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50 - Digit Number (Posted on 2003-11-15) Difficulty: 3 of 5
A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.

  Submitted by Ravi Raja    
Rating: 3.3333 (6 votes)
Solution: (Hide)
3

First, notice that 111111 is divisible by 13 (138547). So, 11111100 is also divisible by 13 (13854700), 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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
answerK Sengupta2008-03-20 09:46:36
SolutionSolutionPraneeth2007-09-11 08:20:38
Hints/Tipsre(2): Re so easy SECOND OPINION att :NKAdy TZIDON2004-02-11 07:24:20
re: Re so easyNK2004-02-10 19:03:12
Re so easyNK2004-02-10 18:13:08
Solution so easyAdy TZIDON2004-02-10 15:08:55
Correction to Another ApproachNK2004-02-10 11:21:08
Another ApproachNK2004-02-10 10:32:01
re: Digit numberRichard2004-01-24 17:51:29
SolutionDigit numberPurna2004-01-24 06:17:38
re: solutionRichard2003-12-31 13:22:37
solutionluminita2003-12-31 11:27:06
re(7): solutionRichard2003-11-23 12:24:27
re(6): solutionSilverKnight2003-11-23 02:11:44
re(5): solutionRichard2003-11-22 21:34:28
No SubjectRichard2003-11-22 21:31:34
re(4): solutionKirk2003-11-22 18:49:19
re: correctionKirk2003-11-22 18:39:12
correctionRichard2003-11-22 12:54:55
re(3): solutionRichard2003-11-21 17:48:09
re(2): solutionKirk2003-11-21 16:33:21
re: solutionRichard2003-11-20 20:35:49
re(6): a little more straightforwardTristan2003-11-20 19:31:37
re: A similar problem but perhaps a bit trickierRichard2003-11-20 17:47:15
re(3): A similar problem but perhaps a bit trickierKirk2003-11-20 17:17:18
re(2): A similar problem but perhaps a bit trickierKirk2003-11-20 17:13:45
re: A similar problem but perhaps a bit trickierCharlie2003-11-20 14:59:49
A similar problem but perhaps a bit trickierKirk2003-11-20 14:53:01
re(5): a little more straightforwardRichard2003-11-19 21:30:45
re(4): a little more straightforwardRichard2003-11-19 21:21:06
re(3): a little more straightforwardTristan2003-11-19 19:06:25
Solutionre(2): a little more straightforwardRichard2003-11-19 01:59:52
re: a little more straightforwardRichard2003-11-19 00:10:42
re: a little more straightforwardTristan2003-11-16 20:52:59
re: a little more straightforwardVictor Zapana2003-11-16 12:57:06
Solutiona little more straightforwardEric2003-11-16 01:07:26
SolutionLess thinking using extended precisionCharlie2003-11-15 11:46:40
SolutionsolutionCharlie2003-11-15 11:32:16
first thoughtsrerun1412003-11-15 11:06:34
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