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.
Sub-group 'obre' consists of the four end-digits of all the other prime options. Two-digit + primes all end in 1, 3, 7 or 9, so 'obre' can only consist of these digits, with 'e' itself limited to 3 or 7. The only unique four-digit prime possibilities for 'obre' to this point are 1973, 3917, 7193, 9137 or 9173. The first fails since 973 ('bre') is not prime. The second fails since 917 ('bre') and 91 ('br') are not prime. The third fails since 93 ('re') is not prime. The final two possibilities, however, do result in primes for all the given combinations from 'obre' down to 'e', so we now know 'o' must be 9 and 'b' must be 1 (i.e. '9ct91re'). <o:p></o:p>
<o:p> </o:p>
Since 'to' is now 't9', the only possible two-digit primes here are 29, 59 or 89. Similarly, 'tobr' (now 't91r') possibilities are reduced to 2913, 2917, 5913, 5917, 8913 or 8917. Of these, only 2917 is prime, so 'r' must be 7, leaving 'e' as 3 and 'obre' as 9173. We now have '9c29173'.<o:p></o:p>
<o:p> </o:p>
For 'octo' we now have '9c29', which as prime could only be either 9629 or 9829. As 629 ('cto') is not prime and 829 is, our 'octobre' product is now 9829173. Working backward, the only possible two and six-digit multipliers (as whole numbers) for this are 29 x 338937 or 87 x 112979. Since 9 x 338937 results in 3050433, when only six digits (asterisks) are possible, the correct solution is:<o:p></o:p>
<o:p> </o:p>
112979 <o:p></o:p>
x 29<o:p></o:p>
790853 <o:p></o:p>
903832_ <o:p></o:p>
9829173<o:p></o:p>
<o:p> </o:p>
Analytic Solution
