Ex. 2. Find the greatest common measure of 891 and 3429. 891)3429(4 3564 135)891(7 54)135(2 In finding the G.C.M. in this example, the work has been shortened by taking the quotient figures 4 and 7, because the products 3564 and 945 differ less from the dividends 3429 and 891, than if 3 and 6 had been taken. Here 27 is the G.C.M. of 891 and 3429. 27)54(2 Ex. 3. Find the greatest common measure of 3852 and 762896. 73. This process enables us to find the G.C.M. in all cases. But in the case of small numbers it can often be seen by inspection, and the student must accustom himself to do so. Thus, since 48 12 X 4, and 60 = 12 X 5, 12 is evidently the greatest common measure of 48 and 60. 74. To find the G.C.M. of three numbers, we first find the G.C.M. of two of the numbers, and then the greatest common measure of the result of the third number; and, similarly, if there were more numbers. EXAMPLES FOR PRACTICE. Find the greatest common multiple of— I. 24 and 36; 36 and 45; 121 and 143; 87 and 145; 289 and 529. 2. 567 and 576; 1001 and 1331; 729 and 999; 175 and 2042; 1878 and 3443. 3. 2314 and 3721; 7512 and 12207; 13731 and 19366; 55328 and 22755. 4. 107358 and 107271; 814407 and 91797; 63404 and 42867; 12679235 and 1864381. 5. 2442431 and 1562009; 1718446 and 4439618; 2158654 and 12130265. 6. 252177 and 3628013; 68590142 and 85044059; 318732033187 and 32033187203. 7. 726221432301 and 432301726221; 318732033187 and 320331873203. 75. To find the Least Common Multiple of two or more numbers. RULE XXI. 1°. Having arranged all the numbers in a row, cancel those which exactly divide any of the others. 20. Select from the numbers which remain that one which appears to contain the greatest number of different factors, and divide all the uncancelled numbers in succession by the greatest number which will divide each of them and the aforesaid selected number, putting down the quotients for a second row. 3°. Treat this second row as the first, that is, cancel every number which exactly divides any of the others. Then, if the uncancelled numbers in the second row be all prime to one another, the operation will be finished; for the continued product of the numbers in this line and the divisor will be the least common multiple. Should the uncancelled numbers in the second row be not all prime to one another, then we may conclude that an unsuitable number has been selected, and another number, taken from the given numbers, must be used instead, always taking that which offers the greatest number of different factors. Ex. 1. EXAMPLES. Find the least common multiple, L.C.M., of 6, 8, 12, 14, 15, 18, 20, 21, and 24. 24)6, 8, 12, 14, 15, 18, 20, 21, and 24 7, 5, 3, 5, 7 3 X 5 X 7 X 24 = 2520 Strike out all the numbers which will divide any of the other numbers exactly, as 6, 8, 12, which divide 24. Selecting 24 as the number containing the most factors, write it outside. Divide the other numbers in succession by the greatest number which will divide each of them and 24; thus, the G.C.M. of 24 and 14 is 2, leaving 7; of 24 and 15 is 3, leaving 5; of 24 and 18 is 6, leaving 3; of 24 and 20 is 4, leaving 5; of 24 and 21 is 3, leaving 7. Then 7 cancels 7, 5 cancels 5, and the remaining figures in the row, viz., 3, 5, 7, being all prime to one another, their continued product and the divisor 24 gives the L.C.M.2520. Reason of the process. We see that 252024 X 3 X 5 X 7. Hence it contains as factors all the given numbers, for 6, 8, 12 are each contained in 24; the part of 14, viz., 7, not contained in the 24 is introduced by the factor 7; the part of 15 not contained in 24 is introduced by the factor 5, and so on for the other numbers. Again, we cannot dispense with any of the factors 3, 5, 7; for, if the L.C.M. did not contain the factor 3, beside the factor 6 in the 24, it would not be a multiple of 18; and, similarly, the other factors, 5 and 7, are necessary. Hence, 2520 is a common multiple of all the given numbers, and if we took away any of the factors composing it, it would cease to contain one or more of the given numbers. It is, therefore, the L.C.M. of the given numbers, Ex. 2. Find the least number that can be divided by 5, 6, 9, 15, 24, 27, and 126. 126)5, 6, 9, 15, 24, 27, 126 15, 4, 8 Therefore, 126 X 15 X 47560, the number required. Strike out 5, 6, 9, as they are contained in 15, 24 and 27 respectively; then the G C.M. of 126 and 24 is 6, leaving 4; of 126 and 27 is 9, leaving 3; cancel the 3 as it is contained in 15. Then the continued product of 126 X 15 X 4 = 7560, the number required. Ex. 3. Required the least common multiple of 7, 12, 15, 27, 35, 40, and 45. 3 divides 45 and 12, leaving 4; 9 divides 45 and 27, leaving 3; 5 divides 45 and 35, leaving 7; 5 divides 45 and 40, leaving 8; cancel 4 as it is contained in 8; and since the remaining figures 3, 7, 8 have no common measure but unity; the continued product of these and 45 gives 7560 as the L.C.M. required. Ex. 4. Required the L.C.M. of 8, 9, 10, 12, 25, 32, 75, and 80. 76. The least common multiple of two or more numbers may also be found by the following process: RULE XXII. 1o. "Write the numbers in a line, and strike out any that are contained in any of the others. Divide those not struck out by any number that will exactly divide one of them; under any which it exactly measures place the corresponding quotient; under any which it partially measures containing some factor common to it, (but not being itself wholly contained in it), place the quotient obtained by dividing it by the common factor; and under any which it does not measure at all, repeat the number itself. 2o. Now treat the new line thus formed in the same manner as the first, and so on until all the numbers left in any line have no common measure but unity. 3°. Then the continued product of the numbers in this line and all the divisors is L.C.M. required of the given numbers." NOTE. It will generally be most convenient to take pretty large numbers, if possible, for divisors, as fewer lines will thus be necessary, especially if such be chosen as contain themselves many simple factors. Thus, 12 contains the factors 2, 3, 4, 6, and 12, and is, therefore, when possible, a very good divisor to be employed. Strike out 6, 4, 8, and 12, as they are contained in 24; having placed 12 outside as divisor, divide the other numbers in succession by the greatest number which will divide each of them and 12; thus, 12 goes into 24 2; the G.C.M. of 12 and 16 is 4, leaving 4; of 12 and 20 is 4, leaving 5; of 12 and 30 is 6, leaving 5; bring down 25. Next, cancel 2 as it is contained in 4; cancel 5 and 5 because they are contained in 25; and the remaining figures, viz., 4 and 25, are prime to one another, having no common measure but unity. Then the continued product of 4 X 25 X 12 = 1200, the L.C.M. required. Ex. 2. What is the least number that can be divided by each of 14, 16, 40, 50, 25, 8, 64. 10)14, 16, 40, 50, 25, 8, 64 Ex. 3. Find the least common multiple of 27, 24, 6, 15, 5, 9, 126. 2. 319 and 407; 333 and 504; 2961 and 799; 8, 9, 12, 18. 3. 6, 15, 27, 35; 3, 9, 7, 15, 18, and 42; 75, 125, and 1000; 54, 96, and 64. 4. 12, 20, 24, 54, 81, 63, and 14; 225, 255, 289, 1023, and 4095. 5. 8, 18, 28, 35, 54, 72, and 90; 30, 60, 90, 150, 180, 300, 450, and 900. 6. 12, 18, 24, 27, 36, and 42; 15, 21, 25, 36, 48, 60, 78, 9, 18, 24, 72, and 144. 7. 9, 16, 42, 63, 21, 14, and 72; 34, 51, 85, 65, 104, and 117. 8. 29, 27, 216, 261, 297, and 783; 343, 5929, 7007, 8281, and 16807. 9. 221, 117, 153, 289, 169, and 81; 529, 437, 361, 161, and 133. 10. 105, 403, 248, 221, 119, and 12648; 3528, 25725, 23625, and 432. 11. 658, 752, 1222, 1410, and 2444; 13792381, 32080621, 33452239, and 5806592401. I. 2. MISCELLANEOUS EXAMPLES IN G.C.M. AND L.C.M. What multiple of 5 results from multiplying 314725 by 9? 3. Of all the numbers beside unity that may be subtracted from 7905 an exact number of times, which is that which may be taken the greatest, and which the least, number of times? 4. What is the least number from which 34, 510, 91, and 16, may each be taken an exact number of times? 5. What is the least number by which 17595 must be multiplied, that 36 may be taken from the product an exact number of times? 6. A school was found to contain such a number of boys that whether arranged in sixes, sevens, nines, or elevens, there were always five over; how many children at least did the school contain ? 7. Find all the numbers which will divide 396, 5184, and 6912, without a remainder, and the smallest number they will all divide without a remainder. In practice the L.C.M. may often be written down at once. Example-8, 12, 15, 16, and 18. The L.C.M. 15 X 16 X 3720, for the numbers 15 and 16 having no common factor, must both remain in the L.C.M, 8 is included in the 16, 12 partly in the 15, partly in the 16, and 18 containing the factor 3 twice, requires the addition of the factor 3, since 15 contains it only once, and 16 not at all. 8. What is the least number of dollars at 48. 2d. each, which is equal to an exact number of sovereigns? 9. Standard gold being coined at the rate of £3 178. 10d. per oz., what is the least number of ounces which can be coined into an exact number of sovereigns? 10. Eight bells begin tolling simultaneously, and they toll at intervals of 1, 2, 3, 4, 5, 6, 7, and 8 seconds, respectively; after what interval of time will they again be tolling at the same instant? II. What is the common unit which will express 10080 minutes, and 15840 minutes, in the smallest possible integral numbers? 12. Find all the common multiples of 27, 36, and 45, which are less than zooo? 13. Two bells, which toll respectively every 4 and 6 seconds, commence tolling together; how often will they be tolling simultaneously in an hour? 14. Three boys start together to run round a ring, and they run round it in 4, 5, and 6 minutes respectively; when will they all be together again, and how many times will they each have been round? 15. A has 5040 shillings, B has 5292, and they wish each to separate their own money into equal heaps, so that A's heaps are as large as B's, and each as large as possible; how many shillings must they put into each heap, and how many heaps will they each have? 16. Two foot rules are divided, one into 120 divisions and the other into 96; if they are placed side by side, how many divisions will coincide? 17. Two cog-wheels containing 32 and 36 cogs, respectively, work together; if the large wheel has made 64 revolutions, how often have the same cogs been in contact? CHAPTER IV. FRACTIONS. DEFINITIONS. 77. A Fraction is a part or parts of unity or one. The nature of a fraction will be understood by supposing that a unit of any kind is divided into several equal parts. One or more of these parts is called a fraction of that unit. A fraction in its simplest form consists of two numbers placed one over the other with a line between them. The number under the line denotes the number of parts into which the unit is divided, and is called The Denominator of the fraction, i.e., the "name given," because it gives the name to the parts. The figure placed above the line is called The Numerator, i.e., the "numberer, because it shows how many of the parts named by the denominator are to be taken. Thus, if the unit be a yard, the fraction signifies that the yard is divided into 8 equal parts, and three of these parts are taken. It is read three-eighths. If the unit be divided into 2, 3, 4, 5, or 6 equal parts, the corresponding fractions with their names are as follows: -, one-half; }, one-third;, one-fourth; }, one-fifth; }, one-sixth. VULGAR FRACTIONS. 78. In treating of the subject of Vulgar Fractions, it is usual to make the following distinctions: 1. A Proper Fraction is one whose numerator is less than its denominator; thus,,, are proper fractions. 2. An Improper Fraction is one whose numerator is either equal to, or greater than, its denominator, as 4, 41, 10, 101. 3. A Simple Fraction has only one numerator, and one denominator, which are called the terms of the fraction, as, t ៖. 4. A Mixed Number consists of a whole number and a fraction; thus, 1, 68, and 15%, which represent, respectively, 1 unit together with of a unit; 6 units together with ths of a unit, and 15 units together with 4th of a unit. NOTE.-It is evident that 1, 68, and 15 is each only a shortened form of 1 + †, 6 + †, and 15, the sign + (plus) being understood. 5. A Complex Fraction is one which has either a fraction or a mixed number in one or both terms of the fraction; thus,— 유 3 4층 5층 2/1/1 6. A Compound Fraction is a fraction of a fraction, that is, two or more simple fractions connected by the word of; thus, of, where is the quantity of which is to be taken, 24 of 4 of 3. |