GNU Emacs Calc 2.02 Manual
3.7.38 Types Tutorial Exercise 6
The full rule for leap years is that they occur in every year divisible
by four, except that they don't occur in years divisible by 100, except
that they do in years divisible by 400. We could work out the
answer by carefully counting the years divisible by four and the
exceptions, but there is a much simpler way that works even if we
don't know the leap year rule.
Let's assume the present year is 1991. Years have 365 days, except
that leap years (whenever they occur) have 366 days. So let's count
the number of days between now and then, and compare that to the
number of years times 365. The number of extra days we find must be
equal to the number of leap years there were.
 1: <Mon Jan 1, 10001> 2: <Mon Jan 1, 10001> 1: 2925593
. 1: <Tue Jan 1, 1991> .
.
' <jan 1 10001> RET ' <jan 1 1991> RET 

 3: 2925593 2: 2925593 2: 2925593 1: 1943
2: 10001 1: 8010 1: 2923650 .
1: 1991 . .
.
10001 RET 1991  365 * 

There will be 1943 leap years before the year 10001. (Assuming,
of course, that the algorithm for computing leap years remains
unchanged for that long. See section 5.9 Date Forms, for some interesting
background information in that regard.)