Three positive integers P, Q and R, each with no leading zeroes and having more than one digit, are such that:

(i) Q = P + sod (P), and:

(ii) R = Q + sod (Q), where: sod(x) = sum of the digits of x, and:

(iii) The digits of R are obtained by reversing the digits of P.

Determine all possible value(s) of P, that satisfy the given conditions.

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

Let n equal the number of digits of R.

The minimum value for sod(x) is 1, thus P < Q < R.

The maximum value for sod(x) is n*9.

As R is the sum of P+sod(P)+sod(Q), [sod(x), twice], the maximum difference between P and R would be 2*n*9 (= n*18).

In order for R to be have the reversed digits of P where neither R or P have a leading zero digit, the difference between P and R must be equal or greater than 9*10^(n-2).

As n*18 must be less than 9*10^(n-2) but for where n is 0, the only solutions possible are for 2-digit numbers.

The maximum possible difference between P and Q, therefore is less than 36 (= 2*18; P and Q cannot sum to be equal 36 as this would make both P and Q have all 9s, for which P being less than Q and also having the same number of digits cannot be possible).

And, as the difference between any number and its mirrored number is a multiple of 9, the maximum difference between P and R is 27.

As it is, there are two values that are valid solutions:
P=12 (Q=15, R=21) and P=69 (Q=84, R=96).  

Edited on September 3, 2008, 10:48 pm

Posted by Dej Mar on 2008-09-03 15:15:35