Weekly Math Question

I don't think the pro question is true, e.g 1+2+4=7,1^3+2^3+4^3=73 which isn't divisible by 7.

@ReChesster said in #10:

Given that x + y + z is divisible by 3 (this means you can divide x + y + z by 3 without getting a remainder). Prove that (x^3) + (y^3) + (z^3) is also divisible by 3.
(x + y + z) = (x^3) + (y^3) + (z^3) = 0 (mod 3)

Case 1:
x = a + 1
y = b + 1
z = c + 1
Where a, b, and c are multiples of 3.
x^3 = a^3 + 3a^2 + 3a + 1
y^3 = b^3 + 3b^2 + 3b + 1
z^3 = c^3 + 3c^2 + 3b + 1
Adding these together results in a sum divisible by 3.

Case 2:
x = a + 1
y = b + 2
z = c
Where a, b, and c are multiples of 3.
x^3 = a^3 + 3a^2 + 3a + 1
y^3 = b^3 + 6b^2 + 12b + 8
z^3 = c^3
Adding these together results in a sum divisible by 3.

Case 3:
x = a + 2
y = b + 2
z = c + 2
Where a, b, and c are multiples of 3.
x^3 = a^3 + 6a^2 + 12a + 8
y^3 = b^3 + 6b^2 + 12b + 8
z^3 = c^3 + 6c^2 + 12c + 8
Adding these together results in a sum divisible by 3.

Case 4:
x = a
y = b
z = c
Where a, b, and c are multiples of 3.
x^3 = a^3
y^3 = b^3
z^3 = c^3
Adding these together results in a sum divisible by 3.

This should prove that if (x + y + z) = 0 (mod 3), then (x^3) + (y^3) + (z^3) = 0 (mod 3).

Nash's Equilibrium Theory - let's discuss ...

Kramnic points out the obvious when he says 'if cheating continues to go uncontrolled, then all competitors will be forced to cheat as a mere matter of survival ...

Comments, observations, objections ?

I think simpler is noting that x^3=x mod 3 (as x(x+1)(x-1)=0 mod 3).

@chess1048576 said in #11:

I don't think the pro question is true, e.g 1+2+4=7,1^3+2^3+4^3=73 which isn't divisible by 7.

So my trick didn't work, huh ;)

It was a trick question bc I couldn't think of anything good xD

@InkyDarkBird said in #12:

(x + y + z) = (x^3) + (y^3) + (z^3) = 0 (mod 3)

Case 1:
x = a + 1
y = b + 1
z = c + 1
Where a, b, and c are multiples of 3.
x^3 = a^3 + 3a^2 + 3a + 1
y^3 = b^3 + 3b^2 + 3b + 1
z^3 = c^3 + 3c^2 + 3b + 1
Adding these together results in a sum divisible by 3.

Case 2:
x = a + 1
y = b + 2
z = c
Where a, b, and c are multiples of 3.
x^3 = a^3 + 3a^2 + 3a + 1
y^3 = b^3 + 6b^2 + 12b + 8
z^3 = c^3
Adding these together results in a sum divisible by 3.

Case 3:
x = a + 2
y = b + 2
z = c + 2
Where a, b, and c are multiples of 3.
x^3 = a^3 + 6a^2 + 12a + 8
y^3 = b^3 + 6b^2 + 12b + 8
z^3 = c^3 + 6c^2 + 12c + 8
Adding these together results in a sum divisible by 3.

Case 4:
x = a
y = b
z = c
Where a, b, and c are multiples of 3.
x^3 = a^3
y^3 = b^3
z^3 = c^3
Adding these together results in a sum divisible by 3.

This should prove that if (x + y + z) = 0 (mod 3), then (x^3) + (y^3) + (z^3) = 0 (mod 3).

Very clean <3

@ReChesster The equation you’ve given is already in the form of y. So, y is equal to x^n + x^(n-1). This equation represents a polynomial where x is the variable, n is the degree of the polynomial, and the terms are x^n and x^(n-1). The values of y will depend on the values you substitute for x and n.

Am I right?

@ReChesster said in #1:

Given the equation y = (x^n) + x^(n-1)
Solve this expression for n.
x and n can be literally anything, and y is just given by (x^n) + x^(n-1)
Given the equation f(x) = (a^g(x)) * (ln(a) + 1)
Solve this expression for a.
Same thing : a, x, and g(x) can be anything, and f(x) is then given by the right hand side of the equation.

Unless of course your question is "find n such that y=0 and find a such that f(x) =0".

In that case, we can write y=(x+1)x^(n-1). Then y=0 if and only if x+1=0 or x^(n-1)=0.
So either x=-1 and y=0 regardless of n, or x=0 and y=0 provided n>1.

Similarly f(x)=0 if and only if either a^g(x) =0 or ln(a) +1=0.
So either a=0 and g(x) >0, or a=1/e.

@Bobby-the-kid said in #18:

x and n can be literally anything, and y is just given by (x^n) + x^(n-1)

Same thing : a, x, and g(x) can be anything, and f(x) is then given by the right hand side of the equation.

---

Unless of course your question is "find n such that y=0 and find a such that f(x) =0".

In that case, we can write y=(x+1)x^(n-1). Then y=0 if and only if x+1=0 or x^(n-1)=0.
So either x=-1 and y=0 regardless of n, or x=0 and y=0 provided n>1.

Similarly f(x)=0 if and only if either a^g(x) =0 or ln(a) +1=0.
So either a=0 and g(x) >0, or a=1/e.

Actually, you were supposed to express n in terms of x and y and express a in terms of f(x) and g(x). Maybe I should have said that more clearly...

@PeynirSever2134 said in #17:

@ReChesster The equation you’ve given is already in the form of y. So, y is equal to x^n + x^(n-1). This equation represents a polynomial where x is the variable, n is the degree of the polynomial, and the terms are x^n and x^(n-1). The values of y will depend on the values you substitute for x and n.

Am I right?

Yes, but the question was to express n in terms of x and y.