How to find the 18th steady-3 number ?

A number n is considered to be a steady-p number if n^p ends in the same digits. For example, the number n=25 is a steady-3 number since 25^3 = 15625. The number n=251 is also a steady-3 number since 251^3 = 15813251.

Determine the 18th steady-3 number that exists starting from the number 1. The 1st steady-3 number is n=1 (1^3 = 1). The 2nd steady-2 number is n=4 (4^3 = 64). You may want to use the num2str() function.

Answer 1

[Jan]

Let's assume the problem can be solved in Matlab without high precision toolboxes. Numbers can be represented exactly up to 2^53. Then searching for cubic numbers works until (2^53)^(1/3), which is 208063. This number looks small enought for a dump brute force approach.

The question looks like a homework. So please post, what you have tried so far and ask a specific question.

The hint of the num2str function is misleading. It is more efficient not to convert the data types. The MOD() function is better. Remember, that the number k has n = floor(log10(k)) + 1 digits. Then mod(k.^3, 10.^n) replies the rightmost n digits.

This is half of the required code already. Please try to finish this and ask again in case of problems.

PS. The result is smaller than 1000.

[Syed Hafiz]

Hello Jan, thanks for your response. It was in fact a previous quiz question I was trying to solve which during the time given I failed miserably to attempt. But I wanted to go forward and try and learn how to write such algorithms for future problems.This is the code I wrote and I was thinking if I can somehow modify this to get to the result i want

[Syed Hafiz]

i know my indexing is messed up, so I was hoping we could do it the way I'm trying to approach

[Syed Hafiz]

Never mind i figured it out, the log10 thing didn't come intuitively to me though, so took some time to wrap my head around it, thanks once again !

[Jan]

Omit the brute clearing header. Especially clear all removes all loaded functions from the memory. Reloading them from the slow disk wastes time only without any benefits. Teachers mention this header, because they have been told to use it when they were students. This is "cargo cult programming".

Here a version without a loop:

k = 1:208063;
n = floor(log10(k)) + 1;
result = k(mod(k.^3, 10.^n) == k);
fprintf('%3d: %8d\n', [1:numel(result); result]);

num2str is a wrapper for sprintf. Then it is faster to call it directly. Here a version with string conversion:

c = 0;
for k = 1:1000
   if endsWith(sprintf('%d', k^3), sprintf('%d', k))
      c = c + 1;
      fprintf('%3d: %6d\n', c, k);
   end
end