50 - Digit Number (Posted on 2003-11-15)

	Submitted by Ravi Raja
Rating: 3.0000 (7 votes)
Solution:	(Hide)
3 First, notice that 111111 is divisible by 13 (13×8547). So, 11111100 is also divisible by 13 (13×854700), and so on. Also, the sum of any multiples of 13 will also be divisible by 13, which includes any string of 6n 1s (where n is any integer), and obviously those numbers followed by any number of 0s. Thus, the number formed by 24 1s is divisible by 13: 111111111111111111111111 as is the number formed by 24 1s followed by 26 0s: 11111111111111111111111100000000000000000000000000. Finally, their sum is a multiple of three: 11111111111111111111111100111111111111111111111111 and if you replace the middle two digits with 13 (literally, add 13000000000000000000000000 to the number), the result will, again, still be a multiple of 13. Thus, 11111111111111111111111113111111111111111111111111 is divisible by 13, and the digit in question is 3.

	Subject	Author	Date
	School Education	Christopher Jaeger	2022-04-16 11:25:12
	Explanation for Puzzle Answer	K Sengupta	2022-01-31 21:06:48
	Puzzle Answer	K Sengupta	2022-01-31 20:49:32
	re: A similar problem but perhaps a bit trickier	hepev92013	2020-09-29 12:22:49
	answer	K Sengupta	2008-03-20 09:46:36
	Solution	Praneeth	2007-09-11 08:20:38
	re(2): Re so easy SECOND OPINION att :NK	Ady TZIDON	2004-02-11 07:24:20
	re: Re so easy	NK	2004-02-10 19:03:12
	Re so easy	NK	2004-02-10 18:13:08
	so easy	Ady TZIDON	2004-02-10 15:08:55
	Correction to Another Approach	NK	2004-02-10 11:21:08
	Another Approach	NK	2004-02-10 10:32:01
	re: Digit number	Richard	2004-01-24 17:51:29
	Digit number	Purna	2004-01-24 06:17:38
	re: solution	Richard	2003-12-31 13:22:37
	solution	luminita	2003-12-31 11:27:06
	re(7): solution	Richard	2003-11-23 12:24:27
	re(6): solution	SilverKnight	2003-11-23 02:11:44
	re(5): solution	Richard	2003-11-22 21:34:28
	No Subject	Richard	2003-11-22 21:31:34
	re(4): solution	Kirk	2003-11-22 18:49:19
	re: correction	Kirk	2003-11-22 18:39:12
	correction	Richard	2003-11-22 12:54:55
	re(3): solution	Richard	2003-11-21 17:48:09
	re(2): solution	Kirk	2003-11-21 16:33:21
	re: solution	Richard	2003-11-20 20:35:49
	re(6): a little more straightforward	Tristan	2003-11-20 19:31:37
	re: A similar problem but perhaps a bit trickier	Richard	2003-11-20 17:47:15
	re(3): A similar problem but perhaps a bit trickier	Kirk	2003-11-20 17:17:18
	re(2): A similar problem but perhaps a bit trickier	Kirk	2003-11-20 17:13:45
	re: A similar problem but perhaps a bit trickier	Charlie	2003-11-20 14:59:49
	A similar problem but perhaps a bit trickier	Kirk	2003-11-20 14:53:01
	re(5): a little more straightforward	Richard	2003-11-19 21:30:45
	re(4): a little more straightforward	Richard	2003-11-19 21:21:06
	re(3): a little more straightforward	Tristan	2003-11-19 19:06:25
	re(2): a little more straightforward	Richard	2003-11-19 01:59:52
	re: a little more straightforward	Richard	2003-11-19 00:10:42
	re: a little more straightforward	Tristan	2003-11-16 20:52:59
	re: a little more straightforward	Victor Zapana	2003-11-16 12:57:06
	a little more straightforward	Eric	2003-11-16 01:07:26
	Less thinking using extended precision	Charlie	2003-11-15 11:46:40
	solution	Charlie	2003-11-15 11:32:16
	first thoughts	rerun141	2003-11-15 11:06:34