Challenge: Absolute Absolute Absolute Values

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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

[sebmata]

If you write out a piecewise interpretation of the absolute value function in the problem, you get that y=x intersects when x=c/2 and when x=c/4 so clearly if c is a multiple of 4 then both equations will yield an integer. There are 504 such multiples of 4 between 0 and 2018.

[fmakofmako]

The answer is 504.

When you process the nested abs function you get y = x-c, when x >= c; y=c-x, when c/2 <= x <c; y = c-3x, when x <= c/3; and y=3x-c, when c/3<x<c/2.

Then notice the cases where y=x you"ll see that x=c/4 is the tightest condition, so from c=4 to 2016 (multiple of 4) results in 504 total. possible c

[nathanielsharabi]

Answer is 504

You can split the first graph into simple lines stuck together like this with simple testing to see it works:
y= x-c when x>=c
-x+c when c/2=<x<c
3x-c when c/3=<x<c/2
-3x+c when x<c/3

Both graphs intersect with those lines at. (c/4, c/4) and (c/2, c/2) therfore if we want integer coordinates we look for numbers divisible by 4 in the bound 0 to 2018 which gives 504.

[gabrieln.]

Answer is 504

If we draw the graph (I did it by hand, there is no good way of showing it in this comment) we get a kind of sawtooth with peaks at x=0 and x=c/2 with heights c and c/2 and falls at x=c/3 and x=c at height 0. The line y=x hits the peak (c/2;c/2) and another point on line y=c-3x. This other point has coordinates (c/4;c;4). So, the only condition is c a multiple of 4. There are 504 of those.

[TheBetterVersion]

504. While trying points, we get that the first function is
c-3x {x < c/3}
3x-c {c/3≤x<c/2}
c-x {c/2≤x<c}
x-c {c≤x}
Using these functions, we get that the intersects are at (c/4, c/4) and (c/2, c/2)
There are floor(2018/4) positive numbers that are divisible by 4, so the amount is 504

[juanpisocampo]

The answer is 504.
we match both equations for find the values of x for which the two graphs intersect. Then we've From this, is abs(abs(x-c)-x) -x=x, then, is (1) abs(x-c) -x = 2x or (2) abs(x-c) -x = -2x. From (1) is x=c/4 or x=-c/2, and from (2) is x=c/2 or c=0. So, the other posibility of (*) is abs(abs(x-c)-x) -x=-x, and we've got c=0 or x=2c. Therefore, as conclusion, we obtain x=c/4 or x=-c/2 or x=c/2 or x=2c, so if 4 divides c, then 2 divides c. Therefore, if we want that x be an integer, then c has a multiple of four, positive and less than 2018, that is 504 posibles values of c.

[minhcongnguyen]

The answer is 504.

We need to find the the condition for c such that the equation |||x-c|-x|-x|=x has only integer solutions (x and c should be positive, and c <2018). Solving this give us x= c/2 and x=c/4 (note that x and c are strictly larger than 0). Therefore, c is {4, 8, 12, ..., 2016}, which has 504 values.

[lucaslorenzo]

The answer is 504.
The function will equal x at some constant times C, lets call that constant k.
|||x-c|-x|-x| can be re-written as |||kC-C|-kC|-kC|.
Further evaluating, the function can be simplified to ||kC-C|-2kC|.
Since C > kC, the function can be re-written as |C-3kC|.
Since C is also > 3kC, the function can be re-written as C-3kC.
Solving for C-3kC=kC, we find k= 1/4.
Since the function can only equal x on values of C divisible by 4, there are 504 values of C that satisfy this since 504 * 4 = 2016 < 2018.

[danielbranscombe]

Answer: 336

notation: as I solve the equation, on the right hand side inside
the curly brackets I am going to keep track of all the possible values
each time you remove an absolute value you branch to a positve/negative value
of each previous possible value

|||x-c|-x|-x|={x}
remove an abs and branch
||x-c|-x|-x={-x, x}
add x to each possible value
||x-c|-x|={0, x}
branch again
|x-c|-x={0, -x, x}
add x
|x-c|={x, 0, 2x}
branch one last time
x-c={-x, x, 0, -2x, 2x}
negate both sides
c-x={x, -x, 0, 2x, -2x}
add x
c={2x, 0, x, 3x, -x}
so solving for x we get the following solutions
x=c/2 or x=c or x=c/3 or x=-c

so for x to always be an integer we need C to be an integer divisble by
both 2 and 3, or in otherwords divisble by 6. the floor of 2018/6 is 336.
Thus there are 336 solutions

[evyatarbaranga]

The answer is 504
If you write the posibilities, each abs you right both uf its positive, negative, 0. You see thet c = 4x => y=x=c/4, so if x, y are integers c/4 is an integer. There are 2018/4 = 504.5 => 504 posibilitis, not including 0. If c = 0, a solotion is just y= x or (0, 0). And y=x isnt only integer points. QED

[cuberking]

y=abs(abs(abs(x-c)-x)-x) and y= x intersect whenever
abs(abs(x-c)-x)-x=-x or x, since we can assume x>0 as the triple abs function will always be positive and thus will not intersect y=x unless x>0
It follows that abs(x-c) = x or abs(x-c)-x = +or- 2x
The first case yields c=0 (wrong) or x = c/2
The second case after a similar deconstruction process yields x=-c/2 (wrong, c must be greater than zero), x=c/4, or abs(x-c) =-x (wrong as + cannot equal -)

Therefore, the curves intersect at (c/2, c/2) and (c/4, c/4), which are lattice points when 4|c.

2018/4=504.5 so there are 504 such values of c.

[margintoosmall]

504, this can be solved easily by just trial and error. First, we can see that x=c/2 is always a solution because |c/2 - c| = c/2 --> |c/2 - c/2| = 0 --> |0-c/2| = c/2. c/4 is also a solution because |c/4 - c| = 3c/4 --> |3c/4 - c/4| = c/2 --> |c/2 - c/4| = c/4. This can be shown by working out a few cases, and/or by graphing.

From x=c on, it's a straight line with a slope of 1 with an x-int of 8. Then it's a slope of -1, to x=c/2. Then it's a slope of 3 to x=c/3. Then it's a slope of -3 to negative infinity.

Then just do 2018/4 to find the answer.

[si]

The Answer is 504
Drawing the Graph and breaking the moduli, we see that intersection point turns out to be (c/4, c/4) which is an integral co-ordinate for 504 values of c

[cobalt]

There are 504 possible values for c.

Let us start by solving the equation x = |||x – c| – x| – x|, on a case by case basis:
We will start by breaking up the first absolute value into -x = ||x – c| – x| – x and x = ||x – c| – x| – x. Looking to the first equation, we see that 0 = |x – c| – x, or x = |x – c|. This equation has two solutions: c = 0, which we can't have, or x = c/2. So far, we require c to be a multiple of 2 for an integer solution.
Next, we look to the (simplified) second equation: 2x = ||x – c| – x|. We can again break this equation up into 2x = |x – c| – x and -2x = |x – c| – x. Looking to the first, we have 3x = |x – c|, meaning either x = -c/2 or x = c/4. The first one would imply either a nonpositive c or a negative absolute value, which we can't have, and the second implies that c is a multiple of 4 for integer solutions. As all multiples of 4 are themselves multiples of 2, it is sufficient that c be a multiple of 4.
Now we look to the (simplified) second equation from above: -x = |x – c|. To be a valid equation, this requires that x be nonpositive, and would result in either c = 0 or x = c/2. Both, however, yield nonpositive c, so this equation has no solutions that we are interested in.
So in summary, to intersect at only integer solutions, we require that c be a multiple of 4. We are only interested in multiples of 4 less than 2018, and there are a total of floor(2018/4) = 504 of them. Thus, there are 504 valid choices for c.

[Quwertyn]

|||x-c|-x|-x|=x
Solving when c>=3x:
C=4x (condition checks out assuming X>0, but for all our solutions it has to be anyway since otherwise one part of the equation is negative)
X<=c<2x:
c=X (checks out)
2x<=c<3x:
2x=c (checks out)
c<X:
c=0
Which means that there are solutions when c<X if and only if c=0 which can't be true since c is positive. From the rest of our results we get that X is a solution if and only if X=c/4, X=c/2 or X=c. Therefore, c/4 has to be an integer. 2017/4=504+1/4 so there are 504 values of c<2018 for which these two equations intersect at integer values(if the X coordinate of the intersection is an integer then y is obviously as well since y=X)

[timurpryadilin]

Theese functions intersect only in points x=c/2 and x=c/4. If 4 devides x, then 2 also devides x, then there are two points. But if x is even, but indevisible (is there such word?) by 4, we get exacly one point. There are only 2016/4 + 1 = 505 of such numbers including 2018

[nicolacaravaggi]

I think it's 504.
Setting up various inequalities reveals that there is just a solution for the equation |||x-c|-x|-x|=x, that is x=c/4.
Therefore the point of intersection is (c/4, c/4) and it has integer coordinates iff 4 | c.
2018/4=504, ... so the answer is 504

[hunghinsun]

The answer is 504.
By solving the inequality with x>0, we have x =c/2 or x=c/4. Since we must have all the solution to be integers, c must be multiples of 4. Hence we have the answer 504, since c is less than 2018.

[moskthinks]

Answer: 504

First: Observation (Take real values and make an observation)

Take c=1
Intersections: 0.25 and 0.5

Take c=2
Intersections: 0.5 and 1

Take c=3
Intersections: 0.75 and 1.5

Take c=4
Intersections: 1 and 2

Observation we want to prove:
Take c=k
Intersections: k/4 and k/2

Second: Proof.

When x=c/4:
When x=c/2:

We showed that when x equals those values, then it intersects with our y=x graph.

How do we prove that there is no other solutions?

When x>c: y=x-c



When c/2<x<c: y=c-x



When c/3<x<c/2: y=3x-c



When x<c/3: y=c-3x



Third: Value

2018/4=504.5

Round down by not including 2017 and 2018.

Answer is 504.

Also, the current answers suggest the answer is 504 but only by observation, which is not really wanted in mathematics.
(Note: The absolute value identities are quiet trivial in their own sense really)

[nikstopics]

the answer is 504

We first need to solve the equation abs(abs(abs(x-c)-x)-x) = x. So I did it like this abs(abs(x-c)-x)-x = x or -x: abs(abs(x-x
c)-x)-x = -x abs(abs(x-c)-x) = 0 abs(x-c)-x = 0 abs(x-c) = x x-c = x or -x: -c = 0 cannot be a solution but x-c = -x c-x = x x = c/2 is a valid solution. Doing a similar deconstruction of the absolute values in the case where abs(abs(x-c)-x)-x = x we get that abs(x-c) = 3x or -x, the case where the abs equals -x cannot happen, x-c = 3x or -3x: -4x = -c x = c/4 is a valid solution but. 2x = -c x= -c/2 is not valid because it is negative.

So we get that the valid solutions are of the form c/4 and c/2. Since all multiples of 4 are also multiples of 2 and the intersections of the graphs are of the form (x, x) we simply need to find how many multiples of 4 are less than 2018. Now we just divide 2016 by 4 to get 504 which is the answer.