CHAPTER XXV. MISCELLANEOUS PROBLEMS. CHRONOLOGY. ACCORDING to the Julian Calendar or OLD STYLE, the solar year was considered as being 365 days and 6 hours. The 6 hours in 4 years amounted to a day, therefore every fourth year was called a Leap Year, and consisted of 366 days. But the true solar year is about 11 minutes less than the Julian year, and on this account, in 1582, it was found that Spring commenced 10 days later than at the establishment of the Julian Calendar. Pope Gregory the XIlIth, therefore, caused ten days to be taken out of the month of October in that year, and to prevent the recurrence of a similar variation, he ordered the centurial years should not be regarded as Leap Years, unless the number of centuries were divisible by 4. This computation, which is called the Gregorian or New STYLE, was soon adopted in the greater part of Europe ; but in England and America, the change was not made until 1752, when the error had amounted to eleven days. It was then ordered that the 3d of September should be called the 14th, and the Gregorian calendar adopted for the future. In Russia and Greece, the Old Style is still retained. One of the first seven letters, A, B, C, D, E, F, G, is attached to every day in the year; thus, A is applied to Jan, 1st, 8th, 15th, &c.; B, to Jan, 2d, 9th, 16th, &c.; C, to Jan. 3d, 10th, 17th, &c. In this manner all days in any year which have the same letter, fall on the same day of the week. The DOMINICAL LETTER for any year is the letter that falls against all the Sundays. Thus, the 5th of January, 1845, will fall on Sunday, and the dominical letter will, therefore, be the 5th letter, or E. But in Leap Year there are two dominical letters, the first for January and February, the second for the remainder of the year. The following are the dominical letters for a few years to come ; 1844, G, F; 1845, E; 1846, D; 1847, C; 1848, B, A; 1849, G; 1850, F; 1851, E; 1852, D, C, &c. PROBLEM I. To find the Dominical Letter for any year, according to the Julian or OLD STYLE. RULE. To the given year add one-fourth of itself, plus 4, and divide the sum by 7. If there is no remainder, the dominical letter is G; if 1 remainder, F; and so on in inverse order. If the given year be Leap Year, the letter thus found will be the dominical letter for the last 10 months, and the next following letter, for the remainder of the year. What was the dominical letter for A. D. 1531 ? To the given year, we add 4 one-fourth of itself, (rejecting the fraction,) and 4. Dividing 7)1917 this sum by 7, we have a re273 + 6 remainder. mainder 6, which indicates that the dominical letter sought is the 6th from G, counting in retrograde order, which is A. What were the dominical letters for A. D. 564 ? 564 The remainder 2, indicates that the dominical letter 141 is the 2d from G, or E. But the year being leap year 4 the dominical letter for January and February, will be 7)709 the next following, or F. The two letters sought are 101 + 2 therefore F, E. If the given year were before the Christian era, the remainder would indicate the direct order of the letter. Thus, 1 denotes A ; 2 denotes B; 5, E, &c. 1. What was the dominical letter for A. D. 769? PROBLEM II. To find the dominical letter for any year, according to the Gregorian or New Style. 3. 2 = RULE. Divide the centuries by 4, and take the remainder from Add twice this remainder to of the odd years, and divide the sum by 7. If there is no remainder, the domi. nical letter is G; if 1' remainder, F, &-c., as in the pre. ceding rule. What is the dominical letter for 1895 ? Dividing 18 centuries by 4, there is 2 re. 3 1 mainder. Taking this remainder from 3, wo 2 x1 have a remainder of 1. Twice 1 added to 95 2 Odd years 95 years plus 4 of 95, (rejecting the fraction,) 23 gives 120, which divided by 7 gives a ro. mainder 1, indicating that the dominical letter 7)120 is the 1st below G, which is F. 17 + 1 8. Find the dominical letter for 1833. 9. Find the dominical letters for 1856. 10. Find the dominical letters for 2040. 11. Find the dominical letter for 1911. 12. Find the dominical letter for 1799. 13. Find the dominical letters for 1876. 14. Find the dominical letter for 1921. PROBLEM III. To find the day of the week corresponding to any given day of the month. RULE. The dominical letter found by one of the preceding rules, will indicate the day on which the first Sunday in January will fall. The day of the week for the corresponding day of each succeeding month, may be found by the initials of the following couplet : At Dover Dwells George Brown Esquire, On what day of the week was the Declaration of Independence signed ? The dominical letters for 1776 were G, F. Therefore the first Sunday in January was the 7th of the month. Then A representing the 7th Jan., D would represent the 7th Feb.; D the 7th March ; G the 7th April; B the 7th May; E the 7th June; and G the 7th July. But 1776 being a Leap Year, the dominical letter after February is one day earlier in the month, and a day of the month which would otherwise be represented by G, will be represented by A or Sunday. The 7th July, therefore, came on Sunday, and the 4th on Thursday. The initials O. S. denote the Old Style. In all cases not thus marked, the New Style is understood. 15. Washington was born on the 22d Feb. 1732. What was the day of the week ? 16. The pilgrims landed at Plymouth, Dec. 11, 1620, 0. S. What was the day of the week ? 17. The Battle of Waterloo was fought June 18, 1815. Is it probable that a letter, purporting to have been written at the time, and dated Friday, June 18, is authentic ? 18. On what day of the week was Oct. 11, 1492, O. S., the day that Columbus discovered America ? 19. On what day of the week did Columbus set sail, Aug. 3, 1492, O. S. 20. On what day of the week will a note, at 90 days, dated Aug. 20, 1844, become due, allowing 4 days grace? 21. On what day of the week will a note, at 60 days, dated May 27, 1844, become due, allowing 3 days grace ? 22. Washington died on the 14th day of the last month of the last year of the last century. What was the day of the week ? 23. On what day of the week will be the 3d of April, 1896? MENSURATION. PROBLEM I. To find the area of any surface bounded by four sides, the opposite sides being equal. RULE. Multiply one of the sides by the perpendicular let fall upon it, from the opposite side. 1. The length of an oblong rectangular field is 40 rods, and the breadth 16 rods. How many square rods does it contain ? How many acres ? 2. What are the contents of a four-sided field, whose opposite sides are equal, the length being 81 rods, and the distance between the longest sides, 13 rods ? 3. The average length of Pennsylvania is about 300 miles, and the breadth 157 miles. What is the area ? 4. What is the area of a rhombus, the side being 73,5 rods, and the breadth 61.25 rods? 5. How many square feet in a board that is 144 ft. long, and 11 inches wide ? 6. How many square yards in a floor 14.3 ft. long, and 103 ft. wide ? PROBLEM II. To find the area of a trapezoid, or figure of four sides, two of which are parallel. RULE. Multiply the sum of the two parallel sides by half the distance between them. 7. The two parallel sides of a trapezoid measure 11 and 15 inches respectively, and the height is 8 inches. What is the area ? 8. What is the area of a field, two sides of which are parallel, and measure 72.5, and 89.25 rods, the distance between them being 39 rods? 9. What is the area of a trapezoid, one of the parallel sides measuring 96 rods, the other 634 rods, and the distance between them being 84.6 rods ? PROBLEM III. |