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I'm trying to create a math equasion puzzle using the numbers 3, 7, 20 and 40. I want to end with an answer of either 7 or 40. Is there any way this can be done?

Thanks for the help

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closed as unclear what you're asking by michaelpri, xnor, Tryth, mdc32, AeJey Jan 27 '15 at 8:43

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If I understand your question correctly, you just want to combine the numbers 3, 7, 20 and 40 with arithmetic operations to get 7 and/or 40: $$(3 \times 7 - 20) \times 40 = 40\\ (3 - 40 \div 20) \times 7 = 7$$

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  • $\begingroup$ that's perfect, and exactly what I was looking for. thank you $\endgroup$ – Kristen Jan 26 '15 at 18:02
  • $\begingroup$ i'm so sorry, but I can't get the second version to come out to 7. I'm getting 12.95 Did I do that right? $\endgroup$ – Kristen Jan 26 '15 at 18:08
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    $\begingroup$ @Kristen: division before subtraction. 40 / 20 = 2 hence (3-2)=1 hence 1x7 = 7. $\endgroup$ – BmyGuest Jan 26 '15 at 18:17
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I wrote a script to go through all possible equations using +, -, /, *. Here is the output:

7 

7 * (3 - (40 / 20))
(3 - (40 / 20)) * 7
7 / (3 - (40 / 20))

40 

40 * ((3 * 7) - 20)
40 * ((7 * 3) - 20)
((3 * 7) - 20) * 40
((7 * 3) - 20) * 40
(20 * (7 - 3)) - 40
((7 - 3) * 20) - 40
40 / ((3 * 7) - 20)
40 / ((7 * 3) - 20)

... Although most of these are trivial rearrangements of earlier ones. The ones I consider "unique" are:

$$7 \times (3 - (40 \div 20)) = 7\\ 7 \div (3 - (40 \div 20)) = 7\\ ((7 - 3) \times 20) - 40 = 40\\ 40 \times ((3 \times 7) - 20) = 40\\ 40 \div ((3 \times 7) - 20) = 40$$

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  • $\begingroup$ I guess I really only found one answer that theosza didn't already mention, if you consider dividing by one trivially similar to multiplying by one. Still, it's interesting to know that no other solutions exist (if my script contains no bugs). $\endgroup$ – Kevin Jan 26 '15 at 18:47

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