5001 being divisible by 3 doesnt feel right

Дата публикации: 19-07-2026 03:04:14

rosszulorzott:unnatural-transformations:unnatural-transformations:noelements-setempty:shinyhappygoth:dajo42:dajo42:5001 being divisible by 3 doesnt feel rightShortcuts to determine if an integer is divisible by:This is a given.If the last digit is divisible by 2 (a.k.a. even), then so is the whole number.If the sum of the digits is divisible by 3, then so is the whole number. The recursivity of this means that if the sum has multiple digits, you can add them up again until you get a single digit and see if it’s 3, 6, or 9.Like the rule for 2, but check if the last two digits are divisible by 4.If it ends in 5 or 0.If the rules for both 2 and 3 apply.No shortcut. Alas.Like the rules for 2 and 4, but check if the last three digits are divisible by 8. (Yes, this pattern keeps going for 16, 32, etc.)Like the rule for 3, but the sum of the digits (or the sum of the sum of the digits, etc.) must be 9.If it ends in 0.11. Put alternating + and - signs in front of each digit such that the last digit has a + sign (so for 5001 it’s - 5 + 0 - 0 + 1). If the answer is divisible by 11, so is the original number. This is recursive like the 3 and 9 rules.7: Subtract twice the last digit of your number from the number you get by ignoring the last digit. If the result of that operation is divisible by 7, so is your original number (eg, 91 -> 9 - 2×1 = 7, so 91 is a multiple of 7). As with the rules for 3, 9 and 11 this rule is recursive (2261 -> 226 - 2×1 = 224; 224 -> 22 - 2×4 = 14; 14 -> 1 - 2×4 = -7 and so 2261 is divisible by 7). It’s not a fast way of checking if a number is a multiple of 7 (you honestly may as well just do the division), but it does work.In fact, I think, something along these lines works for any prime (excluding 2 or 5).Suppose that p is such a prime. Then, by Bezout’s lemma, there exist integers X, Y such that 10X + pY = 1 (and clearly any such X is not a multiple of p). Let N := 10a + b. Then p | N if and only if p | NX. But NX = 10aX + bX = (1-pY)a + bX = (a+bX) + p(aY). So p | N if and only if p | (a+bX).For p=7 we have, for example, (X,Y) = (-2,3). Hence 7 divides 10a+b if and only if 7 divides a-2b. For p=3 we have (X,Y) = (1,-3). So 3 divides 10a+b if and only if p divides a+b (which implies the standard sum of digits test given above). For p=11 we have (X,Y) = (-1,1). So 11 divides 10a+b if and only if 11 divides a-b (which implies the alternating sum of digits test given above). And once you find a suitable (X,Y) you can derive divisibility tests for larger primes as well.well that escalated quickly@matetikus

Схожие новости

#	Наименование новости	Тональность	Информативность	Дата публикации
1	source	0	31.2	08-08-2026
2	FORMIDABLE FOES 🗡️💥	0	0	17-06-2026
3	I’m starting a collection	0	34.61	08-08-2026
4	Sjsjiwjwn	0	38.5	05-08-2026
5	laney	0	39.98	08-08-2026
6	XENAGABRIELLE + height difference	0	25	04-08-2026
7	random lockscreens!!!!! like if save or use	0	29.46	08-01-2026
8	This is a lot to unpack	0	37.7	02-08-2026
9	The original four	0	30.04	13-07-2026
10	Hát, nem. Megtaláltam a blokklistán, de hogy hogy került oda, nem tudom. @drszoszie, Sok…	0	18.82	03-07-2026

Классификация: . Схожих патентов: 0. Схожих новостей: 10. Тональность: 0. Информативность: 27.23. Источник: matetikus.tumblr.com.