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
Solution 9
If is a perfect square, we can write for a positive integer , so The given equation turns into
\begin{align*} \frac{k^2 + 1000}{70} &= k \\ k^2 - 70k + 1000 &= 0 \\ (k-20)(k-50) &= 0, \end{align*}
so or , so
If is not square, then we can say that, for a positive integer , we have \begin{align*} k^2 < &n < (k+1)^2 \\ k^2 + 1000 < &n + 1000 = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\ k^2 + 1000 < &70k < (k+1)^2 + 1000. \end{align*}
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities and . To solve the first one, we have
\begin{align*} k^2 - 70k + 1000 &< 0 \\ (k-20)(k-50) &< 0\\ \end{align*} because the portion of the parabola between its two roots will be negative.
The second inequality yields
\begin{align*} 70k &< k^2 + 2k + 1 + 1000 \\ 0 &< k^2 -68k + 1001. \end{align*} This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are and (they are roughly equal, but this is to ensure that we do not miss any solutions).
Notation wise, we need all integers such that
or
For the first one, since our uppoer bound is a little less than , the that works is . For the second, our lower bound is a little more than , so the that work are and .
total solutions for , since each value of corresponds to exactly one value of .
-Benedict T (countmath1)
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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