perplexus.info :: Word Problems : Phone Keys

Phone Keys (Posted on 2021-11-11)

There are 8 keys on a standard telephone that have letters associated with them. Consider Q included with P, R and S; and Z with W, X and Y.

  1   2   3 
     ABC DEF
     
  4   5   6 
 GHI JKL MNO
 
  7   8   9 
 PQRS TUV WXYZ
  
    *   0   #

How many 8-letter words can you find that use each of these 8 keys exactly once? Using the concept, the 3-letter word KEY would be represented by 539.

Submitted by Charlie

Rating: 3.0000 (1 votes)

Solution: (Hide)

As Steven Lord says, the list depends on your word list. Here's from mine:

23764859 adroitly 25468937 blintzes 25694378 blowiest 27965483 cryolite 34268957 dicotyls 34528679 dilatory 36842579 enthalpy 37426895 epicotyl 36982457 foxtails 48692537 gunwales 42956387 haylofts 43652879 idolatry 46837529 interlay 52768493 jarovize 53978624 keypunch 56489327 knitwear 59264837 kyanites 52894637 lauwines 54762983 lipocyte 73625489 penality 75298463 playtime 76325489 sodality 78326459 steamily 82539746 taleysim 83674259 tensibly 87429653 triazole 89264537 twangler 89264537 twangles 89265437 twankies 86327459 uneasily 86324759 unfairly 86352749 unflashy 82567493 valorize 92837564 waterlog dial={'' 'abc' 'def' 'ghi' 'jkl' 'mno' 'pqrs' 'tuv' 'wxyz'}; words=fileread('c:\words\words8.txt'); clc while ~isempty(words) w=extractBefore(words,9); words=extractAfter(words,8); numstr=''; good=true; for psn=1:8 for button=2:9 found=false; ix=strfind(dial{button},w(psn)); if length(ix)==1 found=true; numstr=[numstr char(string(button))]; continue end if found==false good=false; continue end end end if length(unique(numstr))==8 disp([numstr ' ' w]) end end

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: soln	Dej Mar	2021-11-29 06:24:05
soln	Steven Lord	2021-11-11 11:57:41

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