Prove that in each year, the 13th day of some month occurs on a Friday Prove that in each year, the 13th day of some month occurs on a  Friday.

No clue... please help!
Let us ignore leap years for a moment.

And let us assume that some date (e.g., 13th) falls on Monday in January. 

What happens in February? January has $31=4\cdot 7+3$ days, so the days move by 3 to Thursday.

So if we denote the days in week by numbers $\{0,1,2,3,4,5,6\}$, we only have to consider the number of days in a month, compute modulo 7 and see what happens:

0 January (31 days)
3 February (28 days)
3 March (31 days)
6 April (30 days)
1 May (31 days)
4 June (30 days)
6 July (31 days)
2 August (31 days)
5 September (30 days)
0 October (31 days)
3 November (30 days)
5 December (31 days)
1 January next year 


(Computing day in January in the next year is irrelevant for this question; but comparing this with an actual calendar is a good sanity check.)

Notice, that all numbers 0,1,2,3,4,5,6 appear in the above table.



Now what happens if a year starts not on Monday (=0) but on Tuesday. You simply have to add $+1$ (computing modulo 7) in each row of the table. But since every number appeared at least one, we will get $6+1=0$ in some row. And the same is true for any other day.

So now you only have to create similar table for a leap year, check whether all numbers from 0 to 6 appear there, and you are done.



Somewhat related: Occurrence of Friday the 13th on Wikipedia.
I think this question has already good answers,

It is enough to show that there exist a month  starting on Sunday.

Let $x$ be the starting day of the year then next month's first day is $x+31\equiv x+3$ and next month's first day, $x+3+28\equiv x+3$...

At the and we have, $$x,x+3,x+3,x+6,x+1,x+4,x+6,x+2,x+5..$$ now we reached seven different days, $x,x+1...,x+6$ by pigeonhole principle we must have all days which conclude the result.