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This question is inspired by the Prime to Prime puzzle.

The first 24 Prime Numbers are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89

Using up to 4 prime numbers and the following mathematical operations, get all 24 primes.

+ - × /

No other operators are allowed (note the added restrictions from the previous question).

Other rules

  • You cannot use same prime number more than once in any equation.
  • You can use only prime numbers.
  • You do not have to use all the 4 primes in every equation.
  • You must use the same set of 4 primes in every equation. If you select say 2, 13, 17, 23 then they are the only primes that to appear in every equation to get the 24 primes.
  • Concatenation is forbidden.
  • Parentheses are permitted.

Please refrain from posting partial solutions as there are many sets of primes which will not work.

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  • $\begingroup$ I will be really surprised if there is valid answer exists in this condition. $\endgroup$ – Oray Dec 11 '19 at 12:29
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    $\begingroup$ @Oray according to Peter Taylor's comment on the previous puzzle, there is one unique solution $\endgroup$ – DEEM Dec 11 '19 at 12:40
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    $\begingroup$ @hexomino have you excluded ^ intentionally? $\endgroup$ – DEEM Dec 11 '19 at 12:47
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    $\begingroup$ @DEEM Yes, ^ is excluded. $\endgroup$ – hexomino Dec 11 '19 at 12:48
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    $\begingroup$ @athin I have decided to allow computers for this one, but if somebody comes up with a non-computer solution, that would be very interesting and impressive. $\endgroup$ – hexomino Dec 11 '19 at 14:16
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Here is the answer:

$2=2$

$3=3$

$5=2+3$

$7=\frac{43-3}{2}-13$

$11=43-2\cdot(13+3)$

$13=13$

$17=43-13\cdot2$

$19=13+3\cdot2$

$23=43-2\cdot(13-3)$

$29=43-13-3+2$

$31=43-13+3-2$

$37=43-3\cdot2$

$41=\frac{43+3\cdot13}{2}$

$43=43$

$47=2\cdot43-3\cdot13$

$53=43+13-3$

$59=43+13+3$

$61=43+13+3+2$

$67=2\cdot(43-3)-13$

$71=\frac{3\cdot43+13}{2}$

$73=2\cdot43-13$

$79=2\cdot(43+3)-13$

$83=2\cdot43-3$

$89=2\cdot43+3$

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    $\begingroup$ Great work! As far as I can tell, there is only one other set that works. $\endgroup$ – hexomino Dec 12 '19 at 10:12

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