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3. Multiply 71930082546 by each of the numbers 21, 31, 41, 51, 61, 71, 81, 91, and find the difference between the sum of the products and one hundred quadrillions, nine trillions, twelve billions, one hundred and forty thousand, three hundred and sixty.

Ans. Ninety-nine quadrillions, nine hundred and seventysix trillions, seven hundred and eighty-seven billions, three hundred and twenty-three millions, one hundred and fifty-nine thousand, seven hundred and fifty-two.

4. Multiply 670092099 by each of the numbers 9, 99, 999, and find the sum of the products.

Ans. Seven hundred and forty-one billions, seven hundred and ninety-one millions, nine hundred and fifty-three thousand, five hundred and ninety-three.

129. When the multiplier is the product of two numbers, neither of which is too great to multiply by at once, we may, if we please, first multiply by one of them, and then multiply the product by the other. 1. 24958 X 132 =24958 X 12 X 11= 3294456

24958

12 299496

11

3294456 11 times 12 times, or 132 times. In the same manner perform the following examples : 2. 392877 X 25

5. 367948 X 56 = 3. 864396 X 49

6. 9436599 X 72 = 4. 2988657 X 64

7. 84951645 X 144 = 8. Multiply 239548167 by each of the numbers 27, 28, 36, 45, 72, 84, and find the ar. comp. of the sum of the products.

Ans. Thirty billions, fifty-one millions, nine hundred and thirty-five thousand, two hundred and thirty-six.

130. By a method similar to that given Art. 127, we multiply at once by any of the numbers 112, 113, 114, 115, 116, 117, 118, 119. We have only to observe that instead of adding, besides what we carry, simply the right-hand figure, we must, in this case, add the two nearest figures on the right.

1. 63792 x 118=7527456

63792

118

7527456

Beginning with the units, we say, 8 times 2 is 16 ; six and go 1; 8 times 9 is 72, and 1 is 73, and 2 is 75; five and go 7; 8 times 7 is 56, and 7 is 63, and 9 is 72, and 2 is 74; four and go 7; 8 times 3 is 24, and 7 is 31, and 7 is 38, and 9 is 47; seven and go 4; 8 times 6 is 48, and 4 is 52, and 10 (3 + 7) is 62; two and go 6 to 6 is 12, and 3 is 15; five and go 1 to 6 is 7.

By performing the operation in the usual way, the scholar will easily perceive that we have here taken the multiplicand 8 times plus 10 times plus 100 times, as was required.

Let the following multiplications be performed in the same

manner :

2. 243x111 X112 X113X114=38916212832 3. 612 X 115 116 117 118 119=13412881329120

131. Again, by a method similar to that given Art. 128, we multiply at once by any of the numbers 211, 311, 411, 511, 611, 711, 811, 911. For example, to multiply 59463 by 711, we write the numbers as usual,

59463

711

42278193 and say, 3 is three; 3 and 6 is nine; then, 7 times 3 is 21, and 6 is 27, and 4 is 31; one and go 3; 7 times 6 is 42, and 3 is 45, and 4 is 49, and 9 is 58; eight and go 5; 7 times 4 is 28, and 5 is 33, and 9 is 42, and 5 is 47; seven and go 4; 7 times 9 is 63, and 4 is 67, and 5 is 72; two and go 7; 7 times 5 is 35, and 7 is 42, which we write entire.

The scholar may prove the work by multiplying in the ordinary way, which will sufficiently illustrate the above method.

2. 513 X 211 x 311 X 411 X 511 7070057265033 3. 234 X 611 X 711 X 811 x 911 =75104489687994 132. Also by a combination of the methods given in Arts. 127 and 128, we multiply, in one line, by any of the numbers

121, 131, 141, 151, 161, 171, 181, 191. For example, to multiply 478152 by 151, after placing the numbers

478152

151

72200952

we say, 2 is two; 5 times is 2 is 10, and 5 is 15; five and go 1; 5 times 5 is 25, and 1 is 26, and 2 is 28, and 1 is 29; nine and go 2; 5 times 1 is 5, and 2 is 7, and 5 is 12, and 8 is 20; nought and go 2 ; 5 times 8 is 40, and 2 is 42, and 1 is 43, and 7 is 50; nought and go 5; 5 times 7 is 35, and 5 is 40, and 8 is 48, and 4 is 52; two and go 5; 5 times 4 is 20, and 5 is 25, and 7 is 32; two and go 3 to 4 is seven.

A close examination of Articles 127 and 128, as well as the proof in multiplying by the ordinary method, will be sufficient elucidation.

2. 315 x 121 x 131 x 141 X 151 106307346915 3. 621 X 161 X 171 X 181 x 191 = 591051778821

133. When one part of the multiplier is a multiple of another part of it, the work may often be very much contracted by, first, multiplying by the lesser part, and then multiplying the product thus obtained by the number which shows how many times this lesser part is contained in its multiple. This last is the product of the multiplicand multiplied by the multiple. 1. 66827394X13617=909988624098

66827394

13617
1136065698 17 times the multiplicand.
9088525584 8 X 17 X 100, or 13600 times.

909988624098 and having multiplied by 17, we have 1136065698 for the first partial product. Then, as the part 136, of the multiplier is a multiple of 17, seeing that it contains 17 just eight times, we multiply the first product 1136065698 by 8, and have 9088525584 for 8 times 17 times, or 136 times the multiplicand. The units figure of this last product we place under the hundreds of the first, which makes the last product signify 13600 times the multiplicand. The sum of the two products is therefore 13617 times, as was required.

2. 358974 X 1664 597332736

358974

1664 5743584 sixteen hundred times the multiplicand.

22974336 sixty-four times. 597332736

product, or 1664 times. As 16, the part of the multiplier on the left, is a measure of 64, the remaining part, we first multiply by 16, and by removing the product two places towards the left, so that its unit figure may stand in the place of hundreds, we have 1600 times the multiplicand. Then, as the part 64 is a number of single units, we multiply 5743584, which is 16 times the multiplicand, by 4, and by placing the unit-figure of this product two places farther to the right, we have simply 64 times the multiplicand. The sum of the two products is therefore 1664 times the multiplicand, as was required.

Let the value of the following expressions be found according to the above methods of contraction; also, let them, by way of proof, be multiplied from left to right and from right to left: 3. 243 X9x11x13= 7. 242121 X 7813 X 132115 4. 4214 x 99 x 486= 8. 181111X12999 X4515= 5. 1272 X 9999X1836= 9. 11214 X 99999x17515 6. 1451 X 811 X3913= 10. 6111222 X 19118 X 168143

134. A vinculum is a bar -, or parenthesis (), used to connect several numbers or quantities, so that the whole may, as one quantity, be subjected to any of the cardinal operations. Thus 8 + 4X 2 or (8 +4)2, signifies that the sum of 8 and 4 is to be multiplied by 2. Now 12, the sum of 8 and 4, multiplied by 2 is (59 and 101) 12+ 12, and this is evidently equal to (

84)+(8+4) or 8+8+4+4: that is, to twice 8 plus twice 4; wherefore (8+4)2= 16+ 8. Now if we resolve the number 12 into any other parts as 7+3+2: then (7 + 3 + 2)3, is evidently (101) equal to (7+3+2) +(7 + 3 + 2) + (7 + 3 + 2) or 7 + 7 + 7 + 3 + 3 +3+2+2 + 2; that is, to 3 times 7, plus 3 times 3, plus 3 times 2: wherefore (7+3 + 2)3 = 21+9+ 6. Hence we infer that when a number is multiplied by any other number, the product is equal to the sum of the products when all its parts are separately multiplied by that other number, what

ever may be the parts or the multiplier. Or, as the sum of several numbers is a number whereof the numbers are parts, we say, that when the sum of several numbers is multiplied by any number, the product is equal to the sum of the products when each number is separately multiplied by that number, and conversely. Wherefore 9,3 + 6,3 +5,3 + 2.3=(9+ 6+5 + 2)3. The sign into, made thus [.], though not so commonly used as the other, is, in many cases, very commodious. Also (8

4) 3 signifies that the difference between 8 and 4, or 4, is to be multiplied by 3. Now 4.3 or 4 + 4 + 4

(8 — 4) + (8 — 4)+ (8 4), or 8 + 8 + 8 4 4 – 4 or 3 times 8 minus 3 times 4: wherefore (8 — 4) 3

24 - 12. Hence we infer that when the difference between two numbers is multiplied by any number, the product is equal to the difference of the products when each number is separately multiplied by that number : consequently, if there be several positive and several negative numbers under the vinculum to be multiplied : as the sum of the positive products is one number and the sum of the negative another, it is plain that the above elucidation will apply to them in their combined form, and therefore the value of the expression will equal the several products found in multiplying each number by the multiplier and writing the products in succession, each with its appropriate sign. Thus, (9+8—5—6) 5 = 45 +40 — 25 — 30. When the vinculum is thus removed by writing the several products, the expression is said panded.

To reduce such an expression, or, in other words, to find its value, proceed thus : (9+8-5-6)5=45 +40-25-30=85-55 =

30, the value of the expression. 135. Again 8 -- 4+2, or 8-(4+2), signifies that the sum of 4 and 2 is to be taken from 8. Also 8

Also 8-4-2 or 8—(4-2), signifies that the difference between 4 and 2 is to be subtracted from 8. Now if from the expression 8 - (4 + 2), we remove the vinculum, we have 8 but this last does not mean that the sum of 4 and 2 is to be taken from 8, but that 4 only is to be subtracted from the sum of 8 and 2, and thus the number 2, instead of a number to be subtracted, becomes by its sign, a number to be added. Wherefore, that the expression 8 -- (4 + 2) may retain its value,

be ex

4+2 ;

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