And 299999999 is divisible by 13
And 299999999 is divisible by 13
26 comments
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⅐ = 0.1̅4̅2̅8̅5̅7̅
The above is 42857 * 7, but you also get interesting numbers for other subsets:
7 * 7 = 49 57 * 7 = 399 857 * 7 = 5999 2857 * 7 = 19999 42857 * 7 = 299999 142857 * 7 = 999999Related to cyclic numbers:
142857 * 1 = 142857 142857 * 2 = 285714 142857 * 3 = 428571 142857 * 4 = 571428 142857 * 5 = 714285 142857 * 6 = 857142 142857 * 7 = 999999 -
42857 for those who wonder
And for ops title: 23076923
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Actually disgusting
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I know you opened your calculator app to check it.
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Can we just say it isn't? Like that's an exception, and then the rest of math can just go on like normal
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49 is divisible by 7, so why not?
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Phew, for a moment I worried that 2.9999... was divisible by 7 and I woke up in some kind of alternate universe
26 comments
The divisability rule for 7 is that the difference of doubled last digit of a number and the remaining part of that number is divisible by 7.
E.g. 299'999 → 29'999 - 18 = 29'981 → 2'998 - 2 = 2'996 → 299 - 12 = 287 → 28 - 14 = 14 → 14 mod 7 = 0.
It's a very nasty divisibility rule. The one for 13 works in the same way, but instead of multiplying by 2, you multiply by 4. There are actually a couple of well-known rules for that, but these are the easiest to remember IMO.
Thanks I hate it
This math will not stand man!
If all of the digits summed recursively reduce to a 9, then the number is divisible by 9 and also by 3.
If the difference between the sums of alternating sets digits in a number is divisible by 11, then the number itself is divisible by 11.
That’s all I can remember, but yay for math right?
Well, on the side of easy ones there is "if the last digit is divisible by 2, whole number is divisible by 2". Also works for 5. And if you take last 2 digits, it works for 4. And the legendary "if it ends with 0, it's divisible by 10".
The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)
I think it might be easier just to do the division.