Solving a floor value equation. A challenge in algebra.

This video is about a floor value equation that can be solved with little effort.

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[ashishpradhan]

Really I fall in love with this question. After watching the question I am amezed that how can this be solved but now I am very pleased. Thankyou sybermath.

[skwbusaidi]

It is easier if from the benning we figure that x is integer
Floor(x) = x
Floor (x/2) = 7-2x
7-2x <=x/2< 8-2x
14-4x <= x < 16-4x
14<= 5x <= 16
14/5 <= x < 16/5
But since x is integer, x=3

[ShufengTan]

After figuring out x must be integer, we have floor(x/2) = 7-2x. Floor(x/2) equals x/2 or x/2-1 depending on x being even or odd.

[riskassure]

By inspection, x is a positive integer. There are two cases based on the parity of x. If x is even, say of the form 2m for some positive integer then we get 5m=7, a contradiction. Therefore x is odd, say of the form 2m+1. Solving 5m+1=7 gives us that m=1, or x=3.

[dustinbachstein]

I solved this in 20 seconds in a very simple way:
1. The left-hand side is strictly increasing, so there's at most 1 solution.
2. Checked integers and found x=3
Nice morning puzzle ;)

[BorisRaifler]

x + [x] + [x/2] =7 => [x]+{x} +[x] + [x/2] = 7 => {x} = 0. So, x is an integer. It might be = 2k or 2k+1. In both cases [x/2] = k.
So we can try 2 cases: 1) 2k+2k+k =7 => 5k=7, but k must be integer, so 2k+1+2k+1+k =7 => 5k=5=> k=1 and x=2k+1=3

[mathtv]

Since floor(x/2), floor(x) are integers, x is positive integer. Thus floor(x) =x. Then 2x<7. So x = 1, 2, or 3. Then only x=3 satisfies the equation

[bernhardkiniger]

In the original equation, there are 4 terms, 3 of which are known to be integers (floor(x), floor(x/2), 7), therefore the 4th (x) needs to be an integer as well.

[marklevin]

Since floor values are always integers and value on the right is an integer, therefore x must be an integer...if x is even then we get 2.5 x =7 This is impossible. If x is odd then 2x+(x-1)/2=7
This leads to solution x=3....
Just This simlle

[skwbusaidi]

Another method, after figuring that x is integer
Case 1 x is even
Assume x=2n
Then floor( x) = 2n and floor (x/2) = n
2n + 2n + n =7
5n = 7 ==> n=7/5
Not acceptable because n has to be integer

Case 2 x is odd
x= 2m+1
Floor(x) = 2m+1
Floor (×/2) = m

2m+1 + 2m+1 + m = 7
5m +2 = 7
m=1
x =3 👍

[justinnitoi]

Right away by inspection, x must be an integer since floor(x) and floor(x/2) are both integers. Seeing that x>0, and since floor(x) = x, you can be sure that 0<x<7/2. Then checking x from 1 to 3, you obtain x = 3 as the only solution.

[zeydbahadir]

Since x=7-[x]-[x/2] and RHS is an integer, then LHS i.e. x is an integer. If x≤2, then x+[x]+[x/2]≤5, which is a contradiction. If 4≤x, then x+[x]+[x/2]>x+[x]≥8, which is a contradiction. So x=3.

[damiennortier]

Notice that the floor value of x and the floor value of x/2 are integer. So x needs to be an integer. Since x is a integer, |_x_| = x. When we divide an integer by 2, we only get 2 results :
- we can got an integer and then, |_x/2_| = x/2
- we can got an integer + 0.5 and then, |_x/2_| = x/2 - 0, 5.
Now, we have two equation :
5x/2 = 7 or 5x/2 - 0, 5 = 7.
5x = 14 or 5x = 15
X = 2, 8 or x = 3
But x needs to be an integer so the only solution is x = 3
That's another way 🙂

[angelmendez-rivera]

x + floor(x) + floor(x/2) = 7 ==> x = 7 – floor(x) – floor(x/2) ==> floor(x) = x ==> 2·x + floor(x/2) = 7 – 2·x <==> 7 – 2·x =< x/2 < 8 – 2·x ==> 14 – 4·x =< x < 16 – 4·x ==> 14 =< 5·x < 16 ==> 14/5 =< x =< 16/5 ==> x = 3 immediately. No further case analysis is necessary.

[SimchaWaldman]

07:06 Wow!
P.S. I Never thought I would fall in love with floor/ceiling-valued algebraic problems.

[WolfgangKais]

The floor function has integer values and 7 is an integer, so x must as well be an integer and therefore floor(x)=x…

[redwanekhyaoui]

X equal 2 or 3 because x=7-[x]-[x/2] which means x is integer, because 7, [x], and [x/2] are integers