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
First notice that the graphs of and intersect at 2 points. Then, notice that must be an integer, since it is equal to the floor of . This means that n is congruent to .
For the first intersection, testing the first few values of (adding to each time and noticing the left side increases by each time) yields and , so respectively. Estimating from the graph can narrow down the other cases, being , , , , yielding respectively. This results in a total of 6 cases, for an answer of .
~DrJoyo (edited by eagleye and vaporwave)
Solution 2 (Graphing)
One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of . If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that values of intersection lay closer to the left side of the stair, and values lay closer to the right side of the stair.
With meticulous graphing, you can realize that the answer is .
A in-depth graph with intersection points is linked below. https://www.desmos.com/calculator/e5wk9adbuk
Solution 3
- Not a reliable or in-depth solution (for the guess and check students)
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 use wishful thinking to 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 4
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 5
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 .
~Rekt4
Solution 6
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 7 (Sequences - Rigorous)
Right away, we realize that since the floor function returns only integers, is also an integer.
, so to make the numerator divisible by 70.
Let be the sequence of all positive integers such that , and let be the element of , where
As is an arithmetic sequence with initial term 50 and common difference 70, the explicit formula of the sequence is .
calculates for the element in the sequence, so we can plug in this formula for ;
Now we can simplify the left-hand-side:
We know that for any x, . Therefore:
We can square the inequality to get
We can split this three-part inequality into two inequalities:
Now we can solve for each inequality and bound ;
(by considering signs of and ).
The next inequality cannot be factored so we will have to use the quadratic equation to find j: To find j, we treat the inequality as an equation: this becomes
Now we can turn it back into an inequality. These roots are a bit harder to work with than the first inequality, but we recognize that the factored form becomes , and we can use signs as before to determine that the solutions for this inequality are
and .
Because can only have positive integer values, we can approximate as between 7 and 8, and we can approximate as between 32 and 33.
We can then write the stricter inequalities and .
Overall, , , and , so the only possible solutions for are 6, 7, 33, 34, 35, and 36. Because each valid solution for is a valid solution for , we have 6 solutions for that satisfy the given equation, or .
~KingRavi
Solution 8 (General Equation)
General solution to this type of equation :
1. solve for to get 2. apply , solve to get the domain of 3. get from the domain of because is integer, then get from by Note: function maps to its floor. By solving , we get function , mapping 's floor to
Let ,
Let
If , , , has values.
If , , , has values.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.