The preceding analysis shows that instead of dividing by the composite number 15, whose factors are 3 and 5, we may first divide by one factor, then divide the quotient thus obtained by the other factor.

2. Find the quotient of 37, divided by 14. SOLUTION.—Dividing by 2, the quo- 2)37

tient is 18 twos and 1 unit remaining.

Dividing by 7, the quotient is 2, with 7)18 and 1 over.

a remainder of 4 twos; the whole re

mainder then, is 4 twos plus 1, or 9.

2 and 4 twos left.

Rule for Case I.-Divide the dividend by one of the factors of the divisor; then divide the quotient thus obtained by the other factor.

2. Multiply the last remainder by the first divisor; to the product add the first remainder; the amount will be the true remainder.

NOTE. When the divisor can be resolved into more than two factors, you may divide by them successively. The true remainder will be found by multiplying each remainder by all the preceding divisors, except that which produced it. To their sum add the remainder from first divisor.

ART. 48. To divide by 1 with ciphers annexed; as 10, 100, 1000, &c.

REVIEW.-46. NOTE. When any product is greater than the partial dividend from which it is to be subtracted, what must be done?

47. How may division be performed, when the divisor is a composite number? How is the true remainder found? NOTE. When the divisor can be resolved into more than two factors, how may the division be performed? How is the true remainder obtained?

To multiply 6 by 10, annex by 10, annex one cipher, thus, 60. On the principle that Division is the reverse of multiplication, to divide 60 by 10, cut off a cipher.

Had the dividend been 65, the 5 might have been separated in like manner as the cipher; 6 being the quotient, the remainder.

The same will apply when the divisor is 100, 1000, &c.

Rule for Case II.-Cut off as many figures from the right of the dividend as there are ciphers in the divisor; the figures cut off will be the remainder, the other figures, the quotient.

ART. 49. To divide, when there are ciphers on the right

of the divisor.

1. Divide 4072 by 800. SOLUTION.-Regard 800 as a composite number, the factors 100 and 8, and divide as in the margin.

In dividing by 800, separate the two right hand figures for the remainder, then divide by 8.

OPERATION.

100)40 72
8)40

5 Quo. 72 Rem.

8|00)40172

5 Quo. 72 Rem.

REVIEW. 48. How do you divide by 10, 100, 1000, &c.? On what principle does the rule for case II depend? 49. How do you divide when there are ciphers on the right of divisor, Rule for case III?

2. Divide 77939 by 2400.

SOLUTION. Since 2400 equals 24×100, cut off the two right hand figures, the same as dividing by 100; then divide by 24.

24100)779139 (321188 72

59

48

11

Dividing by 100, the remainder is 39; dividing by 24, the remainder is 11. To find the true remainder, multiply 11 by 100, and add 39 to the product, (Art. 47, Rule); this is the same as annexing the figures cut off, to the last remainder. Hence, the

Rule for Case III.-1. Cut off the ciphers at the right of the divisor, and as many figures from the right of the dividend. 2. Divide the remaining figures in the dividend by the remaining figures in the divisor.

3. Annex the figures cut off to the remainder, which gives the

true remainder.

Exercises in more difficult contractions are in " Ray's Higher Arithmetic.”

PROOF OF MULTIPLICATION BY DIVISION.

ART. 50. Division (Art. 37), is a process for finding one of the factors of a product when the other factor is known therefore,

If the product of two numbers be divided by the multiplier, the quotient will be the multiplicand: Or, if divided by the multiplicand, the quotient will be the multiplier.

REVIEW.-50. What is division? If the product of two factors be ided by either of them, what will be the quotient?

1. What number multiplied by 7895, will give 434225 for a product?

Ans. 55.

2. If 327 be multiplied by itself, the product will be 106929. Give the proof.

3. The product is 10741125; the multiplier 375: what is the multiplicand? Ans. 28643. 4. The product is 63550656, and the multiplicand 60352: what is the multiplier?

Ans. 1053.

For additional problems, see Ray's Test Examples.

REVIEW OF PRINCIPLES.

ART. 51. NOTATION and NUMERATION show how to express numbers by words, by figures, or by letters.

For other scales of notation than the decimal or tens' scale, an interesting subject for advanced students, see "Ray's Higher Arithmetic.”

ART. 52. BY ADDITION,

The aggregate or sum of two or more numbers is found, (Art. 18). Thus, when the separate cost of several things is given, the entire cost is found by addition.

EXAMPLE.-A bag of coffee cost $23, a chest of tea $38, a box of sugar $11: what did all cost? Ans. $72.

ART. 53. BY SUBTRACTION,

The difference between two numbers is found. Thus, if the sum of two numbers be diminished by either of them, the remainder will be the other.

Hence, by Addition, if the difference of two numbers be added to the less, the SUM will be the greater.

REVIEW.-51. What do Notation and Numeration show? 52. What is found by addition? Give an example.

53. What is found by subtraction? Having the sum of two numbers, and one of them, how the other found? When the smaller of two numbers and the difference are given, how is the greater found? When the difference and greater are given, how is the less found?

EXAMPLE 1. The sum of two numbers is 85; the less number is 37: what is the greater?

2. The sum of two numbers is 85: the greater number is 48: what is the less?

3. The difference of two numbers is 48; the less number is 37: what is the greater?

4. The difference of two numbers is 48; the greater number is 85: what is the less?

ART. 54. BY MULTIPLICATION,

Is found the amount of a number taken as many times as there are units in another. Art. 28.

Hence, having the cost of a single thing, to find the cost of any number of things, multiply the cost of one by the number of things.

1. If 1 yard of tape cost 3 cents, what will 5 yards cost?

ANALYSIS.-Five yards are 5 TIMES 1 yard; therefore 5 yards will cost 5 TIMES as much as 1: the entire cost is found by multiplying the price of 1 yard by the NUMBER of yards.

The divisor and quotient being given, the dividend is found by multiplying together the divisor and quotient. Art. 37.

2. A divisor is 15; the quotient is 12: what is the dividend?

3. An estate was divided among 7 children; received $525: what sum was divided?

ART. 55. BY DIVISION,

each child Ans. $3675.

Is found how many times one number is contained in another. Art. 36. This enables us,

1. To divide any number into parts, each part containing a certain number of units.

2. To divide a number into any given number of equal parts. Thus, if the cost of a number of things and the price of one are given, the number of things is found by division.

1. James spent 35 cents for oranges, and paid 5 cents each how many did he buy?

SOLUTION. He got one orange for each time 5 cents are contained in 35 cents; 5 in 35, 7 times; therefore, he bought 7 oranges.