40 nominator. But before reducing them to a common denominator, I observe that some of them may be reduced to lower terms, which will render it much easier to find the common denominator In ģ the nurnerator and denominator may both be divided by two, and it becomes may be reduced to ž, and ito . 1 find also that halves may be reduced to fourths, therefore I have only to find the common denominator of the three first fractions, and the fourth can be reduced to the same. Multiplying the denominators together 3 * 4 * 5 = 60. The common denominator is 60. Now 3 is multiplied by 4 and by 5 to make 60, therefore, the numerator of į must be multiplied by 4 and by 5, or, which is the same thing, by 20, which makes 40, In }, the 4 is multiplied by 3 and 5 to make 6o, therefore these are the numbers by which the numerator 3 must be multiplied. In the fraction , the 5 is multiplied by 3 and 4 to make 60, therefore these are the numbers by which the numerator 1 must be multiplied. = . j = These results may be verified, by taking , , and ; of 60. It will be seen that į of 60 is 20, the product of 4 and 5; 1 of 60 is 15, the product of 3 and 5; and } of 60 is 12, the product of 3 and 4. Now the numbers may be added as follows. 23 23 40 28 45 12 30 Ans. 10770 yards. 127 and 125 = 277 · I add together the fractions, which make * = 277. I write the fraction do, and add the 2 whole ones with the others. = 235 = 28. 28,8 37 3 15 A man having 23{ barrels of flour, sold 8 barrels; how many barrels had he left ? The fractions į and must be reduced to a common denominator, before the one can be subtracted from the other. s = 14 and 1. Therefore 23ž 84 But is larger than and cannot be subtracted from it. To avoid this difficulty, I must be taken from 23 and reduced to 21ths, thus, 2344 = 22 + 11 = 2231 15 2 T Ans. 1421 yards. 1 taken from leaves . Then S from 22 leaves 14. Ans. 14, i yards. From the above examples it appears that in order to add or subtract fractions, when they have a common denominator, we must add or subtract their numerators, and if they have not a common denominator, they must first be reduced to a common denominator. We find also the following role to reduce them to a common denominator: multiply all the denominators together, for a common denominator, and then multiply each numerator by all the denominators except "x XX. This seems a proper place to introduce some contractions in division. If 24 barrels of flour cost 192 dollars, what is that a barrel? This example may be performed by short division. First find the price of 6 barrels, and then of 1 barrel;. 6 barrels will cost 1 of the price of 24 barrels. its own. 192 (4 Price of 1 bar. 8 dolls. Ans. If 56 pieces of cloth cost $7580.72, what is it apiece? First find the price of 7, or of 8 pieces, and then of 1 piece. 7 pieces will cost of the price of 56 pieces. 7580.72 (s Divide $24674 equally among 63 men. How much will each have? First find the share of 7 or 9 men, and then of 1 man. The share of 7 men will be of the whole. The share of 9 men will be of the whole. 24674 (9 Share of 9 men 35244 (9 $3917} Ans. In the first case I divide by 9, and then by 7. In dividing by 7 there is a remainder of 45, which is ; this divided by 7 gives it. In the second case, I divide by 7 and then by 9. In dividing by 9 there is a remainder of 5%, which is 41 ; this divided by 9 gives 43, as before. Divide 75345 dollars equally among 1300 men, how much will each have ? First find the share of 18 men, which will be too part of the whole. To part is found by cutting off the two right hand figures and making them the numerator of a a fraction, thus, 75346 Share of 16 men $753466(18 72 $4113*& Ans. share of 1 man. 33 18 of 1 man. Share of 1 man $41116. Ans. In the last case I find the share of 3 men, and then In dividing by 6 there is a remained 347 which is 145, this divided by 6 gives a fraction 2015 In dividing by 3 there is a remainder 2775, which is equal to , this divided by 3 gives a fraction 1655 and the answer is $411145 each. From these examples we derive the following rule : When the divisor is a compound number, separate the divisor into two or more factors, and divide the dividend by one factor of the divisor, and that quotient by another, and so on, until you have divided by the whole, and the last quotient will be the quotient required. When there are zeros at the right of the divisor, you may cut thein off, and as many figures from the right of the dividend, making the figures so cut off the numerator of a fraction, and 1 with the zeros cut off, will be the denominator ; then divide by the remaining figures of the divisor. XXI. In article XIX. it was observed, that if both the numerator and deno.ninator of a fraction can be divided by the same number, without a remainder, it may be done, and the value of the fraction will remain the same. This gives rise to a question, how to find the divisors of numbers. It is evident that if one number contain another a certain number of times, twice that number will contain the other twice as many times; three times that number will contain the other thrice as many times, &c. that if one number is divisible by another, that number taken any number of times will be divisible by it also. 10 (and consequently any number of tens) is divisible by 2, 5, and 10; therefore if the right hand figure of any number is zero, the number may be divided by either 2, 5, or 10. If the right hand figure is divisible by 2, the number may be divided by 2. If the right hand figure is 5, the number may be divided by 5. 100 (and consequently any number of hundreds) is divisible by 4; therefore if the two right hand figures taken together are divisible by 4, the number may be divided by 4. 200 is divisible by 8; therefore if the hundreds are even, and the two right hand figures are divisible by S, the number may be divided by 8. But if the hundreds are odd, it will be necessary to try the three right hand figures. 1000, being even hundreds, is divisible by 8. To find if a number is divisible by 3 or 9, add together all the figures of the number, as if they were units, and if the sum is divisible by 3 or 9, the number may divided by 3 or 9. The number 387 is divisible by 3 or 9, because 3 + 8 + 7 18, which is divisible by both 3 and 9. The proof of the above rule is as follows: 10 = 9+ 1; 20 = 2 x 9 + 2; 30 = 3 x 9 +3; 52 = 5 x 9 + 5 + 2; 100 = 99 + 1; 200 = 2 x 99 + 2; 3 x 99 + 3 + 8 x 9, + 8 + 7 = 3 x 99 + 8 x 9+ 3 + 8 +7. That is, in all cases, if a number of tens be divided by 9, the remainder will be equal to the number of tens; and if a number of hundreds be divided by 9, the remainder will always be equal 10 the number of hundreds. The same is true of thousands and higher numbers. Therefore, if the tens, hundreds, thousands, &c. of any number be divided separately by 9, the remainders will be the figures of that number, as in the above example 387. Now if the sum of these remainders be 387 = |