Thanks for sharing this, Bob. Best, dick -----Original Message----- From: rtwainwright@optonline.net Sent: Saturday, January 05, 2013 4:18 PM To: Robert Munafo Cc: math-fun Subject: Re: [math-fun] Happy New Year A less emphatic but simple countdown: (10 + 9 - 8)(7 - 6 + (5)(4)(3))(2 + 1 + 0) R. Wainwright ----- Original Message ----- From: Robert Munafo Date: Tuesday, January 1, 2013 2:12 am Subject: Re: [math-fun] Happy New Year To: math-fun Cc: "R. Wainwright"
Hans Havermann's version is brilliant:
10/9!*8!*7!-6!*5!/4!+3!*2!+1! = 2013
No rearrangement is needed, if interpreted in the normal way:
((10/9!)*8!)*7! - 6!*5!/4! + 3!*2! + 1! = (10/362880)*40320*5040 - 720*120/24 + 6*2 + 1 = 5600 - 3600 + 12 + 1 = 2013
On 12/31/12, wrote:
4, 3, 2, 1, 0.!!!
The expression above shows a countdown for the New Year.
Using ALL of the thirteen symbols (substituting parentheses for the commas), rearrange them to form a mathematical expression which equals 2013. Any number of the basic arithmetic operations [+, -, x, /] may be used.
R. Wainwright
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