4
$\begingroup$

Using two 1s, try to come up with the most consecutive positive integers.

Allowed operations:

  • Addition

  • Subtraction

  • Multiplication

  • Division

  • Concatenation

  • Square Root

  • Radical

  • Factorial

  • Floor and Ceiling Functions

  • Decimal Point

Improve this question
$\endgroup$
5
  • 6
    $\begingroup$ Isn't there a conjecture that with only factorial, square root and floor, and starting with any single number >1, you can form all positive integers? If so, this seems like one that could go on for ever... $\endgroup$
    – Gareth McCaughan ♦
    Jan 23, 2017 at 3:52
  • $\begingroup$ @GarethMcCaughan It'd have to be $>2$, because $2!=2$, floor decreases and square root makes closer to 1. $\endgroup$
    – boboquack
    Jan 23, 2017 at 4:34
  • $\begingroup$ @GarethMcCaughan ... and floor yields an integer. $\endgroup$
    – Rosie F
    Jan 23, 2017 at 7:04
  • $\begingroup$ Oops, yes, I meant >2. $\endgroup$
    – Gareth McCaughan ♦
    Jan 23, 2017 at 10:04
  • 1
    $\begingroup$ Do you happen to recall the name of this conjecture? $\endgroup$
    – greenturtle3141
    Jan 24, 2017 at 0:48

1 Answer 1

Reset to default
7
$\begingroup$

Let's try (feel free to add on or correct, this is community wiki):

$1=1\times1$

$2=1+1$

$3=\left\lfloor\sqrt{11}\right\rfloor$

$4=\left\lceil\sqrt{11}\right\rceil$

$5=\left\lfloor\sqrt{\sqrt{\left(\left\lfloor\sqrt{11}\right\rfloor!\right)!}}\right\rfloor$

$6=\left\lfloor\sqrt{11}\right\rfloor!$

$7=\left\lceil\sqrt{\sqrt{11}!}!\right\rceil$

$8=\left\lfloor\sqrt{\sqrt{\sqrt{11!}}}\right\rfloor$

$9=\left\lfloor\sqrt{11}!\right\rfloor$

$10=\left\lceil\sqrt{11}!\right\rceil$

$11=11$
$12=\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{(\left\lceil\sqrt{11}\space\right\rceil!)!}}}}\right\rceil!}}}}}\right\rceil$
$13=\left\lceil\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{(\left\lceil\sqrt{11}\space\right\rceil!)!}}}}\right\rceil!}}}}}\right\rceil}}}\right\rceil$

$14=\sqrt{\sqrt{\left\lfloor\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{(1\div(.1))!}}}\right\rceil!}}\right\rfloor!}}$

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I found a way for 12 but it's just a hassle to write it down. We can go all the way to infinity without much problem. $\endgroup$
    – stack reader
    Jan 23, 2017 at 6:05
  • 1
    $\begingroup$ Getting past 10 was the hard part. $\endgroup$
    – humn
    Jan 23, 2017 at 6:08
  • 2
    $\begingroup$ @humn I simplified my 10 as well as the 7 and 9. Looks much better now. But I guess 11 is a good number to stop this... I am not going any higher lol. $\endgroup$
    – stack reader
    Jan 23, 2017 at 6:17
  • 2
    $\begingroup$ Anyone want to write a program for this...? $\endgroup$
    – greenturtle3141
    Jan 23, 2017 at 6:19
  • 1
    $\begingroup$ This is turning into a ridiculous open-ended mathematics exercise. It seems likely that given an unlimited number of factorials, square roots, floor and ceilings, you could express the values of all positive integers. $\endgroup$
    – wildBillMunson
    Jan 23, 2017 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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