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

Question

0

804

4

Guest Jan 7, 2015

+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

Guest · Answer 1 · 2015-01-07T11:33:58

((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 · Answer 2 · 2015-01-07T10:12:47

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

CPhill · Answer 3 · 2015-01-07T10:14:58

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....!!!)

Melody · Answer 4 · 2015-01-09T11:16:05

I have to think about that one anon - Thanks