Every year admits a Friday the 13th

Theorem:

Every year admits a Friday the 13th

Proof:

Note that a year admits a Friday the 13th if there is a month that begins with Sunday. We denote the days of the week with numbers. Specifically, 0 denotes Sunday, 1 Monday, up to 6, a number that denotes Saturday. If k is a natural number greater than 6 then the day it represents is k mod 7. So suppose that the first of January begins on day n with n a natural number between 0 and 6. To prove the theorem it is sufficient to show that there is a month in which the first day begins with k where k=0 mod7. The month of February will begin on n + 31 mod 7 = n + 3 mod 7, the month of March will begin on (n+3)+28 mod 7 = n + 31 mod 7=n+3 mod 7, the month of April will begin on (n+3)+31 mod 7=n+6 mod 7, the month of May will begin on (n+6)+30 mod 7=n+36 mod 7=n+1 mod 7, the month of June will begin on (n+1)+31 mod 7= n+4 mod 7, the month of July will begin on (n+4)+30 mod 7 = n + 34 mod 7 = n+6 mod 7, the month of August will begin on (n+6)+31 mod 7= n + 2 mod 7, the month of September will begin on:(n+2)+ 31 mod 7=n+5 mod 7. We could go on until December, but that’s enough. In fact, we have shown that for every natural number j between 0 and 6 (extremes included), there is a month of the year that begins on day n+j. Then, if we say j’ is the natural number between 0 and 6 such that n+j’=0 mod 7, we obtain that the month that begins with n+j’ is the one whose 13th day is Friday. Q.E.D