Shift the digits

Question

Here, x (supplied as input) and n (the result of your computation) are both positive integers. n * x = n shifted. Find n.

Here's an example of shifting:

123456789 -> 912345678
abcdefghi -> iabcdefgh (letters = any 0~9 digit)
123       -> 312

Shifting only happens once to the right. Shifting left, e.g.

123456789 -> 234567891 

is not a valid shifting.

Rules

• Preceding zeros count after shifting. If the number is 10 and is multiplied by 0.1 (0.1 isn't a valid input), the result is 1, which isn't equal to 01 (10 after shifting).
• If your number only has one digit, the shifted result is your number:

1 -> 1
4 -> 4
9 -> 9

• Given enough time and resources, your program/function should work for any x input, but you only have to support x in the range [1,9] without timing out on Try It Online.

Test cases

For more test cases, this is OEIS A092697.

1 -> 1 (1 * 1 = 1 shifted.)

9 -> 10112359550561797752808988764044943820224719
(In this test case, x = 9 and n = 10112359550561797752808988764044943820224719.
n shifted = n * x =              91011235955056179775280898876404494382022471)

6 -> 1016949152542372881355932203389830508474576271186440677966
```

Answer 1

Python 2, 31 bytes

lambda x:10**1045044/(10*x-1)*x

Try it online!

Produces enormous million-digit numbers. Uses the formula from the OEIS page, x*(10^m-1)/(10*x-1). But, instead of using the smallest possible m for the given x, we hardcode m=1045044, which works for each x from 1 to 9. (I'm assuming only these x need to be supported, because bigger ones are impossible.)

We need an m that's divisible by the order of 10 modulo 10*x-1. So, to get one that works for all x, we take the least common multiple (LCM) of the orders, giving m=1045044.

x order
-----
1 1
2 18
3 28
4 6
5 42
6 58
7 22
8 13
9 44

We save a few bytes by doing 10^m/(10*x-1) instead of (10^m-1)/(10*x-1), since Python 2's floor division cuts off the remainder and gives the same result, as long we move the *x after the division.

Despite how the massive the results are, TIO can in fact compute them for every x from 1 to 9 without timing out. Validating the rotating property of an output string-wise takes longer, but if you set validate=True in the TIO link and the limit to a single x, this also completes before timing out. I've also validated all the results on my machine.

Answer 2

JavaScript (Node.js),  54 48  44 bytes

Saved 4 bytes thanks to @xnor

Takes input as a BigInt. Based on the formula given in A092697.

f=(x,m=t=10n)=>~-m%(q=t*x-1n)?f(x,m*t):m/q*x

Try it online!

---

JavaScript (Node.js),  165 153  151 bytes

Takes input as a BigInt and performs a recursive search.

f=(x,n='',i=-1)=>+n[[d,...a]=[...x*BigInt(n)+''],s=a.join``+d,0]&&n==s?o=n:n.slice(0,i++)==s.slice(-i,-1)&i<60&&([...2**29+'4'].some(k=>f(x,k+n,i)))&&o

Try it online!

---

NB: Both solutions assume \$1\le x\le9\$.

Answer 3

Wolfram Language (Mathematica), 42 bytes

#(10^MultiplicativeOrder[10,a=10#-1]-1)/a&

Try it online!

xnor's method n-shifted my bytes...

Wolfram Language (Mathematica), 24 bytes

#(10^1045044-1)/(10#-1)&

Try it online!

Answer 4

Jelly, 13 bytes

⁵×’⁵*“ÐṂ+’¤:×

Try it online! (The 1045044 digits do not fit within TIO's output)

An implementation of xnor's LCM method.

⁵×’⁵*58!$¤:× works in theory for 12, but there's no chance of calculating results with \$58!\$ digits within 60s using Jelly.

⁵×’⁵*“ÐṂ+’¤:× - Link: integer,    x
⁵             - 10                10
 ×            - multiply          10x
  ’           - decrement         10x-1
          ¤   - nilad followed by link(s) as a nilad:
   ⁵          -   10              10
     “ÐṂ+’    -   1045044         1045044
    *         -   exponentiate    10^1045044
           :  - integer division  10^1045044//(10x-1)
            × - multiply          10^1045044//(10x-1)*x

Answer 5

05AB1E (legacy), 12 bytes

T*<•GIs•°s÷*

Try it online!

T*             - Multiply input by 10 
  <            - subtract 1 
   •GIs•°      - 10**1045044
         s÷    - Integer divide
           *   - Multiply by input again 

Port of xnor's answer