| James B. Dodd - Arithmetic - 1850 - 278 pages
...remainder ; divide the quotient in like manner ; and so on, until the quotient becomes a prime number. 2. The several divisors and the last quotient will be the prime factors required. 3. If the given number can only be divided by itself, or a unit, without a remainder, it... | |
| James B. Dodd - Arithmetic - 1852 - 410 pages
...remainder ; divide the quotient in like manner ; and so on, until the quotient becomes a prime number. 2. The several divisors and the last quotient will be the prime factors required. 3. If the given number can only be divided by itself, or a unit, without a remainder, it... | |
| Charles D. Lawrence - Arithmetic - 1854 - 336 pages
...obtained by any prime number of which it is a multipleProceed in this manner till a quotient is obtained which is a prime number. The several divisors and the last quotient will be the prime factors required. In the application of this rule, the pupil may find it convenient to consult the following... | |
| Charles Davies - Arithmetic - 1856 - 450 pages
...any prime number that will exactly divide it, and so on, till a quotient is found which is prime ; the several divisors and the last quotient will be...NOTE. — It is most convenient, in practice, to use at each division the least prime number that is a divisor. 1. What are the prime factors of 105 ? ANALYSIS.—... | |
| John Fair Stoddard - Arithmetic - 1856 - 312 pages
...manner. and so continue dividing until the quotient obtained is a prime number. Then, a unit, 2)36 the several divisors,^ and the last quotient will be the prime factors required. Proceeding, thus, we find 2)18 the prime factors of 144 to be 1, 2, 2, 2, 2, 3, and 3. 3)9... | |
| Benjamin Greenleaf - Arithmetic - 1857 - 336 pages
...2 divide this also by 2, and obtain a quotient 6. We divide 6 by 2, and obtain 3 for a 6^ quotient, which is a prime number. The ^~ several divisors and the last quotient, all '" being prime, constitute all the prime factors of 24, which, multiplied together, 2X2X2X3 = 24... | |
| Benjamin Greenleaf - Arithmetic - 1857 - 452 pages
...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 will be the prime factors required. NOTE 1. — The composite factors of any number may be found by multiplying together two... | |
| Benjamin Greenleaf - Arithmetic - 1858 - 458 pages
...; and, since 21 is Q 9 ia composite number, we divide this by 3, and obtain - — for a quotient 7, which is a prime number. The « several divisors and the last quotient, all being prime, constitute all the prime factors of 42, which, multiplied together, they equal. Hence... | |
| Benjamin Greenleaf - Arithmetic - 1858 - 456 pages
...and, since 21 is o -7T a composite number, \ve divide this by 3, and obtain - — for a quotient 7, which is a prime number. The « several divisors and the last quotient, all being prime, constitute all the prime factors of 42, -which, multiplied together, they equal. Hence... | |
| James B. Dodd - Arithmetic - 1859 - 368 pages
...it without a remainder. 2. Divide the quotient in like manner; and so on, until the quotient becomes a. prime number. The several divisors and the last...quotient will be the prime factors of the given number. EXAMPLE. To resolve 210 into its prime factors. 2)210 3)105 5)35 ~T 1 2 3 5 7 11 13 17 19 23 29 31... | |
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