Digit Reversal And Subtraction (Posted on 2008-09-10)

	Submitted by K Sengupta
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The required value of N is 741. For the bonus question, the required minimum value is 7425741 EXPLANATION: Refer to the respective analytic methodologies of Dej Mar and pcbouhid in the comments. I solved the original problem in the following manner: Let R(x) = The number obtained by reversing the digits of x. Then, by the problem: N = R(N/3) -1 -> R(N+1) = N/3 -> N+1 = 3*R(N+1) +1 -> abc = 3*cba + 1, where N = abc-1 -> abc – cba = 2*cba + 1 -> 99(a-c) = 2*cba +1 ………(i) From (i), it follows that: 2*cba + 1 is equal to an at most three digit odd multiple of 99, so that: 2*cba + 1 = 99, 297, 495, 693, 891 -> cba = 049, 148, 247, 346, 445 Checking the five values, we observe that only cba = 247 satisfies (i). Thus, R(N+1) = 247, so that: N = 742 – 1 = 741 Consequently, the required value of N is 741. Bonus Question: For the bonus question, we denote the number of digits in N by d. No facile analytic methodology seem to exist for solving this, apart from checking separately the cases arising out of d= 4,5,6,7; whereby we will obtain the minimum value of N as 7425741.

	Subject	Author	Date
	Bonus Question	ed bottemiller	2008-09-10 17:04:36
	analytical solution	pcbouhid	2008-09-10 15:17:05
	Original	ed bottemiller	2008-09-10 14:20:00
	value(s) of N (partial spoiler)	Dej Mar	2008-09-10 13:45:23