(572)
| The Multiplices of 8, increased by 5, are | 13. | 21. | 29. | 37. | 45. | 53. |
| Those divided by 6, the Remainders are. | 1. | 3. | 5. | 1. | 3. | 5. |
Here you see 21 and 45 for the purpose, and take the Progression, adding the common Difference 24 (which is the least Dividend measured by 6 and 8) and you have 21. 45. 69. 93, 117. 141.
Admit, the Question had concerned these three Divisors:
| 6 | the Remainders being | 3 | Then dividing the former Progression by 9, the Remainders are 3. 0. 6. 3. 0. 6. | ||
| 8 | 5 | ||||
| 9 | 6 |
Wherefore I conclude, that the third and sixth of these Numbers are those sought, to wit, 69 or 141, and so on progressively; whereas, if you had propounded the Remainder of 9 to have been any other Number than 3, 0, 6, the Problem, as concerning all these, had not been possible
Some easie Cases of the Problem are these:
When the Remainder of some Divisor is 0, and of each of the rest of the Divisors, an Unit, or less by an Unit, than the Divisor.
In which Cafes you are to find such a Multiplex of the Produéft or least Dividend measurable by those Divisors that have Remainders, which increas'd or diminish'd by an Unit, may be a just Multiplex of that Divisor that hath no Remainder. These Cases are handled by Tacquet, and Bachet in his Problemes plaisans & delectables.
PROBLEM.
To find the Year of the Julian Period for any year of our Lord proposed.
It is necessary to be furnished with the Sun's Cycle, the Prime Number, and the Number of the Roman Indiction, which the industrious Mr. Street thus performs:
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