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LostInParadise's avatar

Is this a fun math fact?

Asked by LostInParadise (32037points) November 3rd, 2015

I just came across this and felt compelled to share. You do not need to be a math geek to appreciate it, though it probably helps.

Take a prime number other than 2 or 5, 7 for example. Now look at one over the number, 1/7 in our case. As I hope you recall, this will be a repeating decimal. For prime numbers, the sequence will always be of the form .a1 a2…an a1a2…an…, and so on.

1/7 = .142857142857… If the number of digits in the repeating sequence is even, take the last half and add it to the first half, 142 + 857 = 999, and you will always get all nines.

This is known as Midy’s Theorem. For the mathematically inclined, this Wikipedia article has a nice description.

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7 Answers

Rarebear's avatar

I have a better one.

dxs's avatar

I remember talking about this in a thread a while back. With 7ths, I like how the repeating sequence (i.e. 1/7=0.142857…) the next fraction is the same permutation, but starting on the next greatest number.
1/7 = 0.142857 (1 is smallest)
2/7 = 0.285714 (2 is next largest)
3/7 = 0.428571 (4 is next largest)
and so on…

dxs's avatar

I looked it up more, and it works with other primes as well, but sometimes with more than one sequence. For example, 13ths have two: 153846 and 076923. Interestingly, they don’t alternate.
1/13=0.076923…; repetend #1 (r1)
2/13=0.153846…; repetend #2 (r2)
Here’s all 13ths and their sequences:
01/13: r1
02/13: r2
03/13: r1
04/13: r1
05/13: r2
06/13: r2
07/13: r2
08/13: r2
09/13: r1
10/13: r1
11/13: r2
12/13: r1

Interestingly, they don’t alternate, but is it symmetric going up and down!

ARE_you_kidding_me's avatar

There are so many neat tricks
Any two digit number multiplied by 11:
90 X 11 = 9_(9+0)_0 = 990
52 X 11 = 5_(5+2)_2 = 572
25X11 = 2_(2+5)_5 = 275

dxs's avatar

@ARE_you_kidding_me WOAH! Tell us more tricks!

dxs's avatar

@ARE_you_kidding_me I just figured out that if he number you are multiplying with 11 has more than 2 digits (3, for example) you do something similar. With three, you add the first two, and then the last two. So 436*11= 4_(4+3)_(3+6)_2=4792. Originally, if I was presented with a multiplication problem n*11, I’d do n*10 + n, but seems easier.

Am I spamming this thread? If so I apologize—I just haven’t talked about numbers a lot recently.

LuckyGuy's avatar

@LostInParadise Very cool! I had not heard of this . Thanks for the introduction.

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