860. The Roman Notation is the method of expressing numbers by seven capital letters, viz. : 861. To express other numbers, these letters are combined 862. The Roman Notation is based upon the following general principles: 1st. Repeating a letter repeats its value. Thus, I denotes one; II, two; III, three; X, ten; XX, twenty, etc. 2d. Placing a letter of less value before one of greater value, diminishes the value of the greater by that of the less; placing the less after the greater, increases the value of the greater by that of the less. Thus, V denotes five, but IV denotes only four, and VI six. 3d. Placing a horizontal line over a letter increases its value a thousand times. Thus, I denotes a thousand; X, ten thou sand; C, a hundred thousand; M, a million. NOTES.-1. The letters C and M are the initials of the Latin centum, a hundred, and mille, a thousand. 2. The radix of the system appears to be doubtful. Some have supposed that at first it was five (V), and was subsequently changed to ten (X), forming a combination of the quinary and decimal systems. 3. Others maintain, more plausibly, that it proceeds according to the alternate scale of 5 and 2, thus uniting the binary with the quinary scale. That is, Five times one (I) are five (V). Two times five (V) are ten (X). Five times ten (X) are fifty (L). Two times fifty (L) are one hundred (C). Five times one hundred (C) are five hundred (D). ENGLISH NUMERATION. 863. By the English Numeration, numbers are divided into periods of six figures each, and then each period is subdivided. into units, tens, hundreds, thousands, tens of thousands, and hundreds of thousands, as in the following The figures in the Table are thus read: 407692 billions, 958604 millions, 413 thousand fifty-six. Hund. of Thousands. or Tens. Units. CONTRACTIONS. 864. To Multiply any Number containing Two Figures by II. The product of any figures two multiplied by 11 consists of the first figure of the number multiplied, the sum of the two figures, and the last figure. Thus, 34 x 11 = 374, is composed of 3 the first figure, 7 the sum of 3+4, and 4 the last figure. NOTE.-If the sum of the two figures exceeds 9, the first or left-hand figure must be increased by 1. 1. What will 11 tons of iron cost, at $45 per ton? ANALYSIS.-Since 1 ton costs $45, 11 tons will cost 11 times $45; and 11 times $45 are $495. 2. Alexander has 84 marbles, and Richard has 11 times as many; how many has Richard? 3. How many are 11 times 27? 11 times 33? 11 times 26? 11 times 34? 11 times 62? 11 times 54? 11 times 72 ? 865. To Multiply by 13, 14, 15, or 1 with a Significant Figure annexed. If one city lot costs $3245, what will 17 lots cost? ANALYSIS.-17 lots will cost 17 times as much as 1 lot. Placing the multiplier on the right, we multiply the multiplicand by the 7 units, set each figure one place to the right of the figure multiplied, and add the partial product to the multiplicand, The result is $55165. OPERATION. $3245 × 17 22715 $55165, Ans. 866. To Multiply by 21, 31, 41, etc., or I with either of the other Significant Figures prefixed to it. If 21 men can do a job of work in 365 days, how long will it take 1 man to do it? EXPLANATION.-We first multiply by the 2 tens, and set the first product figure in tens' place; then adding this partial product to the multiplicand, we have 7665 for the answer. OPERATION. 365d. × 21 730 7665 days, Ans. 867. To Multiply Two or More Numbers by the Same Multiplier. 1. A grocer sold 4 pounds of tea to one customer, 3 lb. to another, and 5 lb. to another; how much did it all come to, at 7 dimes a pound? SOLUTION. (4+3+5) × 7 = 84. Ans. 84 dimes. Hence, the RULE.-Multiply the sum of the numbers by the multi plier. 868. To Multiply a Mixed Number, whose Fractional Part is }, by itself. 1. What is the product of 31 into 3? SOLUTION. The integral part of the given number is 3, and 3+1 = 4 Now 3 into 4 = 12, and 12+ = 121, Ans. Hence, the RULE.-Multiply the integral part by one more than itself, and to this result annex 1. 869. To Multiply by 9, 99, 999, or any number of 9's. 1. How much will 99 carriages cost, at 235 dollars apiece? EXPLANATION. Since 1 carriage costs $235, 100 carriages will cost 100 times as much, or $23500. But 99 is 1 less than 100; therefore, subtracting the price of 1 carriage from the price of 100 gives the price of 99 carriages. OPERATION. 870. To Divide by 5. 1. A merchant laid out $873 in flour, at $5 a barrel; many barrels did he get? EXPLANATION. We first multiply the dividend by 2, and then divide the product by 10, by cutting off the right-hand figure. The figure cut off is written over the divisor, and the fraction, reduced to its lowest terms, is annexed to the quotient. 873 2 1/0 ) 174/6 how 1748, Ans. NOTE. This contraction depends upon the principle that any given di visor is contained in any given dividend just as many times as twice that divisor is contained in twice that dividend, three times that divisor in three times that dividend, etc. 871. To Divide by 25. 1. A farmer paid $150 for cows, at $25 apiece; how many cows did he buy? EXPLANATION.-We first multiply the dividend by 4, and then divide the product by 100. (Art. 118.) 872. To Divide by 125. OPERATION. 150 4 1100) 6100 Ans. 6 cows. 1. A man bought land for $12150, at $125 an acre; how many acres did he buy? EXPLANATION.-We multiply the dividend by 8, and divide the product by 1000. (Art. 118.) Placing the remainder over the divisor, we reduce the fraction to lowest terms, and annex it to the quotient. The answer is 971 acres. NOTE. This contraction multiplies both the dividend and divisor by 8. (Art. 119, 3°.) OPERATION. 12150 8 1000) 97/200 Ans. 97 A. 873. Demonstration of Finding the Greatest Common Divisor by Continued Divisions. 1. Find the g. c. d. of 30 and 42. Two points are to be proved: 1st. That 6 is a common divisor of the given numbers. 2d. That 6 is their greatest common divisor. Ans. 6. That 6 is a common divisor of 30 and 42 is easily proved by trial. Next, we are to prove that 6 is the greatest common divisor of 30 and 42. If the greatest common divisor of these numbers is not 6, it must be either greater or less than 6. But we have shown that 6 is a common divisor of the given numbers; therefore, no number less than 6 can be the greatest common divisor of them. The assumed number must therefore be greater than 6. By the supposition, this assumed number is a divisor of 30 and 42; hence, it must be a divisor of their difference, 42 - 30, or 12. And as it is a divisor of 12, it must also divide the product of 12 into 2, or 24. Again, since the assumed number is a divisor of 30 and 24, it must also be a divisor of their difference, which is 6; that is, a greater number will divide a less without a remainder, which is impossible. Therefore, 6 must be the greatest common divisor of 30 and 42. |