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III. Any number is divisible by 8, 4, 125, 250, 333], or any other num. ber which will exactly divide 1000, when the number expressed by its three right hand figures is thus divisible.
IV. Any number is divisible by 3 or 9, when the sum of its digits is thus divisible.
V. Any number is divisible by 11, when the sums of its alternate digits ure equal, or their difference is a multiple of 11.
VI. A number which is divisible by each of two numbers which are prime to each other, is divisible by their product.
VII. A number which is divisible by each of two numbers not prime to each other, is not of necessity divisible by their product.
VIII. A number which is not divisible by another is not divisible by any multiple of that other number.
IX. A number which is divisible by any composite number is also divisible by all the factors of that composite number.
105. Definitions of Factors, Powers, &c. (a.) A number is said to be divided into factors when any factors which will produce it are found.
Thus, in 36 4 X 9, 36 is divided into the factors 4 and 9; but in 36 = 2 X 6 X 3 it is divided into the factors 2, 6, and 3.
(5.) A number is said to be divided into its prime factors when it is divided into factors which are all prime numbers. Illustrations. 36 2 X 2 X 3 X 3 8= 2 X 2 X 2 2 X 3 X 5
84 = 2 X 2 X 3 X 7 (c.) When any number is taken more than once as a factor to produce another number, we may express the number of times it is taken as a factor, by placing a small figure above it and a little to the right.
Illustrations. 32 means the same as 3 X 3; i. e., that 3 is to be taken twice as a factor.
34 means the same as 3 X 3 X 3 X 3; i. e., that 3 is to be taken + times as a factor.
33 x 24 means the same as 3 X 3 X 3 X 2 X 2 X 2 X 2; i. e., that the product of 3 taken 3 times as a factor is to be multiplied by the product of 2 taken 4 times as a factor.
NOTE. The student should notice the difference between taking a number as a factor a certain number of times, and merely taking it a number of times.
Thus, 5 taken 3 times as a factor = 5 times 5 times 5 = 125, but 5 taken 3 times = 3 times 5 = 15.
(d.) The product of a number taken any number of times as a factor is called a POWER of the number.
Illustrations. 9 is the second power of 3, because it is the product of 32, i. e., of 3 taken twice as a factor.
25 is the fifth power of 2, because it is the product of 25, i. e., of 2 taken 5 times as a factor.
(e.) The figure indicating how many times a number is taken as a factor is called the EXPONENT of the power to which the number is raised.
Thus, in 25, the exponent is 5; in 52, it is 2; in 64, it is 4; &c.
(f.) The following examples indicate the method of reading numbers expressing powers.
37 is read three seventh power, or three to the seventh power.
Note. — The second power of a number is sometimes called its SQUARE, and the third power its CUBE.
106. Method of Factoring Numbers. 1. What are the prime factors of 2772 ?
Solution. — We see by 104, II., that 2772 is divisible by 4, and therefore by 2 X 2, the prime factors of 4.
Dividing by 4 gives 693 for a quotient, which, by 104, IV., we see is divisible by 9, and therefore by 3 X 3, the prime factors of 9.
Dividing 693 by 9 gives 77 for a quotient, the prime factors of which are 7 and 11. Hence, the prime factors of 2772 are 22, 32, 7, and 11; or, which is the same thing, 2772 = 22 X 32 X 7 X 11.
2. What are the prime factors of 29766 ?
Solution. - We see by 104, I. and IV., that 29766 is divisible by both 2 and 3, and hence by their product, 6. Dividing by 6 gives 4961 for a quotient, which, by 104, V., is a multiple of 11. Dividing by 11 gives 451 for a quotient, which (104, V.) is also a multiple of il Dividing by 11 gives 41 for a quotient, which is obviously a prime number. Therefore, 29766 = 2 X 3 X 112 X 41.
3. What are the prime factors of 3871123 ? Solution. — We readily see that it is not divisible by 2, 3, 5, or 11,
and therefore we must see whether it is divisible by the other prime numbers, beginning, for convenience sake, with the smallest. By actual trial, we find that 7, 13, 17, 19, and 23 each leave a remainder after division, but that 29 is contained in it exactly 133487 times. Therefore, 29 is one of the prime factors of the given number, and the others will be found in 133487. But no number less than 29 can be a factor of 133487, for none is a factor of the original number. We therefore first try 29, which we find is contained exactly 4603 times. Therefore, 29 is again a factor of the original number, and the remaining factors must be found in 4603. But no number less than 29 can be a factor of 4603, for none was a factor of the original number. Beginning with 29, we try in succession 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, and 71, and find that none of them will exactly divide 4603; and we observe, moreover, that the entire part of the quotient of the division by 71 is less than 71.
Now, as the divisor increases, the quotient must decrease ; therefore, if there is any number larger than 71 which will divide it, the quotient must be less than 71. But when a division can exactly be performed, the divisor and quotient must each be a factor of the dividend; and hence if 4603 has a factor greater than 71, it must also have one less than 71. But as we have already found that it has no factor less than :1, we infer that it can have none greater, and that it is a prime number.
Hence, 3871123 = 292 X 4603.
(a.) The labor of testing the divisibility of a number by the various prime numbers may be made much less laborious azd tedious, by considering what must be the last figure of the quotient, if the division by any number is possible.
(6.) Thus, in the last example, in testing the divisibility of 4603, we may observe that if 29 is a factor of it, the last figure of the other factor must be 7, for 7, or some number ending in 7, is the only number which, multiplied by 29, will give for a product a number ending in 3. Dividing by 29, we have, 29 is contained in 46 once, and 17 remainder; in 170, 5 times, and 25 remainder; and we can readily see that it is con tained in 253 more than 7 times.
(c.) Again. If 31 is a factor of 4603, the last figure of the other factor must be 3, for 3, or some number ending in 3, is the only number which, multiplied by 31, will give for a product a number ending in 3. Dividing by 31, we have, 31 is contained in 46 once, and 15 remainder; in 150, 4 times, and 26 remainder; and we can see at a glance that it is contained in 263 more than 3 times.
(d.) Again. If 37 is a factor of 4603, the last figure of the other factor must be 9. But 37 is contained in 46 once, and 9 remainder; in 90, twice, and 16 remainder ; and it is obvious that it cannot be con tained as many as 9 times in 163.
(e.) From the above, we see that to find the prime factors of a number, we may first divide it by any number which will divide it without a remainder, then divide this quotient by any number which will divide it without a remainder, and so proceed with each successive quotient, till we reach one which is a prime number, or which can be readily divided into its prime factors. The prime factors of the several divisors and of the last quotient will be the prime factors required.
(f.) If in any case we cannot discover a divisor of any given number by the tests of 104, we try in succession 7, and all the prime numbers from 13 upwards, till we find one which is a divisor of the given number, or till the entire part of our quotient is less than the number employed as a divisor, in which case the number is prime.
107. Exercises for the Student. (a.) Find the prime factors of the numbers from 1 to 100, writing them as in the following model. 1, prime.
7, prime. 2, prime.
8= 2 X 2 X 2 = 28. 3, prime.
9= 3 X 3 = 32. 4= 2 X 2 = 22.
10 = 2 X 5. 5, prime.
11, prime. 6 = 2 X 3.
12= 2 X 2 X 3 = 22 X 3. (6.) We would recommend that the pupil make out a table of the prime factors of the numbers from 1 to 1000. He will find it a very profitable exercise, and one which will greatly aid him in all his subsequent work. (c.) What are the prime factors – 1. Of 1001 ?
8. Of 1183 ? 2. Of 1025 ?
9. Of 1625 ? 3. Of 1024 ?
10. Of 2057 ? 4. Of 1033 ?
11. Of 16128 ? 5. Of 1096 ?
12. Of 3809? 6. Of 1157 ?
13. Of 6381 ? 7. Of 1067 ?
14. Of 7128 ?
15. Of 7854 ? 16. Of 5989 ? 17. Of 5625 ? 18. Of 9257 ? 19. Of 10917 ? 20. Of 843479 ?
21. Of 444528?
OR MORE NUMBERS
108. Common Divisor, Definitions, and Properties. (a.) A DIVISOR of a number is a number which will exactly divide it.
(6.) A COMMON DIVISOR OF TWO is a number which is a divisor of each of them. (c.) The GREATEST
DIVISOR MORE NUMBERS is the largest number which is a divisor of each of them.
(d.) From these definitions and the principles previously established, it follows that,
1. A divisor of a number can contain only such prime factors as are found in that number.
2. A common divisor of two or more numbers can contain only such prime factors as are common to all the numbers.
3. The greatest common divisor of two or more numbers is the product of all the prime factors common to all the given numbers.
109. Greatest Common Divisor. Method by Factors. 1. What is the greatest common divisor of 819 and 1071?
Solution. — The greatest common divisor of 819 and 1071 is the product of all the prime factors common to those numbers.
819 = 32 X 7 X 13
32 X 7 X 17 From which we see that the only common factors are 32 and 7. Therefore, 32 X 7, or 6:3, is the greatest common divisor required.
What is the greatest common divisor 2. Of 792 and 930 ? 13. Of 1125 and 1575 ?