Drill Table No. 2. Page 44.- In this table, Exercises 178-189, the pupils having lettered themselves a, b, c, etc., the first takes 23+24+26 from 100; the second takes 12 + 14 + 13 from 100, and so on.

In Exercises 238-243 the pupils letter themselves a, c,, 7, i, k. The first pupil takes 5542 from 5848; the second takes 1735 from 5641, and so on.

Drill Tables Nos. and 4. Pages 71, 89. - The teacher who has used the preceding tables will need no explanation of the use of these. There are 12 examples in each exercise, and the pupils will letter themselves from a to 1.

Drill Table No. 5. Page 146. -In each of these exercises the pupils

will number themselves from 1 to 15.

Drill Table No. 6. Page 159. - Here the pupils will number themselves 1 to 14. In Exercise 242-255 the first pupil reads the numbers written in line 1, columns E and G, the second in line 2 and so on. Drill Table No. 7. Page 181.-Needs no explanation.

Contractions in Multiplication. (Page 57.)

2. Some contractions in multiplication and division are given in Arts. 91, 141, 172, 237, and 238. Those in multiplication which follow may sometimes be useful.

Rules.

To multiply by any number whose terms are all 9's: Annex as many zeros to the expression for the multiplicand as there are 9's in that of the multiplier, and from the number thus expressed subtract the multiplicand; thus,

27 × 99=2700-27 = 2673.

1. 28 × 99 = ?

2. 92 × 999 = ?

3. 4868 × 99999 = ?
4. 247 × 998 = ?

3. To multiply by a composite number: Separate the multiplier into convenient factors, multiply the multiplicand by one of the factors, and that product by another factor, and so on, till all the factors have been used; the last product is the answer: thus, 41 × 25 = 41 × 5 × 5.

5. Multiply 374 by 45; by 108.

6. Multiply 369 by 144.

7. Multiply 453 by 132...

4. To multiply when the number of tens is the same in the multiplicand and multiplier, and the sum of the units 27 is ten: Multiply the number of tens by the number of tens plus one; write the product as hundreds; at the right express the product of the units by the units.

23

621

5. To square a number consisting of an integer and 4: Multiply the integer by the integer plus one, and to the 7 product add 4.

7

561

14. 81 × 81=? 15. 111 × 111 = ? 16. 16 × 16 = ?

The Divisibility of Numbers. (Page 106.)

6. 1. Divisibility of numbers by 2, 4, 5, or 8.

Ten is divisible by 2, so any number of tens is divisible by 2. Hence a number is divisible by 2 if the number of its units is divisible by 2.

For a similar reason, a number is divisible by 5 if the number of its units is divisible by 5.

One hundred is divisible by 4, so any number of hundreds is divisible by 4. Hence a number is divisible by 4 if its tens and units together are divisible by 4.

One thousand is divisible by 8, so any number of thousands is divisible by 8. Hence a number is divisible by 8 if its hundreds, tens, and units together are divisible by 8.

The number 3483 is now separated into two parts, one of which is divisible by 9, and the other equals the sum of its digits. Every number can be so separated. Hence a number is divisible by 9 if the sum of its digits is divisible by 9.

3. Divisibility of numbers by 3. Any number that is divisible by 9 is divisible by 3. Therefore, when a number has been separated into two parts, as shown above, the first part is divisible by 3. The other part is equal to the sum of its digits. Hence a number is divisible by 3 if the sum of its digits is divisible by 3.

86427 = (7272+546+36+2) × 11+ (8+4+7)−6−2

(8+4+7)-(6+2)=19-8=11

The number 86,727 is now separated into two parts, one of which is divisible by 11, and the other equals the sum of one set of alternate digits, less the sum of the other set. If the sums are equal, or if their difference is divisible by 11, the entire number must be divis ble by 11.

Greatest Common Factor of Two or more

Numbers. (Page 113.)

7. Illustrative Example. Find the g. c. f. of 52 and 91.

WRITTEN WORK.

52)91(1

52

39)52(1 39

13)39(3

39

Divide the greater number by the less, and then divide the less number by the remainder, if there is any. Continue dividing the last divisor by the last remainder until nothing remains. The last divisor will be the g. c. f. sought.

To find the g. c. f. of more than two numbers, find the g. c. f. of any two of them and then of that common factor and a third number, and so on till all the numbers are taken.

This method of finding the greatest common factor of numbers depends on the following principles:

I. Any factor common to two numbers is also a factor of their sum and of their difference.

Thus 3, which is a common factor of 24 and 18, is a factor of 42(24+18) and of 6(= 24 18). Since 24 is equal to a certain number of 3's, and 18 to a certain other number of 3's, their sum must be a number of 3's, and their difference must be a number of 3's.

II. The greatest common factor of two numbers is equal to the greatest common factor of the smaller of them, and the remainder obtained by dividing one of them by the other.

Thus, the g. c. f. of 143 and 52 is equal to the g. c. f.

52)143(2 of 52 and 39.

104

39

Since 39 = 1435252, all factors common to 143 and 52 are also factors of 39, and hence common factors of 52 and 39. Therefore the g. c. f. of 143 and 52 is a common factor of 52 and 39, and cannot be greater than the g. c. f. of 52 and 39.

=

Again, since 143 52 +52 + 39, all factors common to 52 and 39 are also factors of 143, and hence common factors of 143 and 52. Therefore the g. c. f. of 52 and 39 is a common factor of 143 and 52, and cannot be greater than the g. c. f. of 143 and 52.

The g. c. f. of 143 and 52 and the g. c. f. of 52 and 39 are, then, two numbers, neither of which can be greater than the other; they are therefore equal.

Least Common Multiple of Two or more
Numbers. (Page 118.)

8. Illustrative Example I. multiple of 6, 9, and 15?

WRITTEN WORK.

1. c. m. =

6 = 2 × 3

9=3x3

15 = 3 x 5

2 × 3 × 3 × 5 = 90

What is the least common

Explanation. The least multiple of 6 is 6, which may be expressed in the form 2 x 3.

The least multiple of 9 is 9, which may be expressed in the form 3 x 3. But in 6 we have already one of the factors (3) of

9; hence if we put with the prime factors of 6 the remaining factor (3) of 9, we shall have 2 × 3 × 3, which are all the factors necessary to produce the 1. c. m. of 6 and 9.

The least multiple of 15 is 15, which may be expressed in the form 3 x 5. In the 1. c. m. of 6 and 9 we have one of the prime factors (3) of 15; hence if we put with the prime factors of 6 and 9 the remaining factor (5) of 15, we shall have 2 × 3 × 3 × 5, which are all the prime factors necessary to produce the 1. c. m. of 6, 9, and 15.

The product of these factors is 90, which is the 1. c. m. sought.

NOTE. In finding the least common multiple, the factors of the given numbers seldom need to be expressed, and the written work may be greatly reduced. Thus, in this example the written work may be simply 1. c. m. 2 x 3 x 3 x 5 = 90.

=

Rule.

To find the least common multiple of two or more numbers: Take the prime factors of one of the numbers; with these take such prime factors of each of the other numbers in succession as are not contained in any preceding number, and find the product of all these prime factors.

Addition of Mixed Numbers. (Page 119.)

9. Example 91 (page 119). 251 +673 +14,7 = ?

Division of Fractions. (Page 132.)

10. Illustrative Example. What is the quotient of +?

EXPLANATION 2.

The quotient of

divided by 1 is ; of 4 x 3 by (which is as large as 1) is 3 times or ; and of

divided

divided

5

by is of 3 times or

4 x 3
5 x 2'

which equals 13.

Ans. 13.