Japanese Numbering (Posted on 2003-09-21)

   KYOTO
   KYOTO
 + KYOTO
   TOKYO

	Submitted by DJ
Rating: 4.0769 (13 votes)
Solution:	(Hide)
n=9. Inspecting the first digit from the right in the given equation, we find that 2O is divisible by n. So either O = 0 or O = n/2. In the second case, from the third digit we derive K > n/2, but from the fifth digit we have that 3K ≤ T ≤ n-1, so K < n/3. It follows that O = 0. Now we have these equations: 3T = Kn + Y (from the second and third digits) 3Y = cn (where c is the number carried from the fourth to the fifth digit) 3K + c = T (the fifth digit) Multiplying the first equation by 3 and substituting in the values for 3Y and 3K from the two other equations, we get: 3T = Kn + Y 9T = 3Kn + 3Y 9T = (T-c)n + cn 9T = Tn n = 9 We must also make sure that there's at least one solution for this n. In fact, there are four: KYOTO = 13040, 16050, 23070, or 26080.

	Subject	Author	Date
	Great Puzzle	K Sengupta	2007-03-09 04:57:43
	'The Interested Reader'	Charley	2005-05-14 06:18:48
	re: solution	B	2003-09-30 21:05:34
	solution	B	2003-09-30 21:02:09
	OK	Gamer	2003-09-23 22:09:29
	Hint	Bry	2003-09-23 20:53:39
	HINT! HINT! HINT!	Bry	2003-09-23 20:52:37
	Help!	b	2003-09-23 16:52:29
	o=0	b	2003-09-23 16:52:03
	hard	b	2003-09-23 16:51:32
	Solution	desiree	2003-09-22 22:37:40
	re(3): solution	Gamer	2003-09-22 17:22:04
	re(2): solution	Christian Perfect	2003-09-22 12:31:03
	FULL solution (part 2)	John Reid	2003-09-21 15:53:58
	No Subject	John Reid	2003-09-21 15:11:12
	re(2): Proof of solution continued again	Eradicator	2003-09-21 14:39:12
	re: Proof of solution continued	Eradicator	2003-09-21 14:36:15
	Whole Solution	Gamer	2003-09-21 14:30:18
	Proof of solution	Eradicator	2003-09-21 14:22:14
	Solution	Gamer	2003-09-21 13:37:09
	re: solution	Tristan	2003-09-21 11:39:53
	solution	Christian Perfect	2003-09-21 10:12:19