RULE. Divide the given number by any prime factor; divide the quotient in the same manner, and so continue the division until the quotient is a prime number. The several divisors and the last

quotient will be the prime factors required.

PROOF. The product of all the prime factors will be the given number.

EXAMPLES FOR PRACTICE.

1. What are the prime factors of 2150?
2. What are the prime factors of 2445?
3. What are the prime factors of 6300?
4. What are the prime factors of 21504?
5. What are the prime factors of 2366?
6. What are the prime factors of 1000?
7. What are the prime factors of 390625?
8. What are the prime factors of 999999?

143. If the prime factors of a number are small, as 2, 3, 5, 7, or 11, they may be easily found by the tests of divisibility, (136), or by trial. But numbers may be proposed requiring many trials to find their prime factors. This difficulty is obviated, within a certain limit, by the Factor Table given on pages 72, 73.

By prefixing each number in bold-face type in the column of Numbers, to the several numbers following it in the same division of the column, we shall form all the composite numbers less than 10,000, and not divisible by 2, 3, 5, 7, or 11; the numbers in the columns of Factors are the least prime factors of the numbers thus formed respectively. Thus, in one of the columns of Numbers we find 39, in bold-face type, and below 39, in the same column, is 77, which annexed to 39, forms 3977, a composite number. The least prime factor of this number is 41, which we find at the right of 77, in the column of Factors.

144. Hence, for the use of this table, we have the following RULE. I. Cancel from the given number all factors less than 13, and then find the remaining factors by the table.

II. If any number less than 10,000 is not found in the table, and is not divisible by 2, 3, 5, 7, or 11, it is prime.

1. Resolve 1961 into its prime factors.

OPERATION.

37 53

1961
1961 =

=

37 × 53,

Ans.

ANALYSIS. Cutting off the two right hand figures of the given number, and referring to the table, column No., we find the other part,

19, in bold-face type; and under it, in the same division of the column, we find 61, the figures cut off; at the right of 61, in column Fac., we find 37, the least prime factor of the given number. Dividing by 37, we obtain 53, the other factor.

2. Resolve 188139 into its prime factors.

EXAMPLES FOR PRACTICE.

1. Resolve 18902 into its prime factors. 2. Resolve 352002 into its prime factors. 3. Resolve 6851 into its prime factors. 4. Resolve 9367 into its prime factors. 5. Resolve 203566 into its prime factors. 6. Resolve 59843 into its prime factors. 7. Resolve 9991 into its prime factors. 8. Resolve 123015 into its prime factors. 9. Resolve 893235 into its prime factors. 10. Resolve 390976 into its prime factors. 11. Resolve 225071 into its prime factors. 12. Resolve 81770 into its prime factors. 13. Resolve 6409 into its prime factors. 14. Resolve 178296 into its prime factors. 15. Resolve 714210 into its prime factors.

Ans. 2, 13, 727.

CASE II.

145. To find all the exact divisors of a number.

It is evident that all the prime factors of a number, together with all the possible combinations of those prime factors, will constitute all the exact divisors of that number, (142, II).

1. What are all the exact divisors of 360?

ANALYSIS. By Case I we find the prime factors of 360 to.be 1, 2. 2, 2, 3, 3, and 5. As 2 occurs three times as a factor, the different combinations of 1 and 2 by which 360 is divisible will be 1, 1x2 = 2, 1 × 2 × 2=4, and 1 × 2 × 2 × 2=8; these we write in the first line. Multiplying the first line by 3 and writing the products in the second line, and the second line by 3, writing the products in the third line, we have in the first, second and third lines all the different combina tions of 1, 2, and 3, by which 360 is divisible. Multiplying the first, second and third lines by 5, and writing the products in the fourth, fifth and sixth lines, respectively, we have in the six lines together every combination of the prime factors by which the given number, 360, is divisible.

Hence the following

RULE. I. Resolve the given number into its prime factors.

II Form a series having 1 for the first term, that prime factor which occurs the greatest number of times in the given number for the second term, the square of this factor for the third term, and so on, till a term is reached containing this factor as many times as it occurs in the given number.

III. Multiply the numbers in this line by another factor, ind these results by the same factor, and so on, as many times as this factor occurs in the given number.