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4b4N & after (Posted on 2013-07-17) Difficulty: 3 of 5
Let N be a non-zero natural number composed of n digits.
Let us then both prepend and append the number 4 to create two new numbers, 4N and N4, that are both (n+1) digits long.
For example, if N is 123, then we create two numbers: 4123 and 1234.
The question is to find the smallest value of N, such that the following equation holds true:
4N = 4*N4
Again, using the example above, this would require that 4123 = 4*1234.
This is obviously not true, so N=123 is not a solution.

So, find the smallest value of N and its length n.


Bonus:
A generalized question: For which values of K can we find a value
of N (of length n) that solves the general equation KN = K*NK, as defined above?


source: March issue of Science 2.0

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: Bonus later (spoiler) Comment 3 of 3 |
(In reply to Bonus later (spoiler) by Steve Herman)

The different rearrangement k(k-10^n+10N) = N, when (k-10^n+10N) = M, for some M, and N=kM seems to suggest that for big enough N, there will always be a solution.

Certainly there are solutions for all k up to 30, though some N are big; in the case of 29, for example, n=271, with N, necessarily, a 271 digit number:

10034602076124567474048442906574394463667820069204152/
24913494809688581314878892733564013840830449826989619/
37716262975778546712802768166089965397923875432525951/
55709342560553633217993079584775086505190311418685121/
10726643598615916955017301038062283737024221453287197/
231831

Edited on July 18, 2013, 10:05 am
  Posted by broll on 2013-07-18 09:57:04

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