Wonderful problem (Posted on 2009-06-07)

	Submitted by pcbouhid
Rating: 4.0000 (2 votes)
Solution:	(Hide)
OODDF is the square root of WONDERFUL. “O” cannot be greater than 2 because this would give a square of 10 digits. It cannot be 1 because there is no way that a number , beginning with “11”, can have a square in which the second digit is 1. Therefore “O” must be 2. WONDERFUL must be between the squares of 22,000 and 23,000. The square of 22 is 484, and the square of 23 is 529. Since the second digit of WONDERFUL is 2, we conclude that WO = 52. The square of 229 is 52,441 and the square of 228 is 52,984. Therefore, OODD is either 2,299 or 2,288. Using a concept of digital root, the sum of the nine digits in WONDERFUL (we were told zero is excluded) is 45, which in turn sums to 9, its digital root. Its square root must have a digital root that, when squared, gives a number with a digital root of 9. The only digital roots meeting this requirement are 3, 6, 9, therefore OODDF must have a digital root of 3, 6, or 9. F cannot be 1, 5 or 6, because any of those digits would put an F at the end of WONDERFUL. The only possible completions are 22,998, 22,884 and 22,887. The square of 22,887 is 523,814,769, the only one that fits the code word WONDERFUL.

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-01-14 01:05:12
	Essay Help UK	Sarah John	2022-08-10 02:14:53
	No Subject	Jorja Magrath	2020-04-27 02:11:40
	re(3): Not Quite Pencil and Paper	Ady TZIDON	2009-06-08 14:10:22
	re(2): Not Quite Pencil and Paper	brianjn	2009-06-08 10:10:42
	re: Not Quite Pencil and Paper	Ady TZIDON	2009-06-08 08:00:06
	Not Quite Pencil and Paper	brianjn	2009-06-08 04:09:22
	solution	Dej Mar	2009-06-08 00:08:55
	solution (QuickBasic)	Charlie	2009-06-07 16:25:10
	Solution (Maple)	Harry	2009-06-07 14:46:02