3
$\begingroup$

Write $1681$, using $4$, four times only, and you can use any mathematical operation available within mathematics(except catenation or $4.4$ etc, it should be an operation), like factorial and cube root , $\sqrt{}$, greatest integer etc...

Example- writing $131096=4!+(\sqrt4\times {4^{4!!}})$ (yes for this example I first did the R.H.S!!)

I saw this question in a newspaper as a mathematical puzzle many years ago(in high school), and i solved it then, since then $1681$ is my favourite number, as took me $2$ days then to do it, but now i can't remember how i did it. Any help! I remember I had to use an operation which I was not much familiar with then.

$\endgroup$
  • $\begingroup$ Catenation is an operation. $\endgroup$ – MJD Oct 2 '14 at 22:06
  • $\begingroup$ If you are willing to use catenation, then one possibility is $(4 \operatorname{cat} (4/4))^{\sqrt{4}}$, where $x \operatorname{cat} y$ denotes $x$ followed by $y$ (in base 10). $\endgroup$ – Eric M. Schmidt Oct 2 '14 at 22:18
  • $\begingroup$ nah, i don't think It should be allowed. Other wise things are easier. I read it somewhere all numbers upto $500$ can be written like this, and I am sure almost every number must be, and must be interesting to find the ones which can't, but for that we must fix a domain of operations first.... Thanks though, Nice if catenation is allowed. $\endgroup$ – Bhaskar Vashishth Oct 2 '14 at 22:22
  • $\begingroup$ @EricM.Schmidt may be if you use base $4$ in catenation, we can allow it, but base $10$, it uses $10$ intrinsicly in mathematical formulation of catenation, no? and that is not allowed $\endgroup$ – Bhaskar Vashishth Oct 2 '14 at 22:25
10
$\begingroup$

One way to do it is: $$ \frac{(4 + 4)! + 4!}{4!} $$

$\endgroup$
  • 1
    $\begingroup$ Nice. I am pretty sure, I didn't do it this way. Mine were uglier operations. Thanks. $\endgroup$ – Bhaskar Vashishth Oct 2 '14 at 22:10
  • 2
    $\begingroup$ Another way by the same idea: $\frac{(4!!)!}{4!}+\frac 44$. $\endgroup$ – mathlove Oct 2 '14 at 22:48

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.