Challenge 22: Half Again As Big

[Greg Bernhardt]

What is the smallest integer such that if you rotate the number to the left you get a number that is exactly one and a half times the original number?

(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

 

[Citan Uzuki]

Spoiler

1,176,470,588,235,294

 

[Greg Bernhardt]

Spoiler

1,176,470,588,235,294

How did you come to it?

 

[Simon Bridge]

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

Spoiler

This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e., every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).

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!)

 

[Citan Uzuki]

Sorry for just posting the solution earlier, I was late to dinner. Simon, that's pretty creative :) I actually used a different (and somewhat more boring) method though:

Spoiler

Let a be the first digit of the number, b the number formed from the remaining digits, and n the number of digits in b. Then we are trying to find a, b, and n such that:
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

 

[Simon Bridge]

A good starting place for people who want to learn how to approach this sort of thing from scratch:

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.

 

[Joffan]

The answers rotating the number to the right as not quite as interesting - but just as accessible with Citan Uzuki's method. For the nth largest solution, there is exactly one case where the right-rotation number is larger than the left-rotation number.

 

[PeroK]

Spoiler

I tried this and did it the same way as Citan. For an n+1 digit number I got
$(3 \cdot 10^n - 2)x_n = 17(x_{n-1} ... x_0)$
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.

 

[Joffan]

If you take the set of answers to this puzzle as $a_n$ and the set of answers to the puzzle modified to rotate "to the right" as $b_n$, how often is $b_k > a_k$?

 

[Joffan]

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