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 4b4N & after (Posted on 2013-07-17)
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

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 Bonus later (spoiler) | Comment 1 of 3
The problem requires that 4*10^n + N = 4*(10N + 4)

Rearranging gives
4*10^n - 16 = 39N

So we need to find the smallest n for which 4*10^n -16 is divisible by 39.  This turns out (using Excel) to be n = 5.

N = (400000 - 16)/39 = 10256

Checking:

410256 does equal 4 * 102564

 Posted by Steve Herman on 2013-07-18 01:07:35

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