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I got this one from an 8 old kid who got this at school

Can you get the number in the right by using the numbers on the left with the rules:

  1. Each number should be used 1 time exactly,
  2. The allowed operations are: $(,),+,-,\cdot,/$

A solved exercise as an example:

$12$ $11$ $20$ $2$ $= 9$ solution: $(20/2)-(12-11)$

Any ideas regarding the solutions of these three?

$8$ $2$ $1$ $32$ $= 31$

$11$ $2$ $1$ $9$ $= 111$

For example, to me it seems like the first one has no solution. I can explain it intuitively to myself but I can't prove it formally. Am I right? If yes, is there a way to formally proof it?

Thanks!

Edit exponentials and roots are not allowed (i.e. $8^2$ is not allowed). All the numbers must be used. No concatenation is allowed.

user135172
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  • Can we use exponentiation? – user Aug 16 '21 at 09:52
  • Is it necessary to use all numbers !? – Mithlesh Upadhyay Aug 16 '21 at 09:56
  • Allowing also exponentiation we have: $8^2-1-32=31$, $11^2- 1- 9 = 111$. The third one would also work well for $5$ as a result. – user Aug 16 '21 at 09:57
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    Last one: 5-(9+2)/11=4 – Aman Kushwaha Aug 16 '21 at 10:00
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    Suggestion: you can ask this question in stack overflow. One can definitely make a program with any programming language such as python to calculate all possible expressions and compare them with RHS – Aman Kushwaha Aug 16 '21 at 10:15
  • $(8+1)×3+(2×2)=...$ – Oscar Lanzi Aug 16 '21 at 10:59
  • Can we concatenate digits to make a larger number like making 1000 with eight 8's using 888+88+8+8+8? – Oscar Lanzi Aug 16 '21 at 11:04
  • no concatenation is allowed, also, in the first exercise, the last number is 32 (not 3 and 2) – user135172 Aug 16 '21 at 11:18
  • How about ${11\choose 9}2 +1=111$? Using the parentheses for combination, i.e., $^{11}C_9$. Don't tell me brackets weren't allowed. :) 8 yr old, I got you! – Aman Kushwaha Aug 16 '21 at 11:46
  • :)) its not the brackets but their meaning... Those kids have learned only the basic operations. They didn't study exponentials or "p choose n"-like things. I am curious if this can be done with just basic operations (the brackets are for the right order of operations.. ) – user135172 Aug 16 '21 at 12:12
  • Yeah yeah. I know. I too tried many different possible expressions using the 4 basic operations only for first and second but never got them equal to RHS. I seems to me that they have no solutions (The truth value of this hypothesis can be checked using any computer programming language) and if they actually have no solution then a formal proof for that should exist. You might want to edit or post another question asking for formal proof if the hypothesis is true. – Aman Kushwaha Aug 16 '21 at 12:43
  • I agree with you.. I think there is no solution for the first one. I posted this question in stackoverflow and in mathoverflow and they deleted it (without answering the question). If you are familiar with any math forum or programming forum that could be suitable I'll be happy to know.. thanks :) – user135172 Aug 16 '21 at 12:51
  • Is $11**2-(1+9)=111$ allowed? – Michael Hoppe Aug 16 '21 at 14:46

3 Answers3

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For the third we have: $$5-(2+9):11=4.$$

Michael Rozenberg
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2

I wrote some code to brute-force this instead of doing any hard thinky thinking. Neither {8 2 1 32} = 31 nor {11 2 1 9} = 111 have solutions. If 111 was a typo for 11, $11 + (9+2) \cdot 1$ works.

Evelyz
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A formal way to prove it could be done by exhaustion trying for all the possible allowed combinations of numbers and operations, similarly to what they did for the first proof of the famous "Four colour theorem", using a calculator.

user
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