I 6 BROWN : 



in order to square the record with the fact, Friday, October 

 5th, with letter E, was called Friday, October 15th, with letter 

 A. An advance of ten letters, or, leaving out the seven, 

 which does not influence the result, of three letters. When we 

 reach 1582, then, we have gone forward six letters, three from 

 G to C of the year " o " and three more on account of the 

 change at the calendar reformation. Now our problem sets 

 out to show how many letters the years of a given date will 

 carry us back from our base or bench mark, G. So we go 

 back six letters to G as a first step in the solution. We have 

 seen that the remainder centuries ( if any ) after the Dominical 

 Cycles are thrown out, cany us forward each two letters. So 

 we double and subtract from the backward six just mentioned, 

 leaving always a backward remainder of 6 or 4 or 2 or o. 



The odd years and a quarter of the same, neglecting frac- 

 tions, carry us back each one letter. To these add the back- 

 ward remainder from just above and divide the sum by seven. 

 The remainder shows how many places back from G the 3'ears 

 of our date have carried us, and subtracting this remainder 

 from seven shows how many letters forward from G. The 

 letter reached being the same in both cases. 



In other words, the last subtraction gives the numeral of 

 the required Dominical Letter. 



The rule and the reasons might now take this shape : — 



1 . Divide the centuries by 4. 



Because that throws out the Dominical Cycles which 

 do not affect the result. 



2. Multiply the remainder by 2. 



Because each remainder centuiy gives us two places 

 forward. 

 ^. Subtract the product from 6 and set down the remainder. 

 Because we have moved back six places in order to 

 start from G once more, having gone forward three for 

 the year " o " and three more at the calendar reforma- 

 tion. A backward remainder of o or of an even 

 number less than seven will alwavs be left. 



