 | Thomas Keith - Arithmetic - 1822 - 354 pages
...quotient by the other, for the required quotient. If there be a remainder to each of the quotients, multiply the last remainder by the first divisor, and to that product add the first remainder for the true one. Prop. 3. When the divisor consists of sevtral figures. Rule. Find how many times... | |
 | Francis Walkingame - 1833 - 204 pages
...remainder must be found by the following RULE. Multiply the second remainder by the first divisor : to that product add the first remainder, which will give the true one. (25) 3210473 -ч- 27.§ (26) 7210473 ч- 35. (27) 6251043 4-42 (28) 5761034 -4- 54. • EXAMPLE.... | |
 | Jason M. Mahan - Arithmetic - 1839 - 312 pages
...true one, therefore, to find the true remainder, multiply the first divisor by the last remainder ; to that product add the first remainder, which will give the true one. Examples. 1. Divide 9327 by 108, Here 12 x 9 = 108 Then, 9)9327 12)1036 + 3 _86 +4 Here is 3 for... | |
 | Francis Walkingame - 1855 - 186 pages
...them the true one, it may be found by this RDLE. Multiply the first divisor into the second remainder, to that product add the first remainder, which will give the true one. Thus, to divide 5603782 by 24, since 24 is = 4 times 6, you may divide first by one of these numbers,... | |
 | Daniel Adams - Arithmetic - 1861 - 452 pages
...dividing by TWO or more factors or divisors ; to get the TRUE remainder, — RULE. 1. If there be but TWO divisors. — Multiply the last remainder by the first divisor, and to the product add the first remainder ; the sum will be the true remainder. II. When there are MORE than... | |
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... will give the quotient required ; but as it sometimes happens that there is a remainder to each of the quotients, and neither of them the true one, it may be found by this RDLE. 