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MathGroup Archive 2006

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Re: Arithmetic Puzzle (so simple it's hard)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71590] Re: Arithmetic Puzzle (so simple it's hard)
  • From: "Dana DeLouis" <dana.del at gmail.com>
  • Date: Fri, 24 Nov 2006 01:17:09 -0500 (EST)

Here's a solution similar to Scott's.

Note the unit digit (T) in FEAST.

It is derived from k*t from the other two words.

Therefore, Mod[k*t,10] must equal t.

We can use this simpler test to eliminate choices before we apply a more
complex test.

p = Permutations[{0, 1, 2, 4, 5, 6, 7, 9}]; 

 

Length[p]

40320

 

We can eliminate about 3/4 of the possible solutions based on the unit
digit.

 

p2 = Select[p, Mod[Times @@ #1[[{5, 7}]],10] == #1[[7]] & ]; 

 

Length[p2]

12240

 

m1 = {100, 10, 1}; 

m2 = {10000, 1000, 100, 10, 1}; 

Test[x_] := x[[{2, 8, 7}]] . m1*x[[{1, 6, 5}]] . m1 == x[[{4, 3, 1, 6, 7}]]
. m2

 

 

Select[p2, Test]

{{1, 6, 5, 9, 2, 4, 0, 7}, {4, 0, 7, 2, 1, 9, 6, 5}}

 

We don't want a 0 in positions 1,2, or 4, so we eliminate the second
solution.

-- 

HTH   :>)

Dana DeLouis

Windows XP, Mathematica 5.2

 

"Bruce Colletti" <vze269bv at verizon.net> wrote in message
news:ejc4pv$6nf$1 at smc.vnet.net...

> How would this problem be solved in Mathematica?

> 

> BUT * ASK = FEAST, where each letter is a 1-digit number, no two letters
may stand for the same number, and the letters are in {0,1,2,4,5,6,7,9}.

> 

> Thankx.

> 

> Bruce

> 




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