Weekly Math Question

@ReChesster said in #20:

Yes, but the question was to express n in terms of x and y.
Okay. I will think about that.

@ReChesster said in #10:

Let’s take our equation

y = x^n + x^(n-1)

Factoring out x^(n-1) gives us

y = x^(n-1) * (x+1)

Dividing both sides by (x+1):

y/(x+1) = x^(n-1)

Next, we take the logarithm to the base x and rewrite it with the natural logarithm (log_x(z) means the logarithms to the base x of z):

n-1 = log_x(y/(x+1))

n-1 = ln(y/(x+1))/ln(x)

Finally, rewrite ln(y/(x+1)) by using basic “log”- rules, add one to both sides and we get

n = (ln(y) – ln(x+1))/ln(x)

As we see, because the denominator shouldn’t get zero, we have to include that ln(x) mustn’t be equal to 0, which means x can’t be 1 (test it yourself). So, the final result is:

n = (ln(y) – ln(x+1))/ln(x); (x ≠ 1)

However, anyone who got

n = log_x(y/(x+1)) +1

also got the correct solution. The above solution just shows how I would have solved this problem.

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For our pro question, I’ll have to introduce a new function that many of you probably haven’t heard of: The Lambert W Function. You can go to Wikipedia for details, but it pretty much does this: If you have f(x)* e^f(x) and apply the Lambert W Function on this, you will get f(x):

W(f(x)*e^f(x)) = f(x)

So how does that help us with our problem? First of all, let’s set f(x) = y and g(x) = g for simplicity. In the end, we can substitute f(x) and g(x) back into our solution. So now we have

y = (a^g)*(ln(a)+1)

As we all probably know, e^ln(x) = x, hence x = e^(ln(x)). Applying this fact to a^g gives the equivalent term e^(ln(a^g)). Using a basic property of the logarithm function, we can rewrite this as e^(g*ln(a)). Putting that in our equation, we get

y = (e^(gln(a)))(ln(a)+1)

By multiplying both sides with e^g, we get

y*(e^g) = (e^(gln(a)))(e^g)*(ln(a)+1)

Applying another logarithm property, we get:

y*(e^g) = (e^(gln(a)+g))(ln(a)+1)

y*(e^g) = (e^(g(ln(a)+1)))*(ln(a)+1)

Multiplying by g, we get:

gy*(e^g) = g(ln(a)+1) *e^(g(ln(a)+1))

Now we can apply the Lambert W Function on both sides (yes, this is legal), and we get

W(gy*(e^g)) = g(ln(a)+1)

Now we can simply solve for a:

W(gy*(e^g)) = g(ln(a)+1)

W(gy*(e^g))/g = ln(a)+1

(W(gy*(e^g))/g)-1 = ln(a)

a = e^((W(gy*(e^g))/g)-1)

Substituting f(x) and g(x) back, we get our final answer:

a = e^((W(f(x)g(x)(e^g(x)))/g(x))-1); (g(x)≠0)

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Ok, here's the question for the next week. The answer will again be revealed on Saturday next week.

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.

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Pro Question: For anyone who thinks he's better, try this.

Prove that if x + y + z is divisible by a number, (x^3) + (y^3) + (z^3) is also divisible by that same number.

Oww my brain it hurts!

Another week has passed, and this time, some posted the correct solutions. Congrats to @InkyDarkBird and @chess1048576!
Now to our solution:

Any number must be either a multiple of 3, one below a multiple of 3 or one above a multiple of 3 (if a number is 2 above a multiple of 3, it is automatically also 1 below the next greater multiple of 3). I will call this the “structure” of the number. The structure can either be 3n, 3n - 1 or 3n + 1. To make the some of three numbers divisible by three, we must combine the structures in a way that the number actually gets divisible by three.
What are the structures of the cubes of our structures?

(3n + 1)^3 = (3^3)(n^3) + 3((3n)^2)(1) + 3(3n)(1^2) + (1^3)
= 3((3^2)(n^3) + ((3n)^2)(1) + (3n)(1^2)) + 1
= 3f(n) + 1

where f(n) = (3^2)*(n^3) + ((3n)^2)(1) + (3n)(1^2)

(3n - 1)^3 = (3^3)(n^3) - 3((3n)^2)(1) + 3(3n)(1^2) - (1^3)
= 3((3^2)(n^3) - ((3n)^2)(1) + (3n)(1^2)) - 1
= 3g(n) - 1
where g(n) = (3^2)*(n^3) - ((3n)^2)(1) + (3n)(1^2)

(3n)^3 = (3^3)(n^3) = 3(3^2)(n^3) = 3h(n)
where h(n) = (3^2)(n^3)

As we can see, the cube of a number has the same structure as the original number, which means if we are able to find a trio of structures that is divisible by three, the cubes of those numbers, who have the same structures, also happen to be divisible by three, because they have the same structure that is divisible by three.

We didn’t need to know what combinations of the structures are possible, but for anyone interested, I’ll list them here (n, m and k are all natural numbers):
(3n + 1) + (3m + 1) + (3k + 1)
(3n - 1) + (3m - 1) + (3k - 1)
(3n + 1) + (3m - 1) + (3k)
(3n) + (3m) + (3k)
These structures are all divisible by three, and since the cubes of the numbers preserve their structures, the sum of their cubes are also divisible by three.

The pro question was actually a trick, as you can easily find a combination of numbers for which the statement is not true. Sorry everyone, but I wasn’t able to think of anything better ;)

For this week’s question, I thought of doing some algebra. This time there’s no pro question, as the normal one is tough itself:
Given three real numbers x, y and z. We know that the room diagonal of a cuboid with the measures xy, xz and yz is equal to 7. Furthermore, we know that the numbers satisfy the following equations:

x + y + z = -4
(x^2) + (y^2) + (z^2) = 14

Find all possible solutions for x, y and z.
Have fun!

I won't write all the details, but the main idea is we get (xy)^2+(xz)^2+(yz)^2=49, and some basic algebraic manipulation gives xyz=6,xy+xz+yz=1,x+y+z=-4. By Vieta's formulas, x,y,z are the solutions to the equation X^3+4X+X-6 = 0, which can be factored to
(X-1)(X+2)(X+3), so the solutions are 1,-2 and -3.

By the way, the 'structure' idea in your previous post is effectively modular arithmetic - https://brilliant.org/wiki/modular-arithmetic/#:~:text=Modular%20arithmetic%20is%20a%20system,modulus)%20to%20leave%20a%20remainder.

I'm leaving this site. There are too many algebraic equation questions here.
And when a poor man wants to escape his Algebra class...
He must do so at any cost.

@chess1048576 said in #24:

By the way, the 'structure' idea in your previous post is effectively modular arithmetic - brilliant.org/wiki/modular-arithmetic/#:~:text=Modular%20arithmetic%20is%20a%20system,modulus)%20to%20leave%20a%20remainder.

I know, but I didn't want to confuse anyone because I used the "="-sign as "is equal to" and "is congruent" simultaneously, but thanks for pointing that out :)

@LordSupremeChess said in #25:

I'm leaving this site. There are too many algebraic equation questions here.
And when a poor man wants to escape his Algebra class...
He must do so at any cost.

Want some calculus instead? ;)

@ReChesster said in #27:

Want some calculus instead? ;)
You little son of a -CENSORED- ...

@LordSupremeChess said in #28:

You little son of a -CENSORED- ...
(He)^3

This is great and all, but if we could make it mandatory that every one learn to balance their own personal finances (I.e. checkbook, remember those?) we’d be better off. Once we do that, we’ll work on government officials. Calculate that!