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Properties of Numbers.

152. PRIME FACTORS. 1. An integral factor that is a prime number is a prime factor.

5 is a prime factor of 30.
7 is a prime factor of and
3 is a prime factor of and
2 and 3 are prime factors of and - and

3 and 5 are prime factors of and and
2. Resolve 105 into its prime factors.
Operation.

Explanation. 5)105

Since the prime number 5 is an exact divisor of 105, it 3)21

is a prime factor of 105. Since the prime number 3 is an exact divisor of the quotient (21), it is a prime factor of

21 and 105. Since 3 is contained in 21 exactly 7 times, and since 7 is a prime number, 7 is a prime factor of 21 and of 105. Therefore the prime factors of 105 are 5, 3, and 7.

Observe that if 7 and 3 are prime factors of 21 they must be prime factors of 105, for 105 is made up of 5 21's. 7 is contained 5 times as many times in 105 as it is in 21.

Observe that every composite number is equal to the product of its prime factors.

105 = 5 X 3 X 7. 18 = 3 X 3 X 2. 3. Resolve each of the following numbers into its prime factors: 224, 15, 21, 6. Then prove that the continued product of the numbers is equal to the continued product of all their factors.

Observe that 2 times 3 times a number equals 6 times the number; 3 times 5 times a number equals 15 times the number, etc.

Observe that instead of multiplying a number by 21, it may be multiplied by and the product thus obtained by 7, and the same result be obtained as would be obtained by multiplying the number by 21. Why?

Properties of Numbers. 153. MULTIPLES, COMMON MULTIPLES, AND LEAST

COMMON MULTIPLES. 1. A multiple of a number is an integral number of times the number.

30 and 35 and 40 are multiples of 5.
16 and 20 and 32 are multiples of 4.

2. A common multiple of two or more numbers is an integral number of times each of the numbers.

30 is a common multiple of 5 and 3.
40 is a common multiple of 8 and 10.

is a common multiple of 9 and 6.

is a common multiple of 8 and 12. 3. A common multiple of two or more integral numbers contains all the prime factors found in every one of the numbers, and may contain other prime factors.

48 = 2 X 2 X 2 X 2 X 3. 150 = 2 X 3 X 5 X 5. A common multiple of 48 and 150 must contain four 2's, one 3, and two 5's. It may contain other factors.

2 X 2 X 2 X 2 X 3 X 5 X 5 - 1200.
2 X 2 X 2 X 2 X 3 X 5 X 5 X 2 2400.

2 X 2 X 2 X 2 X 3 X 5 X 5 X 2 X 3 = 7200. 1200, 2400 and 7200 are common multiples of 48 and 150.

4. The least common multiple (1. c. m.) of two or more numbers is the least number that is an integral number of times each of the numbers.

40 and 80 and 120 are common multiples of 8 and 10; but 40 is the least common multiple of 8 and 10.

5. The least common multiple of two or more numbers contains all the prime factors found in every one of the numbers, and no other prime factors.

Properties of Numbers. 36 = 2 X 2 X 3 X 3. 120 = 2 X 2 X 2 X 3 X 5. The 1. c. m. of 36 and 120 must contain three 2's, two 3's, and one 5. 2 X 2 X 2 X 3 X 3 X 5 = 360, 1. c. m. of 36 and 120.

6. To find the 1. c. m. of two or more numbers: Resolve each number into its prime factors. Take as factors of the 1. c. m. the greatest number of 2's, 3's, 5's, 7's, etc., found in any one of the numbers.

EXAMPLE
Find the 1. c. m. of 24, 35, 36, and 50.

OPERATION.

24 = 2 x 2 x 2 x 3.
35 5 x 7.
36 2 x 2 x 3 x 3.

50 = 2 x 5 x 5.
2 x 2 x 2 x 3 x 3 x 5 x 5 x 7 = 12600, 1. c. m.

EXPLANATION.
24 has the greatest number of 2's as factors.
36 has the greatest number of 3's as factors.
50 has the greatest number of 5's as factors.

35 is the only number in which the factor 7 occurs. There must be as many 2's among the factors of the 1. c. m. as there are 2's among the factors of 24 ; as many 3's as there are 3's among the factors of 36; as many 5's as there are 5's among the Sactors of 50; as many 7's as there are 7's among the factors of 35 ; that is, three 2's, two 3's, two 5's, and one 7.

Find the l. c. m.

7. Of 48 and 60. 8. Of 60 and 75. 9. Of 50 and 60. 10. Of 30 and 40.

11. Of 20, 30, and 40.
12. Of 40, 50, and 60.
13. Of 24, 48, and 36.
14. Of 25, 35, and 40.

a

3 a

2 a

Algebra-Parentheses. 154. When an expression consisting of two or more terms is to be treated as a whole, it may be enclosed in a parenthesis. | 12 + (5 + 3) = ?

| 7 a + (3 a + 2 a) = ? 12 + 5 + 3 = ?

7 a + 3 a + 2 a ? Observe that removing the parenthesis makes no change in the results. | 12- (5 + 3) = ?

.J 7 a – (3 a + 2 a) = ?
12
5 - 3 = ?

7

? Observe the change in signs made necessary by the removal of the parenthesis. 12 - (5-3) = ?

7 a - (3 a - 2 a) = ? 12 - 5 + 3 = ?

7 а

3 a + 2a = ? Observe the change in signs made necessary by the removal of the parenthesis.

A careful study and comparison of the foregoing problems will make the reasons for the following apparent:

1. If an expression within a parenthesis is preceded by the plus sign, the parenthesis may be removed without making any changes in the signs of the terms.

II. If an expression within a parenthesis is preceded by a minus sign, the parenthesis may be removed; but the sign of each term in the parenthesis must be changed; the sign + to -, and the sign to +.

155. PROBLEMS. (1) Remove the parenthesis, (2) change the signs if necessary, (3) combine the terms. 1. 15- (6 + 4) =

6. 15 b - (12 b -46) 2. 18+ (4 – 3)

7. 18c+ (9c- 3c) = 3. 27 (8 + 3)

8. 24 d - (5 d + 3 d) = 4. 45 + (12 - 3)

9. 36 x - (5 x + 4 x) = 5. 75 - (18 +27) 10. 45 y + (8 y + 7 y) =

Algebra-Parentheses, 156. Multiplying an expression enclosed in a parenthesis. 6(7 + 4) = ?*

6(7 a + 4b) = ?
6(a + b) = ? Ans. 6 a + 6 6.
a(b + c) = ? Ans. ab + ac.

Observe that in multiplying the sum of two numbers by a third number, the sum may be found and multiplied; or each number may be multiplied and the sum of the products found.

1. In the last three examples given above, let a = 5, b = 3, and c = 2; then perform again the operations indicated, and compare the results with those obtained when the letters were employed. 6(7 – 4) = ?+

6(7a 46) = ? 6(a - b) = ?

= ? Ans. 6a - 66. a(b-c) = ? Ans. ab - ac.

Observe that in multiplying the difference of two numbers by a third number, the difference may be found and multiplied; or each number may be multiplied and the difference of the products found.

2. In the last three problems given above, let a = 5, b = 3, and c = 2; then perform again the operations indicated and compare the results with those obtained when the letters were employed.

157. PROBLEMS. If a = 5, b = 3, and c = 2, find the value of the following:

1. 3(a + b) - 2(6+).
2. 4(a + 2b) - 3(b-c).
3. 2(2 a - b) + 2(2b-c).

* This means, that the sum of 7 and 4 is to be multiplied by six; or that the sum of six 7's and six 4's is to be found.

+ This means, that the difference of 7 and 4 is to be multiplied by six; or that the difference of six 7's and six 4's is to be found.

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