs of integers satisfy the equation
Solutions
Solution 1
Rearranging the terms and and completing the square for yields the result . Then, notice that can only be , and because any value of that is greater than 1 will cause the term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a total of ordered pairs.
Solution 2
Bringing all of the terms to the LHS, we see a quadratic equation in terms of . Applying the quadratic formula, we get In order for to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, must be nonnegative. Therefore, Here, we see that we must split the inequality into a compound, resulting in .
The only integers that satisfy this are . Plugging these values back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .
Thus, the answer is .
~Tiblis
Solution 3: Solve for x first
Set it up as a quadratic in terms of y: Then the discriminant is This will clearly only yield real solutions when , because the discriminant must be positive. Then . Checking each one: and are the same when raised to the 2020th power: This has only has solutions , so are solutions. Next, if : Which has 2 solutions, so and .
These are the only 4 solutions, so our answer is .
~edits by BakedPotato66
Solution 4: Solve for y first
Move the term to the other side to get .
Because for all , then .
If or , the right side is and therefore .
When , the right side become , therefore .
Our solutions are , , , . These are the only solutions, so the answer is
- wwt7535
~ edits by BakedPotato66
Solution 5: Similar to solution 4
Since and are perfect squares, they are both nonnegative. That means plus a nonnegative number equals , which means The only possible integer values for are .
For , can only be .
For , so .
For , can only be as well.
This gives us the solutions , , , and . These are the only solutions, so there is a total of ordered pairs.
- kc1374
Solution 6: (Casework)
We see that has to be , , or , as any other integer would make this value too large. We also know that because is even, both , and for will yield the same value of .
Case 1: . This gives us that . Dividing both sides by gives us . Additionally, we know intuitively that is also a case, which gives us 2 possibilities for this case.
Case 2: or . This gives us that . Bringing the to the other side, we have a simple quadratic. . Factor to get so . Because this works for as and , there are 2 possibilities for this case.
Adding the cases gets us our final answer of ordered pairs. ~iluvme
Solution 7: (Casework)
We can move the term to the other side to obtain , which can be easily factored into . If we raise each side to the power 1/2020, we get . Since we're dealing with integer values, the inside of the square root and thus must be greater than . For this to happen, must be between and inclusive. Testing each integer value, we get the cases:
: When , then is also 0, and since the square root of is , must also be . So our first pair of solutions is .
: When , then is also 1. The possible values for are and , so we get 2 more pairs of solutions: and .
: When , then is , and since the square root of is , must also be - so we get one more pair of solutions,
If we count all these pairs(), we have 4 pairs of solutions, so our answer is ordered pairs.
- youtube.com/indianmathguy
Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)
~Education, the Study of Everything
Video Solution 1 by TheBeautyOfMath
Video Solution by WhyMath
~savannahsolver
Video Solution by Sohil Rathi
https://youtu.be/zfChnbMGLVQ?t=4251
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.