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You may have seen the 24 game where you take four numbers and try to make 24 using the numbers and mathematical operations. However, I present a twist on the 24 game: with an exponent. Here are the modified rules:

  1. You must use each of the four numbers exactly once.
  2. You may use only the following mathematical operations: (), +, -, *, /, ^ (exponent).
  3. You may only use exponents of integral powers, exponents of (1/integer) powers, and not anything that evaluates to -1, 0, and 1.
  4. You must use exactly one exponent.

Here are the problems:

  1. 6 4 1 1 (you should get this one in a few seconds)
  2. 1 1 9 9
  3. 9 8 7 3
  4. 4/7 3 5 11
  5. 3/17 2 4 4
  6. 3/16 4/5 5 2
  7. 1/7 1/6 2/3 2
  8. 3 12 90 105
  9. 2 3 14 39
  10. 3 28 20 16 (the day this was posted)
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  • $\begingroup$ Does "not -1, 0, and 1" mean that those can't be the exponents, or those can't be the bases of exponents? $\endgroup$ – f'' Mar 28 '16 at 19:55
  • $\begingroup$ Those can't be the powers. $\endgroup$ – AMACB Mar 28 '16 at 19:57
  • $\begingroup$ So basically every one of them has to have an exponent? $\endgroup$ – Matthew Lau Mar 28 '16 at 20:12
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    $\begingroup$ Think you should also put whether or not $\sqrt\,$ or $!$ is allowed in the question. $\endgroup$ – Paul Evans Mar 28 '16 at 21:46
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    $\begingroup$ And, what, the exponent can be any new number? If so, you should've said so explicitly. And maybe given an example. $\endgroup$ – Paul Evans Mar 28 '16 at 23:04
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6 4 1 1

$6*4*(1^2)*1$

1 1 9 9

$9^{1/2}*(9-1)*1$

9 8 7 3

$((9+7)/8)^3*3$

4/7 3 5 11

$\frac{4}{7}*(5^2-11)*3$

3/17 2 4 4

$(4^3+4)*\frac{3}{17}*2$

3/16 4/5 5 2

$2^5 * \frac{3}{16} * \frac{4}{5} * 5$

1/7 1/6 2/3 2

$2^2/\frac{1}{7} - \frac{2/3}{1/6}$

3 12 90 105

$(105 - 90 - 12)^3-3$

2 3 14 39

$39 - 14 - (3-2)^2$

3 28 20 16

$\frac{16}{28-20}^3 *3$

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  • $\begingroup$ Almost there, but on number 9, you cannot use the first power. $\endgroup$ – AMACB Mar 28 '16 at 23:51
  • $\begingroup$ Power 2 also works ;) $\endgroup$ – Fabich Mar 28 '16 at 23:53
  • $\begingroup$ How can this be an accepted answer? One of the rules is "You may only use exponents of integral powers" yet in the second answer, the exponent is $\frac{1}{2}$ - clearly not an integer. Don't want to take away from the answer - great work finding solutions for them all. I just think that the puzzle is misrepresented. $\endgroup$ – Trenin Mar 29 '16 at 13:01
  • $\begingroup$ read again the question : You may only use exponents of integral powers, exponents of (1/integer) powers. This is not really clear but the ',' means 'or' $\endgroup$ – Fabich Mar 29 '16 at 13:04
  • $\begingroup$ @Lordofdark Dammit! I didn't see that. Thanks! $\endgroup$ – Trenin Mar 29 '16 at 14:32

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