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3. A floor whose length is 48 ft. 6′ has an area of 1176 ft. 1′ 6"; what is its width? Ans. 24 ft. 3'. 4. From a cellar 38 ft. 10' long and 9 ft. 4' deep, were excavated 275 cu. yd. 5 cu. ft. 1′ 4′′ of earth; how wide was the cellar? Ans. 20 ft. 6'.

CONTRACTED METHOD.

393. Division of Duodecimals may be abbreviated after the manner of contracted division of decimals.

1. Divide 35 ft. 11' 11" by 4 ft. 3′ 7" 3"", and find a quotient correct to seconds.

OPERATION.

4 ft. 3′ 7′′ 3′′ ) 35 ft. 11′ 11′′ ( 8 ft. 4′ 5′′
34 ft. 4' 10"

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ANALYSIS. Having obtained by trial, 8 ft. for the first term of the quotient, we multiply three terms of the divisor, 4 ft. 3′ 7′′, carrying from the rejected term, 3′′ × 8: = 24′′′ = 2′′, making 34 ft. 4' 10", which subtracted from the dividend leaves 1 ft. 7/1 for a new dividend. In the next division, we reject 2 terms from the right of the divisor, and at the last division, 3 terms, and obtain for the required quotient, 8 ft. 4′ 5′′.

EXAMPLES FOR PRACTICE.

1. Divide 7 ft. 7' 3" by 2 ft. 10' 7", extending the quotient to seconds.

Ans. 2 ft. 7' 8"±.

2. Separate 64 ft. 9' 8" into three factors, the first and second of which shall be 7 ft. 2' 4" and 4 ft. 7′ 9′′ 8"" respectively, and obtain the third factor correct to within 1 second.

Ans. 1 ft. 11' 3".

3. What is the width of a room whose area is 36 ft. 4' 8" and

whose length 7 ft. 2' 11"?

SHORT METHODS.

394. Under the heads of Contractions in Multiplication and Contractions in Division, are presented only such short methods as are of the most extensive application. The short methods which follow, although limited in their application, are of much value in computations.

FOR SUBTRACTION.

395. When the minuend consists of one or more digits of any order higher than the highest order in the subtrahend.

The difference between any number and a unit of the next higher order is called an Arithmetical Complement. Thus, 4 is the arithmetical complement of 6, 31 of 69, 2792 of 7208, etc. 1. Subtract 29876 from 400000.

OPERATION.

400000 29876 370124

ANALYSIS. To subtract 29876 from 400000 is the same as to subtract a number one less than 29876, or 29875, from 399999 (Ax. 2). We therefore diminish the 4 of the minuend by 1, and then take each figure of the subtrahend from 9, except the last or righthand digit, which we subtract from 10. Hence the

RULE. I. Subtract 1 from the digit of the minuend and write the remainder, if any, as the first figure in the result.

II. Commencing at the left, subtract each figure in the subtrahend from 9, except the last significant figure, which subtract from 10.

EXAMPLES FOR PRACTICE.

1. Subtract 756 from 1000.
2. Subtract 8576 from 4000000.
3. Subtract .5768 from 10.

4. Subtract 13057 from 1700000.

5. Subtract 90.59876 from 64000.

6. Subtract 599948 from 1000000.

Ans. 244.

Ans. 3991424.

7. What is the arithmetical complement of 271? Of 18365? Of 3401250 ?

FOR MULTIPLICATION.

CASE I.

396. When the multiplier is 9, 99, or any number of 9's.

Annexing 1 cipher to a number multiplies it by 10, two ciphers by 100, three ciphers by 1000, etc. Since 9 is 10 1, any number may be multiplied by 9 by annexing 1 cipher to it and subtracting the number from the result. For similar reasons, 100 times a number — 1 time the number = 99 times the number, etc. Hence,

RULE. Annex to the number as many ciphers as the number contains 9's, and subtract the number from the result.

EXAMPLES FOR PRACTICE.

1. Multiply 784 by 99.
2. Multiply 5873 by .999.
3. Multiply 4783 by 99999.

4. Multiply 75 by 999.999.

CASE II.

Ans. 77616.

Ans. 478295217.

397. When the multiplier is a number a few units less than the next higher unit.

Were we required to multiply by 97, which is 100-3, we could evidently annex 2 ciphers to the multiplicand, and subtract 3 times the multiplicand from the result. Were our multiplier 991, which is 1000-9, we could subtract 9 times the multiplicand from 1000 times the multiplicand. Hence,

RULE.

ciphers.

I. Multiply by the next higher unit by annexing

II. From this result subtract as many times the multiplicand as there are units in the difference between the multiplier and the next higher unit.

EXAMPLES FOR PRACTICE.

1. Multiply 786 by 98.
2. Multiply 4327 by 96.
3. Multiply 7328 by 997.

Ans. 77028.

Ans. 415392.

Ans. 78342.18070.

4. Multiply 7873.586 by 9.95. 5. Multiply 43789 by 9994.

6. Multiply 7077364 by .999993.

CASE III.

398. When the left hand figure of the multiplier is the unit, 1, the right hand figure is any digit whatever, and the intervening figures, if any, are ciphers.

1. Multiply 3684 by 17.

OPERATION.

3684 × 17 62628

ANALYSIS. If we multiply by the usual method, we obtain, separately, 7 times and 10 times the multiplicand, and add them. We may therefore multiply by the 7 units, and to the product add the multiplicand regarded as tens, thus: 7 times 4 is 28, and we write the 8 as the unit figure of the product. Then, 7 times 8 is 56, and the 2 reserved being added is 58, and the 4 in the multiplicand, added, is 62, and we write 2 in the product. Next, 7 times 6, plus the 6 reserved, plus the 8 in the multiplicand, is 56, and we write 6 in the product. Next, 7 times 3, plus the 5 reserved, plus the 36 in the multiplicand, is 62, which we write in the product, and the work is done.

Had the multiplier been 107, we should have multiplied two figures of the multiplicand by 7, before we commenced adding the digits of the multiplicand to the partial products; 3 figures had the multiplier been 1007, etc. Hence the

RULE. I. Write the multiplier at the right of the multiplicand, with the sign of multiplication between them.

II. Multiply the multiplicand by the unit figure of the multiplier, and to the product add the multiplicand, regarding its local value as a product by the left hand figure of the multiplier.

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CASE IV.

399. When the left hand figure of the multiplier is any digit, the right hand figure is the unit, 1, and the intermediate figures, if any, are ciphers.

1. Multiply 834267 by 301.

OPERATION.

834267 × 301 251114367

ANALYSIS. Regarding the multiplicand as a product by the unit, 1, of the multiplier, we multiply the multiplicand by 3 hundreds, and add the digits of the multiplicand to the several products as we proceed. Since the 3 is hundreds, the two right hand figures of the multiplicand will be the two right hand figures of the product; and the product of 3×7 will be increased by 2, the hundreds of the multiplicand.

Had the multiplier been 31, the tens of the multiplicand would have been added to 3 × 7; had the multiplier been 3001 the thousands of the multiplicand would have been added to 3 × 7; and so on. Hence the

RULE. I. Write the multiplier at the right of the multiplicand, with the sign of multiplication between them.

II. Multiply the multiplicand by the left hand figure of the mul tiplier, and to the product add the multiplicand, regarding its local value as a product by the unit figure of the multiplier.

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400. When the digits of the multiplier are all the

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