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How do I get 60 using 7 numbers under 20? Only using a number once

Guest Jan 7, 2015

Best Answer 

 #3
avatar
+5

((a')!+(b')!+(c')!)!! / 12 + d' + e' + f'

 

where letters a - f are distinct natural numbers less than 20 and 12 is not in {a, b, c..., f} 

 

Works every time.

Guest Jan 7, 2015
 #1
avatar
+5

9+10+8+11+7+12+3=60

Guest Jan 7, 2015
 #2
avatar+87333 
+5

Here's one way...using the first 7 integers

(7^3 - 6^3) + (4^3 - 5^3) - (3 * 2 * 1)

(127) - (61) - (6) =

66 - 6   =

60

(I'm sure someone else will come up with something more elegant....!!!)

 

CPhill  Jan 7, 2015
 #3
avatar
+5
Best Answer

((a')!+(b')!+(c')!)!! / 12 + d' + e' + f'

 

where letters a - f are distinct natural numbers less than 20 and 12 is not in {a, b, c..., f} 

 

Works every time.

Guest Jan 7, 2015
 #4
avatar+92806 
0

I have to think about that one anon - Thanks 

Melody  Jan 9, 2015

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