2020 AMC 10B Problems/Problem 24
- The following problem is from both the 2020 AMC 10B #24 and 2020 AMC 12B #21, so both problems redirect to this page.
Contents
Problem
How many positive integers satisfy (Recall that is the greatest integer not exceeding .)
Solution 1
We can first consider the equation without a floor function:
Multiplying both sides by 70 and then squaring:
Moving all terms to the left:
Now we can determine the factors:
This means that for and , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:
For , left hand side but so right hand side
For , left hand side and right hand side
For , left hand side and right hand side
For , left hand side but so right hand side
Now we move to
For , left hand side and so right hand side
For , left hand side and so right hand side
For , left hand side and so right hand side
For , left hand side but so right hand side
For , left hand side and right hand side
For , left hand side but so right hand side
Therefore we have 6 total solutions,
Solution 2
This is my first solution here, so please forgive me for any errors.
We are given that
must be an integer, which means that is divisible by . As , this means that , so we can write for .
Therefore,
Also, we can say that and
Squaring the second inequality, we get .
Similarly solving the first inequality gives us or
is larger than and smaller than , so instead, we can say or .
Combining this with , we get are all solutions for that give a valid solution for , meaning that our answer is . -Solution By Qqqwerw
Solution 3
We start with the given equationFrom there, we can start with the general inequality that . This means thatSolving each inequality separately gives us two inequalities:Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is .
~Rekt4
Solution 4
Let be uniquely of the form where . Then, Rearranging and completeing the square gives This gives us Solving the left inequality shows that . Combing this with the right inequality gives that which implies either or . By directly computing the cases for using , it follows that only yield and invalid from . Since each corresponds to one and thus to one (from and the original form), there must be 6 such .
~the_jake314
Solution 5
Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have ; let . The given equation becomes
Since for all real , we can take with to get We can square the inequality to get The left inequality simplifies to , which yields The right inequality simplifies to , which yields
Solving , and , we get , for values .
Solving , and , we get , for values .
Thus, our answer is
~KingRavi
Solution 8
Set in the given equation and solve for to get . Set ; since , we get The left inequality simplifies to , which yields The right inequality simplifies to , which yields Solving , and , we get , for values .
Solving , and , we get , for values .
Thus, our answer is
Video Solutions
Video Solution 1
On The Spot STEM: https://youtu.be/BEJybl9TLMA
Video Solution 2
https://www.youtube.com/watch?v=VWeioXzQxVA&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9 ~ MathEx
Video Solution 3 by the Beauty of Math
https://youtu.be/4RVYoeiyC4w?t=62
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 20 |
Followed by Problem 22 |
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 |
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