N is a 3-digit positive integer (with no leading zeroes) divisible by 3, such that by reversing the digits of N/3 and subtracting 1 from the result, we will obtain N.

Determine all possible value(s) of N.

Bonus Question:

If the number of digits in N is > 3, with all the other conditions remaining the same, what is the minimum value of N?

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

N = ABC ..... N/3 = DEF ... rev(N/3) = FED .... FED - 1 =  ABC

ABC = 3 * DEF

C = D - 1

D must be =< 3, so C =< 2

C = 2....... D = 3 ........ A = 9 ..... this imply F = 9.... imply C congr 3xF ... C = 7......... contradiction

C = 1 ...... D = 2 ....... A = 6 or 7 or 8 ........ this imply F = 6 or 7 or 8 ....... C congr 3x6 congr 8 (contradiction)..... C congr 3x7 congr 1 (ok)......... C congr 3x8 congr 4 (contradiction)

So, C = 1 ............ A = 7 ...... F = 7

D = 2 ....... so 3 x 2E7 = 7B1

3*E must carry 1....... so E = 4 or 5 or 6

3 * 247 = 741 ......... FED = 742........ OK!!

3 * 257 = 771.......... FED = 752....... no

3 * 267 = 801.......... FED = 762....... no

Unique solution: ABC = 741

Posted by pcbouhid on 2008-09-10 15:17:05