the next product. Eight times 4 tens, are 32 tens, and the 7 tens carried, make 39 tens, which is 9 of the order of tens, to be written under that order, and 3 hundreds to be carried to the next product. Eight times 2 hundreds, are 16 hundreds, and the 3 hundreds carried, make 19 hundreds, which are written down. When the multiplier consists of several orders, another method is adopted. For example, Multiply 324 by 67. The 324 is first to be multiplied by the 7 units, according to the former rule, and the figures stand thus, 324 2268 The 324 is now to be multiplied by the 6; what is the number represented by the 6? Ans. 60 or 6 tens. If 4 is multiplied by 6 tens, the answer is 24 tens, or 240. The 4 is to be written in the order of tens, under the 6, [The cipher is omitted because, by setting it under the 4, we can know what order it is without a cipher,] and the 2 (which is 200) is to be carried to the next product. See below. 324 67 2268 1944 21708 Ans. In the sum above, of what order is each of the figures? What is the product of 4 units multiplied by 6 tens? Why is the cipher omitted? The 2 tens, or (20), are next multiplied by the 6 tens, (or 60) and the answer is 12 hundreds, (1200) and the 2 hundreds to be carried to it, make 1400. The 4 is written in that order, and the 1 carried to the next product. Next the 3 hundreds are multiplied by the 6 tens, and the answer is 18 thousands, (18000) and the 1 to be carried to it, make 19 thousands, which are placed in their orders. Then the two products are added together, and the answer is obtained. Let the pupil answer the following questions on the above sum. What number does the 6 of the multiplier, represent ? What number does the 2 represent? If they are multiplied together, as if they were units, what is the product? How many ciphers must be added, to express the true value of 2 tens, multiplied by 6 tens? How many figures are at the right hand of both the factors, 2 tens and 6 tens? Is the number of ciphers added, the same as the number of figures at the right hand of both the factors? What is the answer if the 3 hundreds be multiplied by 6 tens, as if they were units? How many ciphers must be added, to make the product express the true value? Does the number of ciphers added, correspond to the number of figures, at the right of both factors? By answering the above questions, the pupil will understand the following principle. Figures of any order may be multiplied together like units, and the true value is found, by annexing as many ciphers as there are figures at the right of both the factors. Let the following questions be answered. Multiplicand 869 What number is represented by 61 by 3? If the 6 is multiplied by the 3, what is the answer, if the factors are considered as units? What is the true answer ? If the 8 is multiplied by 3, what is the answer if they are considered as units? what is the true answer? What number is represented by 2? by 8? If the 2 is What is the product of 2 tens multiplied by 6 tens, and how is it set down? How can the true value of any orders that are multiplied together be found? Give examples. multiplied by 8, what is the answer if they are considered as units? What is the true answer? Let the pupil now learn to multiply the above sum, and place the figures in the orders to which they belong; thus, 869 Multiplicand. 237 Multiplier. 6083 2607 1738 205952 Answer. The multiplicand is first multiplied by the 7 of the multiplier, and the product is 6083. Then the 3 tens (or 30) are multiplied into the 9 units, and the answer is 270; which is 7 tens to be set in the order of tens, and 2 hundreds to be carried to the next product.* Then the 6 tens (or 60) are multiplied by 3. tens, and the product is 1800, and the 2 that were to be carried make 2000; which is 2 of the order of thousands to be carried to the next product, and 0 to be set in the or der of hundreds. Then the 8 hundreds are multiplied by 3 tens, and the answer is 24000, and the 2 to be carried make 26000; which is 6 to be set in the order of thousands, and 2 in the order of tens of thousands. Next take the 2 hundred as multiplier, and multiply 9 units by it, and the answer is 1800; which is 8 to be set in the order of hundreds, and 1 to be carried to the next product. Proceed thus, till all the orders have been multiplied by the 2 hundred. Then add the several products and the answer is obtained. The cipher is omitted because, as the figure is set under the 8, we can tell to what order it belongs without the cipher. What is the product of 9 units multiplied by 3 tens, in the above sum? In writing it why is the cipher omitted? What is the product of 6 tens multiplied by 3 tens, and how is it to be written? What is the product of the other orders, and how are they written? RULE FOR SIMPLE MULTIPLICATION, WHEN THE MULTIPLIER HAS SEVERAL ORDERS. Place the multiplier below the multiplicand, so that units of the same order may stand in the same column. Multiply by each order of the multiplier. Write the units of each product, in the order to which they belong, and carry the tens to the next product. Add the products of the several orders, and the sum is the answer. EXAMPLE. 826 234 3304 2478 1652 193284 Multiply by the 4 units according to the other rule. Then multiply each order of the multiplicand by the 3 tens (or 30) thus: 6 units multiplied by 3 tens are 18 tens, which is 8 tens to be written in that order, and 1 of the order of hundreds to be carried to the next product. 2 tens, (or 20) multiplied by 3 tens (or 30) are 600, and the 100 carried, makes 700, which is 7 to be written in the order of hundreds. 8 hundreds multiplied by 3 tens, (or 30) is 24000; which is 4, to be written in the order of thousands, and 2 tens of thousands to be set in that order. Lastly, multiply each order of the multiplicand by the 2 hundreds. 6 units multiplied by 2 hundreds, are 12 hundreds, which is 2 hundred to be written in that order, and 1 thousand to be carried to the next product. 2 tens (or 20) multiplied by 2 hundreds, are 4000, and the 1000 carried makes 5000, which is 5 to be placed in the order of thousands. 8 hundreds multiplied by 2 hundreds, are 160,000, which is 6 tens of thousands, to be written in that order, and 1 hundred of thousands, to be written in the order of hundreds of thousands. What is the rule for Simple Multiplication, when the multiplier has several orders? Add all the orders of the products, by the rule of common addition, and the sum is the answer. If five stands alone (5) of what order is it? If a ci pher is affixed, of what order is it? How much larger is the sum, than it was before? By what number was it multiplied when the cipher was added? If two ciphers are added to the 5, in what order will it stand? How much larger is the sum than it was before? By what number was it multiplied when the ciphers were added? If three ciphers are added to 5, in what order will it stand? How much larger is the sum than it was before? By what number was it multiplied when the ciphers were added? If you wish to multiply 5, by 10, what is the shortest way? If you wish to multiply 5, by 100, what is the shortest way? If you wish to multiply 5, by 1000, what is the shortest way? If you wish to multiply 50, by 2, how would you do it? Would it make any difference if you should multiply the 5 first, and then affix a cipher to the answer? If you are to multiply 5000, by 2, can you begin by multiplying the 5 first? If you are to multiply 35000, by 2; can you multiply the 5 first, and then the 3, and afterwards affix the three ciphers? If you are to multiply 20 by 30, can it be done by multiplying the 3 and 2 together, and then affixing 2 ciphers to the product? Multiply 200 by 20 in the same way. In the above sum, what are the products of the several orders, and how are they written? |