Solution to the Problem:
The answer is six.
Odd Day featuring three consecutive odd numbers happens only six times every 100 years:
1/3/5
3/5/7
5/7/9
7/9/11
9/11/13
11/13/15
Both James Alarie and Chad Fore pointed out that the answer depends upon which standard you use. In the
American Standard order (M/D/Y), you would have six days, but most other nations use a (D/M/Y) format, so there would only be
5 days since 11/13/15 would be the 11th day of the 13th month. But James alarie pointed out that this format is
excluded because the problem stated there would be one next year.
James went on to say that I did not specify whether the odd numbers had to be increasing or if they might be decreasing. So, he went
back to check the decreasing posibility and found:
There will be four M/D/Y format: decreasing:
5/3/1
7/5/3
9/7/5
11/9/7
This is forbidden by the statement that there was 1 last year.
There are five decreasing for the D/M/Y format:
5/3/1
7/5/3
9/7/5
11/9/7
13/11/9
This is forbidden by the statement that there will be 1 next year.
The Y/M/D decreasing format has 5:
5/3/1
7/5/3
9/7/5
11/9/7
13/11/9
This is forbidden by the statement that there will be 1 next year.