Re: For 2014?

• To: mathgroup at smc.vnet.net
• Subject: [mg132145] Re: For 2014?
• From: Ulrich Arndt <ulrich.arndt at data2knowledge.de>
• Date: Thu, 26 Dec 2013 06:24:29 -0500 (EST)
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• References: <20131224071749.C3F1A69D5@smc.vnet.net>

```123 + 45*6*7 - 8 + 9

Generate is maybe a bit wrong - search ;-)

char = CharacterRange["1", "9"]
type = {"+", "-", "*", "/", ""}
RandomSeed[1992]
r = Table[StringJoin[Riffle[char, RandomChoice[type, 8]]], {1000000}];
e = ToExpression[#] & /@ r;
p = Position[e, 2014]
Union[Extract[r, p]]

2015
1*2-3+4*567*8/9, 12*3+45*6*7+89, 12*34*5-6*7+8+9

Am 24.12.2013 um 08:17 schrieb Harvey P. Dale:

> 	There are some nice (very simple) math puzzles using consecutive
integers that produce years.  For example, 10+(9 x 8 x (7/6) x 5 x
4)+321 and 0-12+(34 x 56)+7 x (8+9) both yield 2011, and (10 x 9 x 8) +
7 + 6 - 5 + (4 x 321) yields 2012.
>
> 	Two questions: (1) can anyone generate a similar puzzle yielding
2014 and (2) is there a general Mathematica program that can generate
these?
>
> 	Best,
>
> 	Harvey
>

```

• References:
• For 2014?
• From: "Harvey P. Dale" <hpd@hpdale.org>
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