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

(In reply to Answer by K Sengupta)

acb + bca + bac + cba + cab = 3961 (given)
so, 5(a+b+c)(mod 9) = 1
or,(a+b+c)(mod 9) = 2 ........ (i)

Again, adding abc to each of the sides of the given equation, we have:

acb + bca + bac + cba + cab + abc  = 3961 + abc

or, 222(a+b+c) =  3961 + abc
or, 3961 <= 222(a+b+c) < 4961
or,  17.8423... <=  a+b+c < 22.3468.....(ii)

However from (i), we note that (a+b+c)(mod 9) = 2, and the only integer value of a+b+c satisfying (ii) occurs at a+b+c = 20

Substituting this in the relation: 222(a+b+c) =  3961 + abc, we have:

abc = 222*20 - 3961 = 479, and for this value, we observe that a+b+c = 4+7+9 = 20

Consequently, the required value of abc is 479. 

Edited on September 10, 2008, 5:56 am

Posted by K Sengupta on 2008-09-10 05:48:19