Let abc be a 3-digit number.
Find it, if acb + bca + bac + cba + cab = 3961.

By expanding abc=100*a+10*b+c  and etc...
i got the equation   122*a+221*c+212*b=3961
now c should be an odd number.
if c=1, and if a=9 , b=9  the number that can be acheived is 3227
therefore c!=1
similarly for c=3, the max number is 3669.

By trial and error(!) i got one of the answer as
a=4
b=9 and
c=7
i dont know whether  there are more

Posted by venkatesh on 2005-10-22 07:10:29