Sin(theeta) = square root(3)/2
theeta = sin inverse [square root (3)/2]
theeta = 60 degrees
as seen from trignometric table
We will set a variable, d, to represent the day of the week that January starts on. For instance, if it started on Monday, d + 1 would be Tuesday, d + 2 would be Wednesday, etc. up to d + 6 to represent the last day of the week (in our example, Sunday). The next week would start over at d, and the month would continue. For non-leap years:
If January starts on <u>d</u>, February will start 31 days later. Following our pattern above, this will put it at <u>d</u><u> + 3</u> (28 days would be back at d; 29 would be d+1, 30 would be d+2, and 31 is at d+3). In a non-leap year, February has 28 days, so March will start at <u>d</u><u>+3</u> also. April will start 31 days after that, so that puts us at d+3+3=<u>d</u><u>+6</u>. May starts 30 days after that, so d+6+2=d+8. However, since we only have 7 days in the week, this is actually back to <u>d</u><u>+1</u>. June starts 31 days after that, so d+1+3=<u>d</u><u>+4</u>. July starts 30 days after that, so d+4+2=<u>d</u><u>+6</u>. August starts 31 days after that, so d+6+3=d+9, but again, we only have 7 days in our week, so this is <u>d</u><u>+2</u>. September starts 31 days after that, so d+2+3=<u>d</u><u>+5</u>. October starts 30 days after that, so d+5+2=d+7, which is just <u>d</u><u />. November starts 31 days after that, so <u>d</u><u>+3</u>. December starts 30 days after that, so <u>d</u><u>+5</u>. Remember that each one of these expressions represents a day of the week. Going back through the list (in numerical order, and listing duplicates), we have <u>d</u><u>,</u> <u>d,</u><u /> <u>d</u><u>+1</u>, <u>d</u><u>+2</u>, <u>d+3</u><u>,</u> <u>d</u><u>+3</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u>+5</u>, <u>d</u><u>+5</u>, <u /><u /><u>d</u><u>+6</u><u /><u /> and <u>d</u><u>+6</u>. This means we have every day of the week covered, therefore there is a Friday the 13th at least once a year (if every day of the week can begin a month, then every day of the week can happy for any number in the month).
For leap years, every month after February would change, so we have (in the order of the months) <u></u><u>d</u>, <u>d</u><u>+3</u>, <u>d</u><u>+4</u>, <u>d</u><u />, <u>d</u><u>+2</u>, <u>d</u><u /><u>+5</u>, <u>d</u><u />, <u>d</u><u>+3</u>, <u>d</u><u /><u>+6</u>, <u>d</u><u>+1</u>, <u>d</u><u>+4</u>, a<u />nd <u>d</u><u>+</u><u /><u /><u>6</u>. We still have every day of the week represented, so there is a Friday the 13th at least once. Additionally, none of the days of the week appear more than 3 times, so there is never a year with more than 3 Friday the 13ths.<u />
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).