47. How many tons of hay can be bought for 322 dollars, at 14 dollars per ton?

48. How many acres of land will 18705 dollars buy, at 43 dollars per acre?

49. If 753 bushels of oats cost 26355 cents, what is the price of a bushel?

50. If 5 lbs. of sugar cost 60 cts., what will 12 lbs. cost?

51. If 15 lbs. of sugar cost 165 cts., what will 80 lbs. cost?

52. If 335 lbs. of sugar cost 3015 cts., what will 36 lbs. cost?

53. If 23 tons of hay cost 322 dollars, what will 69 tons cost?

54. If 435 acres of land cost 18705 dollars, what will 87 tons cost?

55. If 753 bushels of oats cost 26355 cents, what will 251 tons cost?

56. If 100 acres of land cost 5625 dollars, what will 1000 acres cost?

57. If 1000 acres cost 78560 dollars, what will 100 acres cost?

REM. 1.-If 1 pound of sugar cost 12 cents, 5 lbs. will cost 5 times 12 cents = 60 cents, or 5 lbs. sugar at 1 cent a pound will cost 5 cts., and at 12 cts. per lb., 12 times 560 cts.; if 5 lbs. of sugar cost 60 cts., and it be divided into five equal parts, 1 lb. will cost 12 cts., and 12 lbs. will cost 12 times 12 cts. = 144 cts.

REM. 2. In the 53d example, 69 is 3 times 23, and will cost 3 times 322 dollars; in the 54th, 435 is 5 times 87, and must be divided into 5 equal parts, that is, divided by 5; and in the 55th, 753 is 3 times 251, and must be divided by 3. The 56th example is solved by adding one zero to the number of dollars, and the 57th by taking away a zero.

PROBLEMS.

1. The sum of two numbers, and one of the numbers given, to find the other number.

The difference between the given number and the sum will be the other number.

2. The difference between two numbers and the smaller number given, to find the larger one.

The sum of the difference and the smaller number is the larger one.

3. The difference of two numbers and the larger one given, to find the smaller one.

Subtract the difference from the larger number; the remainder will be the smaller one.

4. The sum and difference of two numbers given, to find the numbers.

Subtract the difference from the sum, and divide the remainder by 2; the quotient will be the smaller one, to which add the difference, and the sum will be the larger

one.

5. The product of two numbers and one of the numbers given, to find the other.

Divide the product by the given number; the quotient will be the other number.

6. The product of three numbers, and two of the numbers given, to find the third.

Divide the product of the three by the product of the two.

FACTORING.

The Product of two or more numbers is found by multiplication, and the factors of that product are restored by division.

Any number that is the product of two or more numbers is called a Composite Number; as, 12 is the product of 4 and 3, and 4 is the product of 2 and 2; hence, the factors of 12 are 2, 2, and 3; these factors cannot be reduced, they are therefore called Prime Factors; as any number is called Prime which is not formed by other factors than itself and unity.

The Prime Factors are readily found; thus, take the first fifty numbers,

1, 2, 3, 4, 5, 6, 7, 8, 9, 10,

15, 16, 17, 18, 19, 20,

11, 12, 13, 14,

41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

Every even number after 2 is composite, as it is divisible by 2, "strike these;" every third number after 3 is divisible by 3, strike these; every fifth number after 5; every seventh number after 7; the 9's are canceled by 3; every 11th after 11, etc.; in the above there were none after the 7's; when more numbers are taken higher

numbers will be required. The prime numbers in the first fifty are sixteen, viz.,

13,

17,

47.

1, 2, 3, 5, 7, 11,
19, 23, 29, 31, 37, 41, 43,

The following Corollaries enable us readily to discover the Prime Factors of numbers:

1. Every even number is divisible by 2.

2. Every number, whose last two figures express a number which is a multiple of 4, is divisible by 4; for if the number expressed by these two figures is subtracted from the whole number, the remainder will be a certain number of hundreds which are divisible by 4.

3. Every number ending in 5 is divisible by 5.

4. Every number ending with zero is divisible by 10, consequently by 2 and 5.

5. Every number is divisible by 3, when the sum of its figures taken as units is divisible by 3; for if from 1000 one be subtracted, the remainder is divisible by 9; if from 100 one be subtracted, the remainder is divisible by 9; so also if one is subtracted from 10; hence, if from 2000 two be subtracted, the remainder is divisible by 9; so also 2 from 200, or 2 from 20; therefore, in dividing by nine, any number of thousands, hundreds or tens, the remainder will always be the unit of the thousands, hundreds, and tens; consequently if the sum of all these remainders as units, and also of the units of the given number, equals 9 or any number of 9's, then the whole number is divisible by 9, and 9 is divisible by 3.

COR. When a number is resolved into its prime factors, the original number is divisible by all these prime factors, and by all the quotients arising from these factors as divisors of the original number; and when a factor occurs more than once, by the products of the like factors.

DEF. One number is called a Multiple of another number, when it is exactly divisible by that other number.