Reconstruct the multiplication:
* * * * * * where: O C T O
* * C T O
------------- C T O B
* * * * * * T O
* * * * * * T O B E
--------------- T O B E R
O C T O B E R O B E R
B E R
B E
E R
R
are 11 prime numbers.
lets='octber';
idx=combinator(10,6,'p');
digsets=idx-1;
digsets=char(digsets+48);
for i=1:length(digsets)
digs=digsets(i,:);
for j=1:6
cmd=sprintf('%s = ''%s'';',lets(j),digs(j));
eval(cmd);
end
pset=["octo";"cto";"ctob";"to";"tobe";"tober";"be";"er";"r"];
good=true ;
for w=1:length(pset)
ns=char(pset(w));
ns=strrep(ns, 'o', o) ;
ns=strrep(ns, 'c', c) ;
ns=strrep(ns, 't', t) ;
ns=strrep(ns, 'b', b) ;
ns=strrep(ns, 'e', e) ;
ns=strrep(ns, 'r', r) ;
if ~isprime(str2double(ns))
good = false;
break
end
end
if good
disp(digs)
october=str2double([digs(1:3) o digs(4:6) ]);
divisors(october)
end
end
finds the correspondence of letters to digits to be
october
9829173
based upon the primality of the 11 shown numbers,
and october's divisors are
1 3 29 87 112979 338937 3276391 9829173
so candidates for the multiplication are
112979 338937
87 29
------ -------
790853 3050433
903832 677874
------ -------
9829173 9829173
The one on the left fits the set of asterisks shown.
I inadvertently left out two of the primes, but came up with the unique product anyway.
|
|
Posted by Charlie
on 2024-05-03 12:54:38 |
computer solution
