Every now and then Pickover tosses out an extremely obvious puzzle. Given the following sequence of numbers:<\/p>\n
\r\n 1, 27, 125, 343, 729, 1331, 2197, 3375, x\r\n<\/pre>\nwhat is x? Now, when I say “obvious” in this context I should clarify that the solution is relatively straightforward to the math-trained mind. Typically the solutions to such numerical puzzles are along the lines of “The latitude coordinates of the original U.S. colonies multiplied by the longitude coordinates of the same, then add 3.”<\/p>\n
Anyway, that was today’s puzzle. The puzzle for April 4 was to find the numbers, A and B, that satisfy the following equation:<\/p>\n
\r\n AB<\/sup> + BA<\/sup> = A7\r\n<\/pre>\nMy first impulse was there should be some way to solve for A and B, maybe involving logarithms or something, until I noticed that there were 2 variables and only 1 equation which, per my math classes of long ago, proves difficult. So, a brute force program it was. Surprisingly, there is only 1 solution.<\/p>\n
See [more] for the answers, I mean, if you really care.<\/em><\/p>\n
I have a stack of puzzles to work through that should be brute-force-able using a combination of permutation logic and graph techniques. Eventually, I imagine I will be able to assemble a program toolkit to rip through an entire year of these puzzles automatically.<\/p>\n
<\/p>\n
First puzzle: The numbers represent odd numbers, beginning with 1, raised to the 3rd power. Thus, x = 173<\/sup> = 4913.<\/p>\n
Second puzzle: 52<\/sup> + 25<\/sup> = 25 + 32 = 57, as found by this C code:<\/p>\n
\r\n#include <<\/font>stdio.<\/b><\/font>h><\/font>\r\n#include <<\/font>math.<\/b><\/font>h><\/font>\r\n\r\nint<\/b><\/font> main (<\/b><\/font>int<\/b><\/font> argc,<\/b><\/font> char<\/b><\/font> *<\/font>argv[<\/b><\/font>]<\/b><\/font>)<\/b><\/font>\r\n{<\/b><\/font>\r\n int<\/b><\/font> A,<\/b><\/font> B;<\/b><\/font>\r\n\r\n for<\/b><\/font> (<\/b><\/font>A =<\/font> 0<\/font>;<\/b><\/font> A <<\/font>=<\/font> 9<\/font>;<\/b><\/font> A<\/font>+<\/font>+<\/font>)<\/b><\/font> {<\/b><\/font>\r\n for<\/b><\/font> (<\/b><\/font>B =<\/font> 0<\/font>;<\/b><\/font> B <<\/font>=<\/font>9<\/font>;<\/b><\/font> B<\/font>+<\/font>+<\/font>)<\/b><\/font> {<\/b><\/font>\r\n if<\/b><\/font> (<\/b><\/font>(<\/b><\/font>int<\/b><\/font>)<\/b><\/font>pow(<\/b><\/font>A,<\/b><\/font> B)<\/b><\/font> +<\/font> (<\/b><\/font>int<\/b><\/font>)<\/b><\/font>pow(<\/b><\/font>B,<\/b><\/font> A)<\/b><\/font> =<\/font>=<\/font> 10<\/font> *<\/font> A +<\/font> 7<\/font>)<\/b><\/font>\r\n printf (<\/b><\/font>\"solution: %d^%d + %d^%d = %d7\\n\"<\/font>,<\/b><\/font>\r\n A,<\/b><\/font> B,<\/b><\/font> B,<\/b><\/font> A,<\/b><\/font> A)<\/b><\/font>;<\/b><\/font>\r\n }<\/b><\/font>\r\n }<\/b><\/font>\r\n\r\n return<\/b><\/font> 0<\/font>;<\/b><\/font>\r\n}<\/b><\/font>\r\n<\/pre>\n