Prove that every year has at least one Friday the 13th

Question

Everyone knows Friday the 13th is regarded as a day of bad luck.
Why does every year have at least one of this bad day?

Answer 1

A month has a Friday 13th if and only if it begins on a Sunday.

On a regular (non-leap) year, if January begins on day $k$, $0\leq k\leq 6$ (with $k=0$ being Sunday), then we have that:

• January begins on day $k$;
• February begins on day $k+3\bmod 7$ (since January has 31 days, and $31\equiv 3\pmod{7}$;
• March begins on day $k+3\bmod 7$;
• April begins on day $k+6\bmod 7$;
• May begins on day $k+8=k+1\bmod 7$ (since April has 30 days, and $30\equiv 2\pmod{7}$);
• June begins on day $k+4\bmod 7$;
• July begins on day $k+6\bmod 7$;
• August begins on day $k+9 = k+2\bmod{7}$;
• September begins on day $k+5\bmod 7$;

With these, we already have day $k$, $k+1$, $k+2$, $k+3$, $k+4$, $k+5$, and $k+6$, so at least one of these months will begin on Sunday, guaranteeing at least one Friday 13th.

For Leap years, the analysis is similar, except that:

• January begins on day $k$;
• February begins on day $k+3$;
• March begins on day $k+4$;
• April begins on day $k$;
• May begins on day $k+2$;
• June begins on day $k+5$;
• July begins on day $k$;
• August begins on day $k+3$;
• September begins on day $k+6$;
• October begins on day $k+1$.

So at the latest, you will have a Friday 13th by October.

Answer 2

In a common year the 13ths of the month fall at intervals of $31,28,31,30,31,30,31,31,30,31$,and $30$ days, which reduce mod $7$ to $3,0,3,2,3,2,3,3,2,3$, and $2$. The partial sums mod $7$ are $3,3,6,1$, $4,6,2,5,0,3$,and $5$; since these include a complete residue system mod $7$, at least one of those 13ths must fall on a Friday.

In a leap year the corresponding numbers are $31,28,31,30,31,30,31,31,30,31$,and $30$ for the intervals, $3,1,3,2,3,2,3,3,2,3$, and $2$ for their reductions mod $7$, and $3,4,0,2,5,0,3,6,1,4$, and $6$ for the partial sums mod $7$; once again the partial sums include a complete residue system mod $7$, and one of the 13ths must fall on a Friday.

Answer 3

This is not an elegant method, and maybe there is none, but it works. There are 14 possible one-year calendars:

the common year beginning on Sunday, the common year beginning on Monday, the common year beginning on Tuesday, and so on....

and then the leap year beginning on Sunday, the leap year beginning on Monday, and so on....

Look at each of the 14 calendars, and count the number of Friday the 13ths.

Answer 4

You may prefer to focus only on the months May through November inclusive, and show that the 13ths of those seven months fall on seven different days of the week.

Working mod 7, and letting May 13th = X,

May 13th = X

June 13th = X + 31 = X + 3

July 13th = X + 31 + 30 = X + 5

August 13th = X + 31 + 30 + 31 = X + 1

September 13th = X + 31 + 30 + 31 + 31 = X + 4

October 13th = X + 31 + 30 + 31 + 31 + 30 = X + 6

November 13th = X + 31 + 30 + 31 + 31 + 30 + 31 = X + 2

Answer 5

.The maximum number of days in a month is 31 .The number of days of a week is 7 .The minimum number of days in a year is 315 .31*7=217 .217 means there is all kind of combination of days and weeks .217<315 .Thus there is always friday the 13th every year, not just friday the 13th but all kind of dates as well.

This is probably the most easiest way to explain this :D