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 the numerator and denominator may both be divided by two, and it becomes may be reduced to, and to . 1 find also that halves may be reduced to fourths, thereto. fore I have only to find the common denominator of the three first fractions, and the fourth can be reduced to the same. 40 8 45 3 ᄒ = 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 must be multiplied by 4 and by 5, or, which is the same thing, by 20, which makes 40, 4. In, the 4 is multiplied by 3 and 5 to make 60, therefore these are the numbers by which the numerator 3 must be multiplied. 5. 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. = 1}; } = 38. These results may } 12 be verified, by taking,, and of 60. } It will be seen that of 60 is 20, the product of 4 and 5; 4 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. 1 2 30 40 45 12 30 Ans. 107 yards. 127 and 137 28%. = I add together the fractions, which make 27 = 27. I write the fraction, and add the 2 whole ones with the others. A man having 233 barrels of flour, sold 84 barrels; how many barrels had he left? The fractions and must be reduced to a common denominator, before the one can be subtracted from the other. = 14 and 15. Therefore 233/30 8 2 214 821 But is larger than 14 and cannot be subtracted from it. To avoid this difficulty, 1 must be taken from 23 and reduced to 21ths, thus, 231422+114 = 2234 815 Ans. 1420 yards. 15 taken from leaves . Then 8 from 22 leaves 3 20 14. Ans. 1429 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 rule 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 its own. 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 of the price of 24 barrels. 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. 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. In the first case I divide by 9, and then by 7. In dividing by 7 there is a remainder of 45, which is 41; this divided by 7 gives . 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 41, as before. Divide 75345 dollars equally among 1800 men, how much will each have? 1 First find the share of 18 men, which will be part of the whole. part is found by cutting off the two right hand figures and making them the numerator of a a fraction, thus, 753-0. 4 1009 In the last case I find the share of 3 men, and then of 1 man. In dividing by 6 there is a remained 3,45%, which is 35, this divided by 6 gives a fraction 345. In dividing by 3 there is a remainder 2345, which is equal to 1545, this divided by 3 gives a fraction 1345 and the answer is $411845 45 each. 18009 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 them 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 bes 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 be divided by 3 or 9. The number 387 is divisible by 3 or 9, because 3 + 8718, which is divisible by both 3 and 9. The proof of the above rule is as follows: 109+ 1; 20 = 2 × 9 + 2; 30 = 3 × 9 + 3; 52 = 5 × 9+ 5 + 2; 100 = 99 + 1; 200 = 2 × 99 + 2; 387 = 3 × 99 + 3 + 8 × 9+ 8 + 7 = 3 × 99 + 8 × 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 to 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 |