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(To rotate the number left, take the first digit off the front and append it to the end of the number. 2591 rotated to the left is 5912.)

**Please make use of the spoiler tags**

- Thread starter Greg Bernhardt
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- #1

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- 7,947

(To rotate the number left, take the first digit off the front and append it to the end of the number. 2591 rotated to the left is 5912.)

- #2

- 296

- 15

1,176,470,588,235,294

- #3

- 18,264

- 7,947

How did you come to it?1,176,470,588,235,294

- #4

Simon Bridge

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Citan Uzuki simply needed to find a repeating fraction such that the number of repeating digits is one less than the number.

Now you need to find any occurrences where the nth digit is followed by the 1.5nth digit. 'n' will obviously be even.

The repeating digits in 1/7 and their corresponding 'n' where the digits are the first digit after the decimal point are:

142857 - repeating digits

132645 - n

The ordered pairs of adjacent values of n are (1,3), (3,2), (2,6), (6,4), (4,5) and (5,1). None of these has the second number equal to one and a half times the first number.

The next prime number p with a repeating fraction for 1/p containing p-1 repeating digits is 17. The repeating digits for 1/17 are:

.0588235294117647

The easiest way to assign 'n' for each digit is to list all of the digit pairs in order:

XX - n (digit)

05 - 1 (1)

11 - 2 (11)

17 - 3 (12)

23 - 4 (5)

29 - 5 (8)

35 - 6 (6)

41 - 7 (10)

47 - 8 (15)

52 - 9 (7)

58 - 10 (2)

64 - 11 (14)

70 - 12 (16)

76 - 13 (13)

82 - 14 (4)

88 - 15 (3)

94 - 16 (9)

Checking the even-numbered values of n in the above table reveals that ALL even values of n less than 2/3 of 17 are followed by the 1.5n digit. The values of n are and their digit places are:

2,3 (11,12)

4,6 (5,6)

6,9 (6,7)

8,12 (15,16)

10,15 (2,3)

So the five smallest numbers with this property are:

1176470588235294

2352941176470588

3529411764705882

4705882352941176

5882352941176470

Notice that 2352941176470588 can be rotated TWICE, yielding 1.5 times the number each time since the 4->6->9 n digits are in order.

And that's all there is to it. :D

(Probably did it the same way I did!)

- #5

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10b + a = 3/2 * (10^n a + b)

Rearranging this equation a bit, we obtain:

17b = (3*10^n - 2)a

Now, a is at most 9, hence not divisible by 17, therefore we must have that 17 divides 3*10^n - 2. This gives the congruence:

3*10^n ≡ 2 mod 17

Or multiplying both sides by 6 (i.e. 3^(-1) mod 17):

10^n ≡ 12 mod 17

By trial and error, I found that the smallest value of n satisfying this congruence is 15. The smallest possible value for a would of course be 1, and these values yield b = (3*10^15 - 2)/17 = 176,470,588,235,294, which indeed has 15 digits, so it gives us a valid solution

- #6

Simon Bridge

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If a number N has two digits, then N=10a+b where a and b are single digits.

Then the condition N has to satisfy is: 2(10a+b) = 3(10b+a) ... now try to show that this condition cannot be satisfied.

Therefore the correct N has more than 2 digits.

... generalize for any number n of digits.

Say - try doing it for 3 or 4 digit numbers before working out a condition to find the actual number of digits needed.

It's not the most elegant way to go about it - but it will help people learn maths.

Note: maths students are often taught the most elegant historical proofs for things, which ends up being a case of "you just have to know". But very often a theorem was proved by some messy approach and then the elegant approach was worked out. The messy way usually teaches more.

- #7

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- #8

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##(3 \cdot 10^n - 2)x_n = 17(x_{n-1} ... x_0)##

So must have congruence as above:

##3 \cdot 10^n = 2 \ mod \ 17##

As 17 is prime, we get:

##10^{n+1} = 1 \ mod \ 17##

n+1 = 16, 32 ...

So n = 15, 31 ...

Therefore, all solutions are as per Simon's post with the sequence of digits repeated. E.g. the next one is:

11764705842352941176470584235294

Etc.

Last edited:

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- #10

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(Ignore my previous post, I forgot I had already posted on this)

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