Multiply 31416 by 56. Here the multiplier 56 is composed of the two component parts 8 and 7, that is 8 times 7 make 56. Here 112 is composed of 4, 4, and 7, or 4 × 4 × 7 =112 5. Multiply 30917860 by 132. Ans. 4081157520. Ans. 3555555523200. 8. Multiply the 9 digits by the 9 digits inverted; that is, 123456789 by 987654321. Ans. 121932631112635269. 9. What is the difference between twice fifty-six and twice six and fifty? Ans. 50. 12. Multiply twelve thousand, twelve hundred, and twelve, by thirteen thousand, thirteen hundred, and thirteen. Ans. 189103356. SIMPLE DIVISION. DIVISION is the method of finding how often one number is contained in another. The number to be divided is called the dividend. The number by which the dividend is to be divided, is called the divisor. The number of times the divisor is contained in the dividend, is called the quotient. Divide 36 by 4. Here 36 contains 4 nine times. CASE I.-When the divisor does not exceed 9. RULE.-1. Observe how often the divisor is contained in the first figure of the dividend, and place this number, which is called the quotient figure, under the first figure of the dividend. 2. Carry 10 for each unit that remains to the next figure of the dividend, and find how often the divisor is contained in this sum, which place down, and so on, till you have made use of all the figures in the dividend. 3. If the first figure in the dividend be less than the divisor, take the first and second figures, and proceed as above. 4. Proof by Multiplication. In the 1st example, the 7 contains the 3 twice and one over, put down the 2 under the first figure of the dividend; for the 1 which remains carry 10 to the next figure 5, and the sum 15 being divided by 3, the quotient figure will be 5, which put down, and so on to the end. In the 2nd example, the first figure 3 in the dividend does not contain the divisor 5; consequently take the 3 and 6, and find how many 5's there are in 36: here we find it is contained 7 times and 1 over; place this 7 under the 6, and for the 1 add 10 to the next figure 7, which will make 17; then how many 5's in 17, three times and 2 remains; carry 20 to the next figure 8, and so on to the end of the division. In this example we have 1 remaining, which in the proof must be added to the product of the first figure. EXAMPLES FOR PRACTICE. 1. Divide 4567894 by 2. 2. Divide 5673281 by 3. 3. Divide 893246 by 4. 4. Divide 67897628 by 5, by 6, by 7, and by 12. 5. In 567840 shillings, how many crowns ? 6. If a cipher be placed to the right of any given number, and this product divided by 2, the quotient will be five times the given number. Query the reason. CASE II.-When the divisor consists of two or more figures. RULE.-Find how many times the divisor is contained in the same number of figures of the dividend, as in example 1st. But if the same number of figures in the dividend be less than the divisor, take one more, as in example 2nd; and place this figure in the quotient, and multiply the divisor by it; place the product under the figures taken in the dividend, from which subtract this product; to the remainder bring down the next figure in the dividend, and place it to the right of the remainder; then find how often the divisor is contained in this number, and so on. EXAMPLES FOR PRACTICE. 1. Divide 321768430 by 17. Ans. 18927554 7S Ans. 1672940 and 165534 rem. NOTE. This being one of the most difficult of the four leading rules, the pupil ought to practise it with small divisors, as 3, 4, 5, 6, &c., till he thoroughly understands the nature of it; proving each question by multipli cation. CASE III. When the divisor is a composite number, that is, composed of two or more component parts. RULE.-1. Divide the dividend by one of the component parts, and this quotient by the next component part, and so on, for the quotient required. 2. If there be a remainder after dividing by the first component part, place it down with a plus (+) before it, as in example 1. 3. If there be a remainder after dividing by the second component part, multiply this remainder by the previous divisor, to which product add the first remainder, for the true remainder. 4. If there be a remainder after the third division, multiply this remainder by the two preceding divisors, to which product add the last true remainder, and so on for any number of divisions. |