left, and again 5 times from what is left, we have 23 units over, a number which is less than 35. Hence we see that 35 is contained in 9298— 200+60 + 5-i.e., 265 times—with a remainder 23. We might have written down the 35)9298(200 35) 2298 (60 2100 35) 198(5 23 process thus : The quotient is therefore 200 + 60 + 5, or 265, and the remainder 23. 18. The above explanations will sufficiently elucidate the following Rules for Division: (1.) When the divisor contains only one figure, write the divisor on the left of the dividend, with a curved line between them. Beginning at the left hand, divide successively each figure of the dividend by the divisor, and place each quotient figure directly under the figure divided. If there is a remainder after dividing any figure, prefix it to the next figure of the dividend, and divide the number so formed as before. If there occur any figure which does not contain the divisor, place a cipher in the quotient, and prefix this figure to the next one of the dividend, as if it were a remainder, and proceed in the same manner to the last figure. (2.) When the divisor contains more than one figure, beginning on the left of the dividend, find how many times the divisor is contained in the first fewest figures of the dividend which will contain it, and place the quotient figure on the right hand of the dividend, with a curved line between them; then multiply the divisor by this figure, and subtract the product from the figures divided. To the right of the remainder bring down the next figure of the dividend, and divide the number so formed as before. If this number be less than the divisor, annex a cipher to the quotient, and bring down the next figure, continuing this process until the number thus obtained be equal to or greater than the divisor. Proceed in this manner until all the figures of the dividend are exhausted. 19. Tests of Correctness of Division : (1.) Multiply the divisor by the quotient, and add the remainder to the product. This should, as already explained, give the dividend. (2.) Subtract the remainder, if any, from the dividend, and divide the difference so obtained by the quotient. The result should be equal to the divisor, if the working be correct. EXERCISE VIII. EXAMPLES IN LONG DIVISION. (1.) 4783942. (2.) 75043 52. (3.) 93840÷63. (8.) 1912500 ÷ 425. (12.) 653585478232789. (13.) 908070605040 ÷ 654321. (15.) 10000000000000000÷IIII. 30 CHAPTER IV. ABRIDGED METHODS OF MULTIPLICATION AND DIVISION. 1. THE methods of multiplication and division explained in the previous chapters are those ordinarily employed; and the learner must make himself perfectly familiar with them before proceeding further. These processes, however, in particular cases, can often be materially facilitated by various artifices. Some of these shorter methods we subjoin, not only because they are useful in themselves, but because they are valuable as exercises, in explaining the fundamental principles of arithmetic. 2. Any number which is formed by multiplying two or more numbers or factors together is called a Composite number. It has already been explained (Chapter III. 3) that the same numbers multiplied together will give the same product, in whatever order the multiplication is effected. Hence, to multiply any number by one which is composite-i.e., which is composed of several factorswe have only to multiply the number first by one factor, the result by another factor, and so on. 3. Similarly, it will be seen that to divide by any composite number, we have only to divide by one factor, then divide the quotient by another factor, and so on. Thus, to divide 9856 by 28, arrange the process as indicated in the margin. 28 In this case there is no remainder. be required to divide 9873 by 28. Proceeding as before, we get a remainder after the division by 7, 3 and a remainder 2 after the division by 4. The first remainder 3 means 3 units; and the 2 which remains 7) 9856 4 1408 352 Answer. But suppose it 9873 28 1410-3 352-2 after dividing 1410 by 4, means 2 sevens of the 1410 sevens which are contained in 9873. Hence the whole remainder will be 2 sevens + 3 units-i.e., 17. The process may be exhibited analytically thus:9873 = 1410 X 7+ 3 = 352 X 4 X 7 + 2 × 7+ 3 = 352 × 28 +14 + 3 =352 × 28 + 17 Therefore 9873 divided by 28 has a quotient 352, and remainder 17. Hence, when there are two factors, to find the whole remainder, multiply the second remainder into the first divisor, and add the first remainder. 4. If there are more than two factors, similar considerations will show that the following rule will give the whole remainder :-Multiply the last remainder into the continued product of all the divisors but the last, the last but one remainder into the product of all the divisors except the last two, and so on. Add all these results and the first remainder together; the sum will be the whole remainder. EXAMPLE.-To divide 17285 by 84. 84 = 7 X 4 X 3 7) 17285 4) 2469-2 3) 617-1 205-2 And the whole remainder is 2 X 4 × 7 + 1 × 7 + 2, that is, 56+7+2, or 65. It does not follow that this process is in all cases simpler than the method of Long Division; sometimes, however, it is more convenient. 5. Multiplying and dividing by powers of 10, and by numbers ending in any number of ciphers. The products of two tens, three tens, four tens, &c., are called respectively the second, third, fourth, &c., powers of 10. They are 100, 1000, 10000, &c. Thus, the second power is I followed by two ciphers, the third I followed by three ciphers, and so on; the number of the ciphers in each case being the same as that of the power. Ι It has been already explained that to multiply by 10, or any power of 10, we have only to annex to the multiplicand the number of ciphers corresponding to the power. Thus, 345 multiplied by 1000 is 345000. If any number of the right hand figures in the multiplier be ciphers-as, for instance, in 75000-then, as we have already seen (Chap. III., Art. 5), we need only multiply the multiplicand by 75, and annex to the product the same number of ciphers, in this case three. 6. To divide by 10, or any power of 10. If the dividend have more ciphers for its right hand figures than occur in the power of 10 by which it is to be divided, we need only take away from it the number of ciphers in the divisor to obtain the quotient. Thus, 873000 divided by 100 and 1000 respectively, gives quotients 8730 and 873. But suppose that the dividend has no ciphers for its right hand figures. Take, for instance, the case of 87346 divided by 100. Cut off the two right hand figures-viz., 46-from the dividend; then 873 will be the quotient and 46 the remainder. This is evident by exhibiting the process analytically, thus:— Therefore 873 is the quotient, and 46 the remainder. |