Proof as follows.
Assume that 13th Jan is on day 1 (of 7) - it doesn't matter which. Jan has 31 days, so 4 weeks & 3 days, so 13th Feb will be on day 4 and so on - table below.
| Month | Normal Year | Leap Year |
| Jan | 1 | 1 |
| Feb | 4 | 4 |
| Mar | 4 | 5 |
| Apr | 7 | 1 |
| May | 2 | 3 |
| Jun | 5 | 6 |
| Jul | 7 | 1 |
| Aug | 3 | 4 |
| Sep | 6 | 7 |
| Oct | 1 | 2 |
| Nov | 4 | 5 |
| Dec | 6 | 7 |
As you can see from this, in both cases, all 7 numbers / weekdays are used. Therefore regardless of which day of the week the year starts (or 13th of Jan is on - these conditions are equivalent for this argument), there will always be at least one Friday 13th. Also, you can see that the most there can be is 3 in a year - if Feb 13th is a Friday (for normal years) or if Jan 13th is (for leap years).
So this year July 13th will be a Friday (although you could have got this trivially from any calendar) as 2007 is a 7 = "Friday" year (and is not a leap year).
Inspired by posting the TerryWatch poll for this week. Now back to being rude about things :)
S-E
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