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Combine the three numbers in each group to get the same result in each of the three groups. You can use addition, subtraction, multiplication, division, and exponentiation. Here's an example of a solved puzzle:

Puzzle
Group 1: 15, 19, 24
Group 2:11, 30, 36
Group 3:20, 22, 36

Solution
Group 1: 24 / (19 - 15) = 6
Group 2:(30 + 36) / 11 = 6
Group 3: 20 + 22 - 36 = 6

Now help me solve these

1

Group 1:              3, 17, 21
Group 2:              16, 20, 34
Group 3:              27, 30, 39

2

  Group 1:              2, 5, 14
  Group 2:              10, 29, 38
  Group 3:              31, 37, 43

3

Group 1:              2, 3, 20
Group 2:              2, 20, 24
Group 3:              21, 33, 44 
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This might be an inappropriate question, but: Why has this 5 kviews? –  k.stm May 7 '13 at 9:09
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4 Answers 4

up vote 2 down vote accepted

Well for group $2$ you can do $\frac{14}{(5+2)}$, $\frac{38}{(29-10)}$, and $\frac{(31+43)}{37}=2$

For group $3$, you can do $2^3 + 20$, $2*24-20$, and $\frac{21*44}{33} = 28$

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thnx @mathguy... and for group 1 and 3 ?? –  Azzy Dec 28 '12 at 7:30
    
@Azzy These aren't easy, but they're kind of fun to pass the time with. so I am still trying...Safe to assume these are all integers? –  mathguy Dec 28 '12 at 7:31
    
yupp they are... and we dont want fraction answers –  Azzy Dec 28 '12 at 7:37
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Group1 answer 17 * 21 + 3 = 360,20 * (34 - 16) = 360,30 * (39 - 27) = 360

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Group 1: 20/(2+3)=4 Group 2: (20^2)/((2^2)*24)=4 Group 3: (33+44)/21=4

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I have $\frac{20^2}{2^2\cdot 24}=\frac{25}{6}$ –  user47805 May 7 '13 at 9:39
    
$\frac{20}{5}=4$ is right but $$\frac{20^2}{2^2 \cdot 24}=\frac{2^2\cdot 10^2}{2^2 \cdot 24}= \frac{100}{24}\neq 4$$ and $$\frac{33+44}{21}=\frac{77}{21}\neq 4$$ –  Dominic Michaelis May 7 '13 at 10:06
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2.(14/2)-5=2,group 2: 38/(29-10)=2,group 3: (43+31)/37=2

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