[Guest]

Let's consider a 50 digit number ---

11111.........111*(25th from left)*(26th from left)111.........11111

where all digit's are 1 except the 25th and 26th (from the left). They are missing.

---------- what will be the 25th and 26th digit if the number is didvisible by 13?

[Guest]

1 and 3 or

2 and 6 or

3 and 9 or

5 and 2 etc...

If the the two digits are 1 and 3 the divided number will be 854700854700854700854701008547008547008547008547.

If the the two digits are 2 and 6 the divided number will be 854700854700854700854702008547008547008547008547.

etc...

Edited June 24, 2009 by The Zealous Zebra

[Guest]

The 25th and 26th digits are either of these (2,5) or (5,6) or (8,7)... All 3 combinations make the number divisible by 13

Div rule for 13 is that sum of ordered multiplication of digits of numbers and the numbers (1,-3,-4,-1,3,4) is divisible by 13.

[Guest]

DeeGee... U r not correct...

[Guest]

number will 120

25th digit 2

26th digit 0

[Guest]

The rule to check the divisibility by 13 is:

Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.

Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

111,111 is divisible by 13 (13 x 8547)

the same with 111,111,111,111 is divisible by 13 (13 x 8547008547)

following the same pattern, we know the fisrt twentyfour 1's are divisible by 13 and the last 24 1's also

11,111,111,111,111,111,111,111,1xx,111,111,111,111,111,111,111,111 can be expresssed as:

(111,111,111,111,111,111,111,111 * 10^26) + (xx * 10^24) + (111,111,111,111,111,111,111,111)

Since the first and the last numbers in parenthesis are divisible by 13, we are looking for those numbers "xx" divisble by 13 (different than 1),

the only chances are:

26

39

52

65

78

[Guest]

Don't forget 00.

[Guest]

The rule to check the divisibility by 13 is:

Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.

Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

111,111 is divisible by 13 (13 x 8547)

the same with 111,111,111,111 is divisible by 13 (13 x 8547008547)

following the same pattern, we know the fisrt twentyfour 1's are divisible by 13 and the last 24 1's also

11,111,111,111,111,111,111,111,1xx,111,111,111,111,111,111,111,111 can be expresssed as:

(111,111,111,111,111,111,111,111 * 10^26) + (xx * 10^24) + (111,111,111,111,111,111,111,111)

Since the first and the last numbers in parenthesis are divisible by 13, we are looking for those numbers "xx" divisble by 13 (different than 1),

the only chances are:

26

39

52

65

78

That is pretty good but why do you skip middle numbers with a "1" in them.

Let's say I wanted to determine if a number was divisible by three.

These all are:

222

222 222

222 222 222

These are not:

2

22

2 222

22 222

But the fact that four "2's" in a row are not divisible by 3, does exclude numbers like:

12 222

2 202 222

2 222 112 222

All of these are obviously divisible by two as well. So the list of numbers that can go in the middle of the ones are:

0

13

26

39

52

65

78

91

[Guest]

04, 17, 30, 43, 56, 69, 82, 95

[Guest]

Well luckily I wrote a long division program a while ago. I knew it would come in use someday .

13,26,39 etc

If you are interested when using the digits "13" 13 divides 8547008547008547008547008547008547008547008547 times

Don't you just love computers.

[CaptainEd]

The rule to check the divisibility by 13 is:

Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.

Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

As an aside, I'm excited to hear "the rule to check the divisibility by 13". I've always marveled that anyone ever came up with the rules for 7 and 11, but assumed that they were extremely special cases. Now that you refer so calmly to the existence of "the rule" for 13, I'm beginning to suspect that there's a place where all these rules can be found. And I want to see 'em! Where do you go to learn these divisibility rules?

[CaptainEd]

As an aside, I'm excited to hear "the rule to check the divisibility by 13". I've always marveled that anyone ever came up with the rules for 7 and 11, but assumed that they were extremely special cases. Now that you refer so calmly to the existence of "the rule" for 13, I'm beginning to suspect that there's a place where all these rules can be found. And I want to see 'em! Where do you go to learn these divisibility rules?

Never mind. The answer is, of course, "go to the internet", which is what I tell folks about any other question.