Ways to write "50"

Question

A really good friend of mine is an elementary school math teacher. He is turning 50, and we want to put a mathematical expression that equals 50 on his birthday cake but goes beyond the typical "order of operations" problems. Some simple examples are $$e^{\ln{50}}$$ $$100\sin{\frac{\pi}{6}}$$ $$25\sum_{k=0}^\infty \frac{1}{2^k}$$ $$\frac{300}{\pi^2}\sum_{k\in \mathbb{N}}\frac{1}{k^2}$$

What are some other creative ways I can top his cake?

I should note that he is an elementary school teacher. Now he LOVES math, and I can certainly show him a lot of expressions. I don't want them so difficult that it takes a masters degree to solve, but they should certainly be interesting enough to cause him to be wowed. Elementary functions are good, summations are also good, integrals can be explained, so this is the type of expression I'm looking for...

EDIT:: I would make a note that we are talking about a cake here, so use your judgement from here on out. Think of a normal rectangular cake and how big it is. Hence, long strings of numbers, complex integrals, and long summations are not going to work. I appreciate the answers but I need more compact expressions.

Answer 1

We can use only two famous numbers in mathematics, $\large\pi$ and $\large e$, to produce number $50$.

$$\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\Large \lfloor e^\pi \rfloor + \lfloor \pi^e \rfloor + \lfloor \pi \rfloor + \lfloor e \rfloor = 50}} $$

^{Click the box to see Wolfram Alpha's output to confirm the result}

Answer 2

binary code : 50 is given by 110010

Answer 3

$\displaystyle \left(4!-5!-4\right)\int_0^1\frac{\operatorname{Li}_{-1}(t) }{t}\mathbb{d}t$

Although this next one may be too big to fit on a cake, it's certainly beautiful:

$\displaystyle \frac{\pi\operatorname{Li}_1(\frac12)(\Gamma(5)+1)}{\left[\Gamma(\frac32)\right]^2\left(\frac{\Gamma'(1)}{\Gamma(1)}-\frac{\Gamma'({\frac12})}{\Gamma({\frac12})}\right)}$

Here's another one that may or may not fit on a cake:

$\displaystyle\frac{\displaystyle\left(1+2\sqrt{4!+5!}\right)\left[\prod_{n=1}^3\Gamma\left(\frac{n}3\right)\right]^2}{\displaystyle{e}^{2\operatorname{Li}_1(\frac12)}\operatorname{Li}_2\left(\frac{\Gamma'(1)}{\Psi(1)}\right)}$

Answer 4

Quoting Wikipedia

Fifty is the smallest number that is the sum of two non-zero square numbers in two distinct ways: $50 = 1^2 + 7^2 = 5^2 + 5^2$.

So you could write something like \begin{align} 50 = \min_{n\in \mathbb{N}}\{n =p_i^2+p_j^2=p_k^2+p_l^2 \quad | \quad p_i,p_j,p_k,p_l\in\mathbb{N} \quad \wedge\quad p_k \not =p_i \not = p_l \} \end{align}

I like it, because it doesn't involve some sort of scaling and is not obvious (at least not for me).

Answer 5

Whenever one of my friends has a birthday, I find out how old they get and then I visit their number on Wikipedia

For your friend, I would write something like this:

50 is the smallest number that is the sum of two non-zero square numbers in two distinct ways: $1^2 + 7^2$ and $5^2 + 5^2$. It is also the sum of three squares, $3^2 + 4^2 + 5^2$. It is also a Harshad number and a nontotient and a noncototient. 50 is the aliquot sum of 40 and 94. 50 is also the atomic number of tin and fifth magic number in nuclear physics.

While many of the things on Wikipedia are not Mathematical expressions, and some of it is way too long to write on a cake, I am certain that this will brighten his day if you tell him this stuff!

As for a mathematical expression, I'd go with either $3^2 + 4^2 + 5^2$ because I find it simple but elegant, or fill the cake with stuff like "Harshad number, 5th magic number in nuclear physics, nontotient", etc. and see if he can figure out how old he is.

Answer 6

$$-\frac{12!! - 3^{10}}{705 \text{ mod } 101}-\int_0^3 4x^3\;dx = 50$$

Or, if the double factorial is too weird:

$$\frac{9! - 2^{15}}{10000000_2}-\sqrt[5]{243}^2 = 50$$

Answer 7

$$\frac{2^{\frac{(2\cdot2)!}{2+2}}+22+2^{2^2}-\sqrt 2^2}{2^{2-\frac{2}{2}}}$$

Answer 8

I hope he doesn't calculate it by adding all numbers ;)

\begin{align} 50 = \sum_{k=0}^{100} (-1)^k k \end{align}

And a last one, involving only $4$s and $9$s:

\begin{align} 4^9 \mod 49 + \sqrt{49} \end{align}

Answer 9

$$ \begin{align} &50 = \frac{5}{24} \zeta(-7)\\ &50 = \frac{ 1600 \sqrt{2} }{3 \pi ^3}\int_0^{\infty } \frac{x^2 \log ^2(x)}{x^4+1} \, dx \end{align} $$

Answer 10

$$50 = 2\cdot(2\varphi - 1)^4$$ where $\varphi$ is the Golden Ratio.

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$$50 = \sum_{i=0}^{+\infty} (0.98)^i$$

(Geometric series)

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$$50 = \left(\left(\frac{5^5-5}{5}+5^0\right)\cdot\left(5-5^0\right)\right)^{0.5}$$

$$50 = 0.5 \cdot (5+5)^{\frac{5^0}{0.5}}$$

$$50 = 5\cdot\left(\frac{5}{0.5}+5^0\right)-5$$

(Using only the digits "5" and "0")

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$$50 = \frac{3^{3!}-3^{3-3^0}}{3^{3-3^0}}-30$$

(Using only the digits "3" and "0")

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$$50 = \frac{(10i)^2\log(i^i)}{\pi}$$

(Using imaginary unit $i$)

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$$50 = 3 + 47$$ $$50 = 7 + 43$$ $$50 = 13 + 37$$ $$50 = 19 + 31$$

(As sum of two prime numbers)

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$$50 = (7+11)\frac{11}{11-7}+\frac{7+11}{11-7}-11+7$$

(Using only the two next prime numbers of $5$)

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$$ 50 = 7+3+ (7-3)\cdot(7+3)$$

(Using only the previous and next prime numbers of $5$)

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$$50 = 3\cdot(2^3+3^2)-(2\cdot 3)^{3-2}+3+2$$

(Using only the two previous prime numbers of $5$)

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$$ 50 = (1^6-2^5+3^4-4^3+5^2-6^1)^2\cdot(4^1 - 3^2 + 2^3 - 1^4)$$ $$ 50 = 3 - (1^9-2^8+3^7-4^6+5^5-6^4+7^3-8^2+9^1)$$ (Using bases/powers in reverse order)

Answer 11

We also have \begin{align*} 50 &= 11+12+13+14 \\ &= (8+4)+(8-4)+(8\cdot 4)+(8/4) \\ &= 4^2 + 4^2 + 3^2 + 3^2\\ &= 6^2 + 3^2 + 2^2 + 1^2\\ &= (7+i)(7-i) \\ &= (10-\color{red}{5})(10-\color{red}{0})\\ &= 10(\color{red}{5}+\color{red}{0})\\ &= \sqrt{30^2+40^2}\\ &= \sqrt[3]{170^2+310^2}\\ &= \sqrt[3]{146^2+322^2}\\ &= \sqrt[3]{50^2+350^2}\\ \end{align*} Finally $50= 2 + 4 + 8 + 12 + 24$ and $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = 1$.

Answer 12

No fancy math here, but if you want to emphasize how old your friend is getting, nothing says it better than implying he's halfway to the century mark:

$$100\over2$$

Answer 13

Some solutions: $$\dfrac{50(-1)^{-i}}{i^{i^{2}}}=50$$ $$50=\sum_{n=1}^{10}(2n-1) \mod 10$$ $$50=1212_3$$ $$\left(4+\frac{4}{4}\right)\dfrac{\binom{4}{4-\frac{4}{4}}}{0.4}=50$$

Answer 14

The Complex Answers

I'd suggest $4.471527458208!=50$, but this isn't readily solvable. You suggested summations, and I thought "Hey! Why not nested summations?" The product of my "why not" statement: $$\sum _{n=0}^4\sum _{i=n}^7n=50$$

If you're okay with floor functions: $$\left\lfloor\prod _{n=1}^5\frac{\pi +e}{7}n\right\rfloor=50$$

A slightly more complicated one: $$-1\left(\sum _{n=-3}^{11}-n\right)-10=50$$

If you want to complicate that (for a bit of fun), try, using Euler's identity for $-1$ and $n-2n$ for $-n$: $$e^{i\pi}\left(\sum _{n=3e^{i\pi}}^{11}n-2n\right)-10=50$$

The Somewhat Easy Answers

Using only 5's and 0's: $$5.5\frac{505}{5}-5.55=50$$ Something kind of neat, using a pattern 1...7,1: $$1\cdot 2+3\cdot 4+5\cdot 6+7-1=50$$ Where $x_n$ denotes $x$ in radix $n$: $$302_4=50$$ $$200_5=50$$ $$32_{16}=50$$ $$62_8=50$$

Set Theory

If $\alpha=\#A$ states that $\alpha$ is the cardinal of $A$, then: $$\#\{x\in\mathbb{N}:5<x\leq55\}=50$$

Answer 15

I went to visit him while he was lying ill at the hospital.I had come in taxi cab number $50$ and remarked that it was a rather dull number. "No" he replied, "it is a very interesting number. It's the smallest number expressible as the product of $25$ and $2$ in two different ways." https://mathoverflow.net/a/2197/15296

Answer 16

$$50 = \int_{\ln 1}^{3!+2^2}\frac{\Gamma(\frac{1}{2})^4}{6\sum_{1}^{\infty}\frac{1}{n^2}}\sqrt{3^2+4^2}dx$$

can make it more complicated but it wont fit on a cake :)

Answer 17

Prime factorisation

Simple and elegant: $50=2\times5^2$

Answer 18

You could go Roman with $$\Huge{L}$$

Or with $$\begin{array}{c} \\ \\ \end{array}$$

$$\begin{array}{ccccccccccccc} \Huge{X}&&&&&&&&&&&&\Huge{X}\\ \\ \\ \\ &&&&&&\Huge{X}\\ \\ \\ \\ \Huge{X}&&&&&&&&&&&&\Huge{X} \end{array}\\ $$

Answer 19

Compute the smallest positive integer $n$ such that the first two digits of $n^2$ is $\frac{n}{2}$.

Answer 20

How about this

$$50=12+18+20$$

Reference on this page

Now, this would have your teacher scratch his head a little bit initially, but worth a shot! :)