Simple division
- How much is 210526315789473684 divided by 2?
- Let me think.
- Excellent idea!
- 2 / 2 = 1, 10 / 2 = 05, 52 / 2 = 26, … What is going on here?!
- You tell me.
- I get 105263157894736842. The number I started with where the first digit has been added on as the last.
- Amazing!
- Are you pulling my leg?
- Find the smallest number that behaves like that.
- Like what?
- If you move its first digit, let’s call it k, to the end you’ll get a number that is k times smaller than the original.
- 1!
- Now you are pulling my leg!
Problem source: http://juegosdeingenio.org/
April 6th, 2009 at 1:35 pm
k=3: 3103448275862068965517241379
k=4: 410256
k=5: 510204081632653061224489795918367346938775
k=6: 6101694915254237288135593220338983050847457627118644067796
k=7: 7101449275362318840579
k=8: 8101265822784
k=9: 91011235955056179775280898876404494382022471
so k=4 gives the smallest solution (except for 1 which you pointed out).
Working the digits out is just a matter of starting with k as the first digit, then dividing by k using short division. For example, with k=4:
4 / 4 = 1.
1 / 4 = 0 remainder 1.
10 / 4 = 2 remainder 2.
22 / 4 = 5 remainder 2.
25 / 4 = 6 remainder 1.
16 / 4 = 4 which is where we started.
April 9th, 2009 at 4:28 am
Related story: http://tierneylab.blogs.nytimes.com/2009/04/06/freeman-dysons-4th-grade-math-puzzle/
October 18th, 2009 at 11:21 pm
We want 2 * N:2 = 2:N (: string concatenator)
The problem may be written as:
2*(10*N + 2) = 2*10^n + N (N is an n digit number)
=> 20*N + 4 = 2*10^n + N
=> 19*N = 2*10^n – 4 (equation 1)
Using modulo 19 arithmetic.
=> 0 = 2*10^n – 4 (mod 19)
=> 2*10^n = 4 (mod 19)
=> 10^n = 2 (mod 19)
=> 10^(n+1) = 1 (mod 19)
Fermat little theorem: if a is relatively prime to prime p
then a^(p-1) = 1 (mod p). And p-1 is the smallest power
that does that trick.
In our case 10^(19-1) = 1 (mod 19) is the useful result
Comapring powers => n+1 = 19-1 => n = 17
and this is the smallest value of n which satisfies equation 1
Substituting n = 17 into equation 1 =>
N = 10526315789473684
Therefore the required (initial) number is 105263157894736842
You can add multiples of 18 to 17 (n).
e.g. n=35 => 105263157894736842105263157894736842
n= 53 => 105263157894736842105263157894736842105263157894736842.
October 19th, 2009 at 4:29 am
Although can go directly from 2*10^n = 4 to 10^n =2, I had multiplied both sides by 10 and reduced afterwards. I accidentally deleted that line when editing.
October 19th, 2009 at 4:38 am
I just re-read the question. I haven’t answered it. Sorry. I hope you like my “solution” anyway.