# Thread: Most complicated equation you can come up with that equals this number?

1. ## Most complicated equation you can come up with that equals this number?

I'm trying to come up with a complicated equation that equals 53.

5x - 2x + x * 7x + 23 * x / 78652

x = 53

That one is pretty lackluster though.

I'm looking forward to seeing what the math wizards here can come up with.

2. Originally Posted by Jerick
I'm trying to come up with a complicated equation that equals 53.

5x - 2x + x * 7x + 23 * x / 78652

x = 53

That one is pretty lackluster though.

I'm looking forward to seeing what the math wizards here can come up with.
I mean one can make this as arbitrarily complicated as you want.

For example, even the coefficents in your above equation can be complicated amazingly.

$\displaystyle 2=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\s um_{n=0}^{\infty}e^{n\ln\left(\frac{1}{2}\right)}= \sum_{n=0}^{\infty}\sum_{\ell=0}^{\infty}\frac{\le ft(n\ln\left(\frac{1}{2}\right)\right)^{\ell}}{\el l!}\cdots$ ad infinitum

3. Originally Posted by Jerick
I'm trying to come up with a complicated equation that equals 53.

5x - 2x + x * 7x + 23 * x / 78652

x = 53

That one is pretty lackluster though.

I'm looking forward to seeing what the math wizards here can come up with.
Unless you clearly state exactly what mathematical background can be drawn upon, this question is just going to cause confusion and waste time. Note: Posting this question in Pre-algebra and Algebra implies a background severly limit what you might otherwise have in mind.

4. I should've written it differently.

I'm wanting to say: Y + X = 53

5x - 2x + x * 7x + 23 * x / 78652

Something like that.

I guess shouldn't have left the question so open. I'm looking for something that is algebra and can be solved in say 10 minutes by someone with a good grasp of algebra.

5. Hello, Jerick!

I'm trying to come up with a complicated equation that equals 53.

So we given with: .$\displaystyle x \:=\:53$

Subtract 3 from both sides: .$\displaystyle x - 3 \:=\:50$

Double both sides: .$\displaystyle 2(x-3) \:=\:100$

Subtract 9 from both sides: .$\displaystyle 2(x-3) - 9 \:=\:91$

Divide by 7: .$\displaystyle \boxed{\frac{2(x-3) = 9}{7} \:=\:13}$

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If $\displaystyle x = 5$ can be one of the roots of the equation . . .

Start with: .$\displaystyle x \:=\:53$

Subtract 3: .$\displaystyle x - 3 \:=\:50$

Divide by 5: .$\displaystyle \frac{x-3}{3} \:=\:10$

Cube both sides: .$\displaystyle \left(\frac{x-3}{5}\right)^3 \:=\:10^3 \quad\Rightarrow\quad \frac{(x-3)^3}{125} \:=\:1000$

. . $\displaystyle x^3 - 9x^2 + 27x - 27 \:=\:125,\!000 \quad\Rightarrow\quad\boxed{ x^3 - 9x^2 + 27x - 125,\!027 \:=\:0}$

6. if x=53? one of the roots.
lol you used math text for that

7. $\displaystyle 53 = \sqrt{2+\sum^{\infty}_{r = 0}\frac{1}{r!} + 52 \sqrt{2+\sum^{\infty}_{r = 0}\frac{1}{r!} + 53 \sqrt{2+\sum^{\infty}_{r = 0}\frac{1}{r!} + 54 \sqrt{2+\sum^{\infty}_{r = 0}\frac{1}{r!} + \cdots}}}}$

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# complicated sums that equal 9

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