Divisibility - Revision history

Admin: /* Proof */

2019-03-10T02:58:36Z

Proof

@@ Line 427: / Line 427: @@
 Reduce mod 13 as you go. If the final result is 0 or divisible by 13, then the entire number is divisible by 13.
 Quick example: 325
@@ Line 435: / Line 437: @@
 Bingo! 325 is divisible by 13. In fact it is 13*25.
 ==Divisibility by 37==

2019-03-10T02:50:13Z

Proof

← Older revision		Revision as of 02:50, 10 March 2019
Line 423:		Line 423:

	2. Add the next digit		2. Add the next digit

			3. Repeat step 1, then step 2, on the result

	Reduce mod 13 as you go. If the final result is 0 or divisible by 13, then the entire number is divisible by 13.		Reduce mod 13 as you go. If the final result is 0 or divisible by 13, then the entire number is divisible by 13.
Line 433:		Line 435:

	Bingo! 325 is divisible by 13. In fact it is 13*25.		Bingo! 325 is divisible by 13. In fact it is 13*25.


	==Divisibility by 37==		==Divisibility by 37==

2019-03-10T02:46:31Z

Divisibility by 37

← Older revision		Revision as of 02:46, 10 March 2019
Line 451:		Line 451:
	2. Add the next digit over		2. Add the next digit over

	3. Repeat step 1 on the result		3. Repeat step 1, then step 2, on the result

	Continue until all digits have been added. Reduce results modulo 37 as you go. If the result of this calculation is 0 or is divisible by 37, then the original number N is divisible by 37.		Continue until all digits have been added. Reduce results modulo 37 as you go. If the result of this calculation is 0 or is divisible by 37, then the original number N is divisible by 37.

2019-03-10T02:44:49Z

Divisibility by 11

2018-01-14T09:25:56Z

Flags

2018-01-14T09:25:32Z

Flags

2018-01-14T09:23:55Z

Divisibility by 7, 11, 13

← Older revision		Revision as of 09:23, 14 January 2018
Line 665:		Line 665:

	Can you demonstrate the test in general, on your own, two different ways? One can be based on the solution to the problem just shown. The other can be based on the division by 11 trick (mod 11, except use mod 1001 instead of mod 11).		Can you demonstrate the test in general, on your own, two different ways? One can be based on the solution to the problem just shown. The other can be based on the division by 11 trick (mod 11, except use mod 1001 instead of mod 11).

			=Flags=

	[[Category:Math]]		[[Category:Math]]
	[[Category:Crypto]]		[[Category:Crypto]]
	[[Category:Programming]]		[[Category:Programming]]

2018-01-14T09:23:46Z

Divisibility by 7, 11, 13

← Older revision		Revision as of 09:23, 14 January 2018
Line 648:		Line 648:
	==Divisibility by 7, 11, 13==		==Divisibility by 7, 11, 13==

	7, 11, and 13 are 3 consecutive prime numbers. Their product is 1001, and when a product is that close to a power ~~oof~~ 10, a test of divisibility is not far away.		7, 11, and 13 are 3 consecutive prime numbers. Their product is <math>7 \times 11 \times 13 = 1001</math>, and when a product is that close to a power of 10, a test of divisibility is not far away.

	The test is: separate the number from right to left in groups of 3 digits (the commas conventionally printed in a number do this for you.) Add the groups in even positions and the groups in odd positions. If the difference between the sums is divisible by 7, 11, or 13, the entire number is divisible by 7, 11, or 13, respectively. If the difference is 0, the number is divisible by 7, 11, and 13.		The test is: separate the number from right to left in groups of 3 digits (the commas conventionally printed in a number do this for you.) Add the groups in even positions and the groups in odd positions. If the difference between the sums is divisible by 7, 11, or 13, the entire number is divisible by 7, 11, or 13, respectively. If the difference is 0, the number is divisible by 7, 11, and 13.

2018-01-14T09:22:23Z

test for composite numbers

2018-01-14T09:21:50Z

test for composite numbers