8. How much will 5 bushels of apples come to at 22 cents per bushel? Ans. 110 cents. Solution-22 is 2 tens and 2 units, and 5 times 2 units are ten units or 10; and 5 times 2 tens are 10 tens, and 10 tens and 1 ten make 11 tens 110, the answer. 9. There is an orchard containing 5 rows of trees, in each row are 23 trees; how many trees does the orchard contain? In the first row 23 trees second 23 third 23 fourth 23 fifth 23 In the whole 115 trees Multiplicand 23 Product 5 115 By this we see that we may obtain the whole number of trees by setting down 23, 5 times and adding it up, making 115 trees in the whole. But we need only set down the number of trees contained in 1 row, and as 5 rows will contain 5 times as many, we may place the 5 underneath for a multiplier ; we know that 5 times 3 units are equal to 15 units or 1 ten and five units; we set down the 5 units and reserve the 1 ten; then 5 times 2 tens are 10 tens, and 1 ten which we reserved makes 11 (tens) which set down, and the answer is 115, as before. Hence we find that Multiplication is a short way of performing Addition. SÍMPLE MULTIPLICATION Teaches to repeat the greater of two simple numbers as many times as there are units in the less or multiplying number, or it is a compendious method of performing many additions. The number to be multiplied is called the multiplicand. The number you multiply by, is called the multiplier. The number found by the operation, is called the product. Both multiplier and multiplicand are called factors. Note. It will not make any difference in the result of the multiplication, whether we make the greater number the multiplicand, and the less number the multiplier, or the less number the multiplicand and the greater number the multiplier; for the product of any two factors will be the same in either case: (thus 5 taken 3 times is the same as 3 taken 5 times.) But in practice it is more convenient to make the greater number the multiplicand and the less, the multiplier. CASE I. When the multiplier is not greater than 12. RULE. Place the multiplier under the multiplicand, and multiply each figure in the multiplicand by the multiplier, setting down and carrying as in Simple Addition. EXAMPLES. 1. There are 365 days in one year; how many days are there in three years? Illustration-It is evident that if one year contains 365 days, three years will contain 3 times 365 days. Operation. We write down the multiplicand and 365 multiplicand place the multiplier under it, making the greater number the multiplicand; the less the multiplier. 3 multiplier 1095 Product We then say 3 times 5 are 15 setting down the 5 (units) and carrying 1 (ten,) as in Simple Addition; then 3 times 6 are 18 and 1 to carry makes 19, setting down 9 and carrying 1 to the next, we say 3 times 3 are 9 and 1 makes 10, setting down the whole product. Hence we find that 3 times 365 is equal to 1095. We might have obtained this same answer by setting down 365 the multiplicand 3 times and adding it up. 11. If a man travel 48 miles a day, how many miles will he travel in 9 days? Ans. 432 miles. 12. What is the product of 956 multiplied by 12? Ans. 11472. 13. What will 325 barrels of flour come to at 7 dollars a barrel ? Ans. $2275. 14. 320 rods make a mile; how many rods are there in 8 miles ? CASE II. Ans. 2560. When the Multiplier consists of several figures. RULE. 1. Write the multiplier under the multiplicand, placing units under units, tens under tens, &c. 2. Then multiply by each significant figure in the multiplier, separately, and place the first figure in each product, directly under its multiplier. 3. Add together the several products in the same order as they stand, and you will have the total product. Proof.*—Multiply the multiplier by the multiplicand. EXAMPLES. 1. There are 365 days in one year; how many days are there in 25 years? Operation. Days in 1 year 365 25 1825 730 Illustration. We first multiply each figure in the multiplicand by the 5 (units) in the multiplier, which gives the product of 5 times 365 or the number of days in 5 years. We then multiply each figure in the multiplicand by the 2 tens in the Days in 25 yrs. 9125 multiplier: placing the first figure in * Another method of Proof is as follows, viz.: cast out the 9's in both factors, and set down the remainder at the right hand; then if the excess of 9's in the two remainders when muitiplied together be equal to the excess of 9's in the total product, the work is supposed to be right. Thus, 365-5 excess of 9's. 29-2 excess of 9's. 3285 10 excess=1. 10585 excess-1. the product directly under its multiplier; and as the value of any number is increased ten times by being moved one place to the left, it is evident that the product found by multiplying by 2 in the ten's place is equal to 20 times 365 or the number of days in 20 years; and the product of 5 years and 20 years added together are equal to 9125 days or the number of days in 25 years. 37687 205 188435 75374 7725835 4102735 When there are ciphers between the significant figures in the multiplier as in this example; omit the ciphers, and multiply by the significant figures only; placing the first figure in each product under its multiplier, as before. 11. What number is equal to 3027 times 82164973 ? 12. Multiply 27501976 by 271. Ans. 248713373271. Ans. 7453035496. 13. Multiply 8496427 by 874359. Ans. 7428927415293. 14. Multiply 95644796 by 8000009. Ans. 765159228803164 15. Multiply 562916859 by 49007. Ans. 27586866509013. CASE III. When ciphers stand at the right hand of either or both of the factors, neglect them, and place the significant figures under one another and multiply by them only; then place as many ciphers on the right hand of the product, as were neglected in both factors. 8. 930137000 X 9500=8836301500000. When the Multiplier is 10, 100, 1000, &c. Since numbers increase in a tenfold proportion from the right hand to the left, it is evident that if we place one cipher on the right hand of any figure it increases the value of that figure ten times by removing it to the place of tens; and if we place two ciphers on the right of any figure it increases its value a hundred fold by removing it to the place of hundreds. Hence the following RULE. Place the ciphers of the multiplier on the right of the multiplicand and it makes the product required. |