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

**Jerick** · November 18th 2009, 03:30 PM

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.

**Drexel28** · November 18th 2009, 03:38 PM

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.

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

**mr fantastic** · November 18th 2009, 03:55 PM

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.

**Jerick** · November 18th 2009, 03:58 PM

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.

**Soroban** · November 18th 2009, 03:59 PM

Hello, Jerick!

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

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

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

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

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

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

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

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

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

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

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

**Krahl** · November 18th 2009, 04:04 PM

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

**I4talent** · November 18th 2009, 07:42 PM

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

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