2
$\begingroup$

Sometimes a clock is posted on Facebook and the likes where digits are substituted by mathematical expressions. A friend of mine asked me what is the smallest numbers of "nine" which let one express numbers from 1 to 12, using the four basic operations, square root, factorial, decimal point and the notation .(9) = 1. I managed to come up with a solution using 21 nines; is it possible to get 20?

EDIT: the notation .(9) means $.{\bar 9}$ (I forgot that puzzling allows MathJaX). For example, 11 may be expressed as 99/9, while 12 is $9 + \sqrt(9)$. So you must have twelve expressions, one for each number from 1 to 12, and using in total the least number of 9.

$\endgroup$
  • 1
    $\begingroup$ I don't understand the question. Perhaps a visual would help? $\endgroup$ – warspyking Oct 5 '14 at 14:34
  • $\begingroup$ This may belong on mathematics. $\endgroup$ – warspyking Oct 5 '14 at 14:35
  • $\begingroup$ Or is the idea to have 12 different expressions, and we want to minimize the total number of nines among all those expressions? $\endgroup$ – Ben Aaronson Oct 5 '14 at 14:35
  • $\begingroup$ I think Ben's idea is what the question means : use 12 expressions with only nines, and minimize the total number of nines. $\endgroup$ – kevin Oct 5 '14 at 14:41
  • $\begingroup$ Are we allowed to compose the result of a calculation with another number? i.e. We're allowed to compose 9 and 9 to make 99, but could we compose 9 with the square root of 9 to make 93? $\endgroup$ – Ben Aaronson Oct 5 '14 at 15:07
7
$\begingroup$

How many numbers can you represent using only one 9?

  • $9$ itself
  • The four basic operations require at least two numbers, so they're out
  • $\sqrt 9 = 3$
  • $9! = 362880$ and $\sqrt 9! = 6$ and $\sqrt{9!} = 602.395...$
  • Decimal point requires at least two numbers
  • $.{\bar 9} = 1$ and $\sqrt{.{\bar 9}} = 1$ and $.{\bar 9}! = 1$

Therefore the only numbers on a clock face that can be represented with only one 9 are 1, 3, 6 and 9. All the rest require at least two nines. That gives us a minimum of $4 + 2 \times 8 = 20$ nines.

  1. $.{\bar 9}$
  2. $.{\bar 9} + .{\bar 9}$
  3. $\sqrt 9$
  4. $\sqrt 9 + .{\bar 9}$
  5. $\sqrt 9! - .{\bar 9}$
  6. $\sqrt 9!$
  7. $\sqrt 9! + .{\bar 9}$
  8. $9 - .{\bar 9}$
  9. $9$
  10. $9 + .{\bar 9}$
  11. ???
  12. $9 + \sqrt 9$

Is it possible to get 11 from only 2 nines?

  • The square of 11 is 121. If we can get 121 using two nines, we could take its square root.
  • 11 or 121 is not a factorial of any integer, so that's out. ($5! = 120$ which is very close but not enough.)
  • The basic operations won't help (none of numbers 1, 3, 6, 9, 362880 and 602,395... add up to 11 or 121, they're too small or too big for subtraction or division, and 11 is a prime and $121 = 11 \times 11$ so multiplication won't do it.)

Based on this it looks like 21 is the minimum (4 numbers using 1 nine, 7 numbers using 2 nines and one number using 3 nines.)

But if we relax the requirements a bit and allow any mathematical notation, as long as it doesn't use any other numbers than 9? Then you could do:

$$\lceil \sqrt{ (\sqrt 9! - .{\bar 9})!}\rceil$$

That is, the square root of the factorial of 5 (=120), rounded up.

$\endgroup$
  • $\begingroup$ Given how hard it appears to find an expression for 11 that uses less than 3 9's, I like the idea of opening it up to more mathematical notations. Of course, I'm not the OP. $\endgroup$ – Joel Rondeau Oct 5 '14 at 15:47
  • 2
    $\begingroup$ With rounding allowed I think we may express all twelve numbers with just one 9... $\endgroup$ – mau Oct 5 '14 at 15:51
  • 1
    $\begingroup$ Yes, I expect that is very likely. $\endgroup$ – Joel Rondeau Oct 5 '14 at 16:33
  • $\begingroup$ 720 (6!) Can also be made with one 9. Obviously there are others (9!)!, but 720 seems to have the best chance of being useful. $\endgroup$ – Joel Rondeau Oct 5 '14 at 17:14
  • $\begingroup$ For one number, could you count $.\overline{\sqrt{9}} = \frac{1}{3}$? $\endgroup$ – wchargin Oct 6 '14 at 14:37
3
$\begingroup$
  1. $.{\bar 9}$
  2. $.{\bar 9}$ + $.{\bar 9}$
  3. $\sqrt(9)$
  4. $\sqrt(9)$ + $.{\bar 9}$
  5. $(\sqrt(9))!$ - $.{\bar 9}$
  6. $(\sqrt(9))!$
  7. $(\sqrt(9))!$ + $.{\bar 9}$
  8. 9 - $.{\bar 9}$
  9. 9
  10. 9 + $.{\bar 9}$
  11. 99/9
  12. 9 + $\sqrt(9)$

In terms of proving this is minimal, it's easy to see that for only one 9, you have a very limited numbers of possibilities: 1, 3, 6, 9, each included above. The only operators you can use on these which don't require a second number would be factorial or square root. Square rooting any of those either gives you a non-integer, or another number in the same set. Factorial gives a number too high, or another number in the same set. The too high factorials aren't squares, so they can't be square rooted. So that means there's no combination of square roots and factorial that would get you to another number in the 1-12 range.

Since 1 is only possible for those 4, all the rest must be at least 2. So now it's only 11 which remains unproven, and I don't know how I'd go about proving that. It's hard to rule out the case where you factorial several times, add or remove a 1,3,6 or 9, then square root many times to get back to 11.

$\endgroup$
  • $\begingroup$ That's basically my answer, I was stuck with 11 too $\endgroup$ – mau Oct 5 '14 at 15:48
1
$\begingroup$

You can get almost exactly 11 with one 9 only by applying a long series of ! and sqrt functions, as long as you accept factorials of non-integer values.

I.E where x! is gamma(x+1) for non integer values of x.

In this case the string...

s!!ss!ss!s!s!s!s!s!ss!ss!!ss!s!s!ss!s!s!s!ss!!s!ss!s!s!s!!sss!!s!ss!s!s!!ss!s!ss

...applied to a single 9 will equal 11.00004619 (where, reading from left to right, you apply 's' (sqrt) once then '!' (factorial) twice, then 's' (sqrt) twice, then ! once, then sqrt twice etc....

There will be some finite combination of ! and sqrt functions that will get you arbitrarily close to 11. For sufficiently large N, you should be able to get ≥N decimal places accuracy by 'applying' a string of s's and !'s kN characters long (where k is an unknown constant which I estimate to be less than about 50).

A similar argument for the other numbers that need 2 9's would allow you to generate all 12 numbers with only 12 9's.

$\endgroup$
0
$\begingroup$

To get 11 You can by doing so {(√9)!÷√9}+9=11 != Factorial Square root of 9 is equal to 3 then it's factorial is 6 which then divided by square root of nine will be 2 which added by 9 will get the final answer of 11........

$\endgroup$

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.