[ I0 1 ] 
numbers in the calendar, and the pafchal full 
moon, and the Sunday letter, anfwering thereto for 
that century (which hand as in tabic II), I conftrudt 
table- III, for finding Eafter Day during that century;, 
and obferve it never happens on the 22d of March, 
but when the golden number C, and the dominical 
letter D . 
v And the dominical letter D happens only in the 
4th, 9th, 15th, and 26th years of the folar cycle in 
that century, as appears by table I. 
Now the queftion is reduced to this, viz.. 
What number is there between 2200 and 2300 ^ 
which, being divided by 28, leaves either 4, 9, 15,, 
or 26 ; and being alfo divided by 19, leaves 6 ? 
SOLUTION. 
In the general theorem above, viz. ^txd — e \ d 
are given rt=28,/ =: i,r==2,as before ; and to find 
the values of d and e , 
We have ^+9 — 28 =: D = 4 ; therefore d ~ 
And becaufe e -f 1 — E — 6 ; therefore e 
in the theorem 
viz. 28 X 2 X I 8 + 23= I0 3 1 i and 10 3 1 + 53 2 
532 = 2095 : fo that this cyclar number will 
not do, the year falling either below or beyond the 
century required. 
2. Let D 9 ; the reft as before. Then fince 
d -]- 9 — 28 = 9 ; therefore d — 28, and ^ = 5 
as before; and 28 X 2 X 23 +28 = 1316; and 
2316 532 532 = 2380. This cyclar num- 
ber will not do, for the lame reafon as the laft. 
3 
. Let 
