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Month, product and prime alphametics (Posted on 2010-02-20) Difficulty: 2 of 5
Solve the following alphametic multiplication puzzle, where each of the small letters denotes a different base ten digit from 0 to 9. However, each asterisk represents a base ten digit from 0 to 9, whether same or different. None of the numbers can contain any leading zero.

    * * * * * *
               * *
----------------
    * * * * * *
 * * * * * *
---------------
 o c  t o b r e

where each of octo, cto, ctob, to, tobr, tobre, obre, bre, br, re and e is a prime number.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution | Comment 1 of 2
 112979
     87
-------    
 790853
903832
-------
9829173

based on
list
   10   for A=100000 to 999999
   20   for B1=1 to 9
   30     I2=B1*A:if I2>999999 then cancel for:goto 170
   40     for B2=1 to 9
   50        I1=B2*A:if I1>999999 then cancel for:goto 160
   60        Prod=A*(B1*10+B2)
   70        if Prod>1000000 and Prod<9999999 then
   71          :S=cutspc(str(Prod))
   72          :if mid(S,1,1)=mid(S,4,1) then
   90          :if prmdiv(val(left(S,4)))=val(left(S,4)) then
  100           :if prmdiv(val(mid(S,2,3)))=val(mid(S,2,3)) then
  110           :if prmdiv(val(mid(S,2,4)))=val(mid(S,2,4)) then
  120           :if prmdiv(val(mid(S,3,2)))=val(mid(S,3,2)) then
  130           :if prmdiv(val(mid(S,3,4)))=val(mid(S,3,4)) then
  140           :if prmdiv(val(mid(S,3,5)))=val(mid(S,3,5)) then
  150           :if prmdiv(val(mid(S,4,4)))=val(mid(S,4,4)) then
  151           :if prmdiv(val(mid(S,5,3)))=val(mid(S,5,3)) then
  152           :if prmdiv(val(mid(S,5,2)))=val(mid(S,5,2)) then
  153           :if prmdiv(val(mid(S,6,2)))=val(mid(S,6,2)) then
  154           :if prmdiv(val(mid(S,7,1)))=val(mid(S,7,1)) then
  155              :print A:print B1;B2:print I1:print I2:print Prod:print
  159     next
  160   next
  170   next A
OK
run
 112979
 8  7
 790853
 903832
 9829173
OK

 


  Posted by Charlie on 2010-02-20 15:49:47
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