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 3
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 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 that
Solving 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
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 (General Equation)
To solve an equation of the type , we proceed as follows.
.
.
.
.
.
.
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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