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.
If that was too easy, can we generalise the problem? Let K be our single-digit number, between 1 and 9 inclusive, such that KN = K*NK has a solution number N.
There is a very simple solution for K=1. In this case we seek 1N = 1*N1 and the smallest solution is N=1 with n=1. Notice that there are an infinite number of solutions in this case: N can be 1, 11, 111, 1111 and all further ‘repunits’, with lengths n equal to 1, 2, 3, 4 and so on, respectively.
So, the follow-up question is: 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?
Feel free to comment below. An answer will be posted next Sunday, with a new question on Monday.
The Math-e-Monday Puzzle: Peculiar Periodic Numbers
By Richard Mankiewicz | March 11th 2013 06:09 AM | Print | E-mail