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 Month, product and prime alphametics (Posted on 2010-02-20)
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 No Rating

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 computer solution | Comment 1 of 2
` 112979     87-------      790853903832-------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 AOKrun 112979 8  7 790853 903832 9829173`
`OK`

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