Determine all possible triplet(s) (a, b, c) satisfying the equation given below where each of the small letters in bold represents a base-11 digit from 0 to A (whether same or different) and a is nonzero.(No computer programs.)

a!+ b!+ c!= abc – a

Note: abc represents the concatenation of the three digits.

rather than write abc as a concatenation, write the equation in base 10 as

a!+b!+c! = 121a + 11b + c - a
a!+b!+c! = 120a + 11b + c
120 of course stands out as 5! so letting a=6 we have
720+b!+c!=720 + 11b + c
b!+c! = 11b + c

Suppose b=c, then we have
2b! = 12b
b! = 6b
b(b-1)!= 6b
(b-1)! = 6
so b=4

I found a triplet: a=6, b=c=4.

There may be others tho.

Posted by Jer on 2011-08-21 01:35:05