For May 2020, try to create $\dfrac{5}{2020}$ using the least possible number of integers in the set $\{1,3,4,6,7,8,9\}$.

$2$, $5$ and $0$ are not allowed.



Uses $1,1,3,3,3,4,6$ and $7$ $\implies$ $8$ numbers. You must do better than $7$ numbers.

You are allowed to use any operation as long as you can find a wikipedia page created before 2020, that's why there is a lateral thinking tag.

Improving puzzle - thanks to @athin's answer and @Daniel Mathias' comments Any mathematical constant apart from $\{1,3,4,6,7,8,9\}$ is not allowed!

$$\dfrac{\lfloor \phi \rfloor}{\lceil e^{\lfloor \pi + \pi\rfloor}\rceil} $$

is equal to $\frac{5}{2020}$ but it's not a valid solution!


Using the good ol' functions like:

ceiling and exponential functions.

Here is a way with only just two digits!

$$\frac{1}{\lceil \exp(6) \rceil} = \frac{1}{404} = \frac{5}{2020}$$

  • 2
    $\begingroup$ Might as well use $\lfloor\pi+\pi\rfloor$... $\endgroup$ – Daniel Mathias May 14 '20 at 1:25
  • 2
    $\begingroup$ I'm not sure if using constant is allowed, as we can just have five $\pi$s divided by two thousand twenty $\pi$s. $\endgroup$ – athin May 14 '20 at 9:54
  • $\begingroup$ 5 $\pi$s divided by 2020 $\pi$s is not equal to $5$ divided by $2020$ but the spirit of your comment is true! I'm editing my question accordingly to your answer and remark! $\endgroup$ – JKHA May 15 '20 at 0:15
  • 3
    $\begingroup$ @JKHA huh? 5 𝜋s divided by 2020 𝜋s by definition is equal to 5/2020. $\endgroup$ – Quintec May 15 '20 at 0:22
  • $\begingroup$ @Quintec oh yeah, I should go to sleep, I was thinking of $\pi^5$, forget about that x) $\endgroup$ – JKHA May 15 '20 at 0:28

Here is a way to do it with four digits

$$\frac{5}{2020} = \frac{1}{8!! + 4! - 4} $$ where we have used double factorial

Here is a way to do it with three digits

$$ \frac{5}{2020} = \frac{1}{8!! + \sigma(\sigma(8))} $$ where $\sigma$ is the Divisor sum function


Improved solution:

See Home prime.

Straightforward solution:

Using five digits, $\frac{4}{1616}=\frac{5}{2020}$

  • 1
    $\begingroup$ Nice ;) Straightforward and optimal? $\endgroup$ – JKHA May 13 '20 at 12:50
  • $\begingroup$ @JKHA Likely optimal. Though I wouldn't call it yet. Could be a creative solution with fewer digits. $\endgroup$ – Daniel Mathias May 13 '20 at 12:53
  • $\begingroup$ I was asking you knowing there is better ;) $\endgroup$ – JKHA May 13 '20 at 13:16
  • $\begingroup$ @JKHA See improved solution. $\endgroup$ – Daniel Mathias May 13 '20 at 13:19
  • 1
    $\begingroup$ By the way, I like your straightforward solution a lot for its simplicity! $\endgroup$ – JKHA May 15 '20 at 0:39


string concatenation (here denoted by $\otimes$, e.g. $2\otimes5=25$), Wikipedia page about it was created in 2002 and even last modified in 2019

we can get

$$\frac1{4\otimes(6-6) \otimes 4}=\frac1{404}=\frac5{2020}$$.

  • 2
    $\begingroup$ I believe the actual symbol is $||$ (ex. $2||(3+4)=27$). $\endgroup$ – merrybot May 13 '20 at 18:16

This uses four digits as well:

$$\frac{1}{4 \cdot p(13)}$$


$p(n)$ is the number of distinct integer partitions of $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.