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Properties of Numbers. 149. Every integral number is prime or composite.

1. A prime number is an integral number that has no exact integral divisors except itself and 1; as, 23, 29, 31, etc.

2. Is two a prime number? three ? nine ?

3. Name the prime numbers from 1 to 97 inclusive*. Find their sum.

4. A composite number is an integral number that has one or more integral divisors besides itself and 1; as, 6, 8, 9, 10, 12, 14, 15, etc.

5. Name the composite numbers from 4 to 100 inclusive. Find their sum.

6. Is eight a composite number? eleven ? fifteen ?

(a) Find the sum of the results of problems No. 3 and No. 5.

150. To find whether an integral number is prime or composite. 1. Is the number 371 prime or composite? Operation.

Explanation. 2)371 5)371 Beginning with 2 (the smallest prime 185+ 74+

number except the number 1) it is found by

trial not to be an exact divisor of 371. 3)371 7)371 3 is not an exact divisor of 371. 123+ 53

5 is not an exact divisor of 371.

7. is an exact divisor of 371. Therefore 371 is a composite number, being composed of 53 sevens, or of 7 fiftythrees. +

Observe that we use as trial divisors only prime numbers. If 2 is not an exact divisor of a number, neither 4 nor 6 can be. Do you see why?

* There are 26 prime numbers less than 100.
+ Note the similarity of the words composed and composite.

Properties of Numbers. 2. Is the number 397 prime or composite ? Operation.

Explanation. 2)397 3)397

By trial it is found that neither 2, 3, 5, 198+ 132+

7, 11, 13, 17 or 19 is an exact divisor of

397. 5)397 7)397

No composite number between 2 and 79+ 56+

19 can be an exact divisor of 397: for 11)397 13)397 since one 2 is not an exact divisor of the 36+ 30+

number, several 2's, as 4, 6, 8, 12, etc.,

cannot be; since one 3 is not an exact 17)397 19397

divisor of the number, several 3's, as 6, 9, 23+ 20+

12, etc., cannot be; since one 5 is not an

exact divisor of the number, several 5's, as 10 and 15 cannot be; since one 7 is not an exact divisor of the number, two 7's (14) cannot be.

No number greater than 19 can be an exact divisor of the number; for if a number greater than 19 were an exact divisor of the number the quotient (which also must be an exact divisor) would be less than 20. But it has already been proved that no integral number less than 20 is an exact divisor of 397. Therefore 397 is a prime number.

Observe that in testing a number to determine whether it is prime or composite, we take as trial divisors, prime numbers only, beginning with the number two.

Observe that as the divisors become greater, the quotients become less, and that we need make no trial by which a quotient will be produced that is less than the divisor.

3. Determine by a process similar to the foregoing, whether each of the following is prime or composite : 127, 249, 257, 371.

151, Any divisor* of a number may be regarded as a factor of the number. An exact integral divisor of a number is an integral factor of the number.

* NOTE.—The word factor is often loosely used for integral factor.

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 7 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 3 and the product thus obtained by 7, and the same result be obtained as would be obtained by multiplying the number by 21. Why?

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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.

35

24 = 2 x 2 x 2 x 3.

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 aniong 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 factors 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.

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