Find the smallest value of hexadecimal positive integer N which becomes 4*N when the last digit of N is moved to be the first. For example, turning (2345)₁₆ into (5234)₁₆. How about 2*N? 7*N? 8*N?

(In reply to 7*n, analytically by Steve Herman)

I had also been working on an analytical solution, with some use of inspection. The answer I found was a little smaller.

A hexadecimal number for N can be represented as
16ⁿa₀+16^n-1a₁+...+16¹a_n-1+16⁰a_n.

Thus, what is sought is where 
7×(16ⁿa₀+16^n-1a₁+...+16¹a_n-1+16⁰a_n)
= 16ⁿa_n+16^n-1a₀+...+16¹a_n-2+16⁰a_n-1.

Let D = 16^n-1a₀+...+16¹a_n-2+16⁰a_n-1, then
7×(16D + a₀) = 16ⁿa₀ + D, which can be expressed similar to your expression as D = a_n×(16ⁿ-7)/111

Using Excel to iterate through 1 to 15 to find n (which will result in an integer), it is found that n = 8, (or n = 12 : a₁₂ = 6), which gives a₈×(38693399).
Thus 7×N = a₈×(16⁸ + 38693399) = a₈×(4333660695).
Iterating values 1 to 15 for a₈ finds the first value where the second digit of 7×N in hexadecimal is not the digit 0 (as a leading 0 digit would not be permitted) is where a₈ = 7.
7×N = 30335624865 = 71024E6A1₁₆, and
N = 4333660695 = 1024E6A17₁₆.

Edited on December 24, 2014, 8:06 am

Posted by Dej Mar on 2014-12-24 08:06:23