A Reversal Problem (Posted on 2006-12-07)

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
Let abcd be the required number.Then: 2(1000a+100b+10c+d0 - 81 = 1000d+100c+10b+a or, 2d - a = 81 - 10U; where U = 200a+19b-8c-100d......(1) Since, maximum of 2d-a = 2*9-1 = 17 and minimum of 2d-a = 2*1 -9 = -7, it follows that: Either, U = 8, whenever 2d - a =1 Or, U = 7, whenever 2d - a =11 Let us assume U = 8. then, from (1) we obtain: 8 = 200(2d-1)+19b-8c-100d or, 300d = 208+8c-19b........(2) Since, the maximum value of 8c-19b = 8*9 - 19 = 53, it follows that equatiion (2) does not provide any integer solution for d Thus U must be 7. then: 7 = 200(2d-11) + 19b - 8c- 100d or, 300d = 2207 + 8c - 19b This will yield a solution for d only if 8c - 19b = -107, 193 However, the maximum possible positive value of 8c - 19b is 53. Thus, 8c - 19b = - 107 or, b = 5 + (8c+12)/19 Now, (8c+12)/ 19 is an integer for c=8 (mod 19), so that: only c = 8, is possible in the context of the given problem. This gives (b,d,U) = (9,7,7) so that 2*7-a = 11, giving a=3 Consequently, the required number is 3987. ---------------------------------------------------------------- A slightly different methodology leading to the same solution is provided by Dennis in this location.

	Subject	Author	Date
	a solution	Dennis	2006-12-08 08:55:44
	Analytically lacking (partial spoiler)	Dej Mar	2006-12-07 15:43:23