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FACTORING.

178. Factoring is the process of resolving a quantity into its factors.

179. Every number that is not prime is composed of prime factors, since all numbers are either prime or composite; and, if composite, can be separated into factors, which, if themselves composite, can be further separated into those that shall be prime.

180. To resolve a composite number into its prime factors. Ex. 1. It is required to find the prime factors of 42.

Ans. 2, 3, 7.

Operation. We divide by 2, the least prime number greater

2 4 2 than 1, and obtain the quotient 21; and, since 21 is

3 2~1 a comPos'te number, we divide this ^'y 3, and obtain for a quotient 7, which is a prime number. The

7 several divisors and the last quotient, all being prime, constitute all the prime factors of 42, which, multiplied together, they equal. Hence

Divide the given number by any prime number that will exactly divide it, and the quotient, if a composite number, in the same manner; and so continue dividing, until a prime number is obtained for a quotient. The several divisors and the last quotient ivill be the prime factors required.

Note 1. — The composite factors of any number may be found by multiplying together two or more of its prime factors.

Note 2. — Such prime factors as two or more numbers may have alike, are termed prime factors common to them; and these may be readily determined after the numbers are resolved into their prime factors.

Examples.

2. What are the prime factors of 105? Ans. 3, 5, 7.

3. Resolve 220 into its prime factors.

4. What are the prime factors of 936?

Ans. 2, 2, 2, 3, 3, 13.

5. What are the prime factors of 1953?

6. Resolve 12462 into its prime factors. Ans. 2,3, 31, 67.

7. Resolve 19987 into its prime factors. Ans. 11,23,79.

8. What are the prime factors common to 225, 435, and 540? Ans. 3, 5.

9. What are the prime factors common to 960, 1568, and 5824?

10. What are the prime factors common to 2340, 11934, 12987, and 14859? Ans. 3, 3, 13.

11. A man has 105 apples, which he wishes to distribute into small parcels, each of equal numbers; what are the smallest whole numbers, greater than 1, into which they may be exactly divided? Ans. 3, 5, and 7.

DIVISIBILITY OF NUMBERS.

181i One number is said to be divisible by another, when the latter will divide the former without a remainder. Thus, 9 is divisible by 3.

182. One number is divisible by another, when it contains all the prime factors of that number. Thus, 12, which contains all the factors of 4, is divisible by 4.

183i All even numbers, or such as terminate with 0, 2, 4, 6, or 8, are divisible by 2, since each of them contains 2 as a factor. Thus, 10, 24, 36, 58, are each divisible by 2.

184i All numbers which terminate with 0 or 5 are divisible by 5, since each of them contains 5 as a factor. Thus, 20, 25, 50, are each divisible by 5.

185. Every number is divisible by 4, or any other number that will exactly divide 100, when its two right-hand figures are divisible by the same. For any figure on the left of the two right-hand figures must express one or more hundreds, and a factor of one hundred is a factor of any number of hundreds; so, if the sum exactly divides the units and tens of a number, the entire number will be divisible by it. Thus, 116 is divisi ble by 4; 140, by 20; 225, by 25; and 450, by 50.

186. Every number is divisible by 8, or any other number that will exactly divide 1000, when its three right-hand figures are divisible by the same. For any figure on the left of the three right-hand figures must express one or more thousands, and a factor of one thousand is a factor of any number of thousands; so, if the sum exactly divides the units, tens, and hundreds of a number, the entire number will be divisible by it. Thus, 1824 is divisible by 8; 1840, by 40; 3375, by 125; 2750, by 250; and 4500, by 500.

187i Every number the sum of whose digits 3 or 9 will exactly divide, is divisible by 3 or 9. For 10, or any power of 10, less 1, gives a number, as 9, 99, 999, &c, which is divisible by 3 and by 9. Hence, any number of tens, hundreds, thousands, &c, less as many units, must be divisible by 3 and by 9; and if the excess of units denoted by the significant figures, in the aggregate, is likewise divisible by 3 and by 9, it follows that the entire number is thus divisible. For example, 7542 is a number, the sum of whose digits is divisible by 3 and by 9; and separated into tens, hundreds, and thousands, it is equal to 7000 -f- 500 -4- 40 -4- 2- Now, 7000 = 7 X 1000 = 7 x (999 + 1) = 7 X 999 + 7; 500 = 5 x 100 = 5 X (99 +1) = 5 X 99 + 5; and 40 = 4 X 10 = 4 X (9 + 1) = 4 X 9 + 4. Therefore, 7542 = 7 X 999 + 5 X 99+ 4X9 + 7 +5 + 4+ 2. The remainders 7 + 5 + 4 + 2, corresponding with the significant figures of the number, added together, equal 18, which sum being divisible by 3 and by 9, it is evident that 7542 is divisible in like manner.

Note. — Upon the property of 9 now explained depends the method of proving, by excess of nines, multiplication (Art. 68), and division (Art. 75).

The same method of proof may be resorted to in addition and in subtraction. Thus,

To prove Addition. Find the excess of nines in each number added, and then the excess of nines in the sum of these results; which, if the work be right, will equal the excess of nines in the answer.

To prove Subtraction. Find the excess of nines in the subtrahend, and also in the remainder, and then the excess of nines in the sum of these results; which, if the work be right, will equal the excess of nines in the minuend.

188i Every number occupying four places, in which two like significant figures have two ciphers between them, is divisible by 7, 11, and 13. Thus, 9009, 1001, 3003, 4004, &c. are each divisible by 7, 11, and 13

189i Every number is divisible by 11, in which the sum of the digits in the odd places is equal to the sum of the digits in the even places, or in which the difference of their sums can be exactly divided by 11. Thus, 8305, in which 3 -4- 5 = 8 -f- 0, and 628001, in which 2 + 0+1 and 6-4-8 + 0 differ by 11, are each divisible by 11.

190. Every number divisible by two or more numbers, which are prime to each other, is divisible by their product. For, being prime to each other, dividing by one of the numbers does not cancel the others as factors. Thus, 770, being divisible by 2, 5, and 7, which are prime to each other, is divisible by 70, their product.

191 i Every even number, the sum of whose digits 6 will exactly divide, is divisible by 6. For being even, it is divisible by 2 (Art. 183), and its digits being divisible by twice 3, or 6, are evidently divisible by once 3, so that the number is also divisible by 3; and as the 2 and the 3 are prime to each other, the number is divisible by their product, or 6 (Art. 190). Thus, 174, 6312, are each divisible by 6.

192. Every number terminating with 0 or 5 that 3 will exactly divide, is divisible by 15, and every number that 9 will exactly divide, is divisible by 45. For, terminating with 0 or 5, it is divisible by 5 (Art. 184), and, as 3 and 9 are each prime to 5, if it can be exactly divided by 3 or by 9, it must be divisible by 5 x 3 = 15, or by 5 X 9 = 45. Thus, 75, which 3 will exactly divide, is divisible by 15; and 90, which 9 will exactly divide, is divisible by 45.

Divisors Or Measures.

193. A divisor or measure of a number is any number. that will divide it without a remainder. Thus, 3 is a divisor or measure of 6, and 5 a divisor or measure of 10.

194. ° To find all the divisors or measures of a number.

Ex. 1. Required all the divisors of 60.

Ans. 1, 2, 3, 4,5, 6,10,12,15, 20, 30, and 60.

OPERATION.

Divisors

Resolving the

60 = 2X2X3X5- number into its

1 2 4=2X2 Prm'e .fafic'ors'

x wo find 60 =

3 6 12=2X2X3 2X2X3X5.

5 10 20 = 2X2X5 Now,anynum

15 30 60 = 2X2X3X 5. £er TM

Ans. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. ^composite

number by its

prime factors, and every product these factors can form. Now 1, being a divisor of every number, is a divisor of 60, and, since 2 enters twice as a factor into 60, it is evident that 2, and 4 = 2x2, are also divisors of 60. These divisors we arrange on a horizontal line, and determine the other divisors by multiplying those on this line by the factor 8, for the second line of divisors, by 5 for the third line, and by 3 x 5 for the fourth line, and thus obtain all the possible divisors of the given number.

The whole number of divisors is 12, which corresponds to the product arising from multiplying together the exponents, each increased by 1, of the different prime factors of 60; thus, of the different prime factors, since 2 enters twice, its exponent is 2, -\- 1 = 3; 3 enters once, its exponent is 1, -f- 1 = 2; 5 enters once, its exponent is 1, + 1 = 2; and 3 x 2 X 2 = 12, the number of divisors. The same holds true in all cases. Hence, in any composite number,

To Find The Number Op Divisors. Multiply together the exponents, each increased by 1, of the different prime factors of the given number, and the product will be the number of divisors required. And

To Find The Several Divisors. Form from the prime factors of the number all the products possible, and these factors (including 1) xnd products will be the divisors required.

Examples.

2. What are the divisors of 72?

Ans. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.

3. Required the divisors of 105?

4. How many divisors has 1764? Ans. 27.

5. How many divisors has 3528?

6. How many divisors has 5880? Ans. 48.

Common Divisors Or Measures.

195. A common divisor or measure of two or more numbers is any number that will divide them without a remainder; thus, 2 is a common divisor of 4, 8, and 10

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