 | Nathan Daboll - 1817 - 252 σελίδες
...measure, by dividing the greater term by the less, and this divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains; the last divisor is the common measure.* 2. Divide both of the terms of the fraction by the common measure,... | |
 | Nathan Daboll - 1818 - 246 σελίδες
...measure, by dividing the greater term by the less, and this divisor by the remainder, and so on, always dividing the last divisor' by the last remainder, till nothing remains, the last divisor is the common measure.* 2. Divide both of the terms of the fraction by the common measure,... | |
 | George G. Carey - 1818 - 602 σελίδες
...Divide the greater number by the less, and this divisor by the remainder. Proceed in this manner, always dividing the last divisor by the last remainder, till nothing remains; the last divisor is the greatest common measure. EXAMPLE. Required the greatest common measure of 84 and... | |
 | James Ryan - 1827 - 290 σελίδες
...other, problems VII. &c. If there be a remainder, divide the less by it ; and thug proceed, always dividing the last divisor by the last remainder, till nothing remains. The divisor which leaves no remainder, is the common measure required. If the divisor which leaves DO remainder... | |
 | Frederick Emerson - 1833 - 198 σελίδες
...FIND THE GREATEST COMMON DIl'ISOR of two numbers, — Divide the greater number by the smaller, then divide the divisor by the remainder; and thus continue...divisor of 91 and 117. 91)1 17(1 This operation is perform91 ed according to the direction above, and 13 is found to be tne greatest common divisor; or... | |
 | Frederick Emerson - 1834 - 300 σελίδες
...will divide them both without a remainder. R ULE. Divide the greater number by the smaller, and this divisor by the remainder, and thus continue dividing...last remainder, till nothing remains. The divisor last used will be the number required. When the greatest common measure of more than two numbers is... | |
 | James Thomson (LL.D.) - 1837 - 296 σελίδες
...number by the less. (2.) If there be a remainder, divide the less by it ; and thus proceed, always dividing the last divisor by the last remainder, till nothing remains. The divisor Avhich leaves no remainder is the common measure required. If in the operation any divisor be a prime... | |
 | Nathan Daboll - 1837 - 262 σελίδες
...measure, by dividing the greater term by the less, and this' divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains ; the last divisor is the common measure.* 2. Divide both of the terms of the fraction by the common measure,... | |
 | Frederick Emerson - 1839 - 300 σελίδες
...will divide them both without a remainder. R ULE. Divide the greater number by the smaller, and this divisor by the remainder, and thus continue dividing...last remainder, till nothing remains. The divisor last used will be the number required. When the greatest common measure of more than two numbers is... | |
 | Frederick Emerson - 1839 - 300 σελίδες
...will divide them both without a remainder. RULE. Divide the greater number by the smaller, and this divisor by the remainder, and thus continue dividing the last divisor by the last remainder, (ill nothing remains. The divisor last used will fre the number required. common measure of the number... | |
| |
Divide the greater number by the less, and that divisor by the remainder, and so on ; always dividing the last divisor by the last remainder, till nothing remains ; the last divisor is the greatest common divisor required. 
