AGTSSP - Stage 5 - Two's Company
Click here for the cache page. This is a simple numbers puzzle with difficulty = 2. The part we need is:
If A = the first day of this year, B was 9 days later, C the day after that, and D, E, F are the same days next year, then the cache is at:
N51 (B-A-1).(D-A)+(E-B) W000 (E-D+2).(F-A) (F-C)
Hint - I'm British
If A = the first day of this year, B was 9 days later, C the day after that, and D, E, F are the same days next year, then the cache is at:
N51 (B-A-1).(D-A)+(E-B) W000 (E-D+2).(F-A) (F-C)
Hint - I'm British
Solution
I tell you it is a simple numbers puzzle, so the title would imply a binary puzzle.
The date of publication (DOP) is 03/08/2010, so it was published in the year 2010. The first day of the year was 01/01/10, 9 days later was 10/01/10 (British way of writing it) and the day after that is 11/01/10. If we write those as binary numbers we would get 010110, 100110 and 110110, which give A = 22, B = 38 and C = 54.
If we do the same thing for next year we get D = 010111 = 23, E = 100111 = 39 and F = 110111 = 55.
Plug those into the formula and you get N51 15.002 W000 18.331
The date of publication (DOP) is 03/08/2010, so it was published in the year 2010. The first day of the year was 01/01/10, 9 days later was 10/01/10 (British way of writing it) and the day after that is 11/01/10. If we write those as binary numbers we would get 010110, 100110 and 110110, which give A = 22, B = 38 and C = 54.
If we do the same thing for next year we get D = 010111 = 23, E = 100111 = 39 and F = 110111 = 55.
Plug those into the formula and you get N51 15.002 W000 18.331