Solving a Floor and Ceiling Equation

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FloorAndCeilingEquations

[MathElite]

I love equations with floor now, thanks

[SyberMath]

If you like floor equations, here is a cool video from @Math Elite
Check it out!

[theultimatereductionist]

Inspired by your question, I gave myself this question:
find all solutions to Ceil(Floor(x)*y)=88 and x*Floor(Ceil(y)*x)=288.
After 5 pages of checking & elimination, I found exactly one solution:
(x, y) in {288/159} x (87, 88].
There were several near-solutions, however.

I then gave myself this combined differential equation with a floor function:
dx/dt = 3*Floor(x)+1. Given any real number, C, the constant of integration,
I found for any t<0, there exists a positive integer, n, such that 1-C - sum of (1/(3*i-1)) from i=1 to n <= 1-C-sum of (1/(3*i-1)) from i=1 to n-1.
x=(3*n-1)(2/3 - t -C - sum of 1/(3*i-1) from i=1 to n-1) - 1/3 on this domain of t
while for t>0 there exists a positive integer, n, such that 1 - C + sum of 1/(3*i-1) from i=1 to n-1 <= t < 1 - C + sum of 1/(3*i-1) from i=1 to n
x=(3*n+1)*(t+C - sum of 1/(3*i+1) from i=1 to n) - 2*n on this domain of t

[Qermaq]

Watching you "ceiling" the deal really floored me! :D

[hybmnzz]

I love the way you solve floor/ceiling equations with inequalities. Before your videos I would either just eliminate cases (like in this problem x>2 has no sols) or write x = floor(x) + r, 0<r<1 after checking integer solutions. Your method is much more fun.

[DanielWalvin]

3:10 From the inequalities you derived that should be n ≤ (9 - 5n)/3, with a non-strict inequality, and therefore the inequality resulting from that is n ≤ 9/8... not that it ended up mattering, since 9/8 isn't an integer, but if the result had been an integer then it would have mattered.

[raystinger]

I did it differently. I usually solve floor problems using the following identity:
x = floor(x) + a, where 0≤a<1
Applying that to our problem, we have:
ceil(3*(floor(x) + a) + 5*floor(x)) = 9
=> ceil(8*floor(x) + 3a) = 9
Here we can see that if floor(x)≥2, our final result will be at least 16. Therefore, floor(x)=1. So we get:
ceil(8 + 3a) = 9
Since 8 is already really close to 9, so what we want:
3a≤1
=> a≤ 1/3
Also, a≠0, otherwise we get ceil(8)=8. Therefore:
0<a≤1/3
From that, we get:
x = floor(x) + a
=> 1 < x ≤ 1+1/3
=> 1 < x ≤ 4/3

[manojsurya]

that inequality is literally used in all the floor ceiling video😁

[sahilsinghbhandari]

I like this question and solution .Floor value is also known as greatest integer function

[deepjyoti]

Thnq after so much of floors i became good at that types of problems

[armacham]

3:08 The "less than" sign should be "less than or equal to."

n <= x <= (9-5n)/3

both of those signs are "less than or equal to" which means that the LHS can equal the RHS and it would still be true. So you get:

n <= (9-5n)/3

[yogamulyadi]

3x+5x-5(0, ...)=8, ... -> x>1
3x=8, ...-5=3, ... -> x<4/3

[math]

Trying to understand these floor problems. Understand your method, it was very helpful.

[attila]

3:09 That should be smaller than or equal to

[kimsanov]

Why at 3:07 inequality strict and not <= ?

[supergaming]

where can I find more math material like this?

[haricharanbalasundaram]

Really great video, as always.

Why don't you try doing a collab with blackpenredpen (I've seen him commenting on your videos) or Dr. Peyam (he does similar stuff, a little bit more advanced than what you do, though)? I'm sure it's worth a try, and it'll help your channel...

[brightjovanny]

Have you ever drawn the graph of a floor and ceiling equation? It looks weird.