or 316 times; and since in 87 it is contained 3 times, with a remainder of 12, in 7987 it must be contained 316 +3, or 319 times, with a remainder of 12. Therefore, 7987 = 3191 H. Two numbers whose tens are alike, may be multiplied together on the principles involved in the following explanation of the multiplication of 38 by 34: = 34 times 3834 times 30+34 times 8; but as 34 times 8 = 8 times 34 = 8 times 30+ 8 times 4, 34 times 38 = 34 times 308 times 30 + 8 times 4 42 times 30+8 times 4, or it equals *40 times 30 + 2 times 30 +8 times 4 1200+60 +32=1292. = Now, as the same principles will apply to the multiplication of any two numbers similar to the above, we infer that to find the product of two numbers whose tens are alike, we may add the units of one factor to the other factor, multiply the sum by the tens of either factor, and to this product add the product of the units. Applications illustrated. 1. 49 X 4453 X 40+9-X 450 X 40 +3 X 40+ 9 X 4. 2. 83 X 83 83286 X 80 + 3 x 380 x 80 + 6 x 80 + 3 x 3. 3. 67 X 6370 x 60 +7 X 3. 4. 98 X 92 100 x 90 +8 x 2. 5. 35 40 X 30 +5 x 5. *This form is usually regarded as more convenient than the one preceding. In practice, results only should be named. Thus, in multiplying 38 by 34, we may say 1200, 1260, 1292. A little practice in this way will enable the pupil to give the final result at once. He should write no figures in performing the work. 8 x 8. 6. 782 86 X 70 +8 x8 = 80 x 70+6 x 70 + I. When the units' figure is the same in both factors, we may apply the principles illustrated in the following multiplication of 43 by 83: = 43 times 83 40 times 80+ 40 times 3 + 3 times 80, or 80 times 3+3 times 3 = 40 times 80+ 120 times 3, or 3 times 1203 times 33200 + 360 +9 = 3569. As the same principles will apply to the multiplication of any numbers similar to the above, we infer that the product of any two numbers having the same units' figure, must equal the product of the tens by the tens, plus the product of the sum of the tens by the units, plus the square of the units. Show the truth of the following equations : — = = 67 times 37 60 times 307 times 90 +7 times 7. 85 times 45 80 times 405 times 120 +5 times 5. 47 times 67 =40 times 60+ 7 times 100 + 7 times 7. 86 times 26 =80 times 20+ 6 times 100+ 6 times 6. 84 times 84 =80 times 80 + 4 times 160 + 4 times 4. Apply these principles to the following examples: — J. The following illustrates a method that can often be advantageously applied, when one factor is the sum of two numbers of which the other is the difference. 1. How many are 28 times 32 ? Since 28 = times 32 =30+2, it follows that 28 30 times 30+30 = 302, and 32: 30 times 32 - 2 times 32 times 2 2 times 30 2 times 2. But it is evident that if we add one number to another, and from the sum subtract the number which we added, we shall have, as a result, the number to which the addition was made. In the above example, we have to add 2 times 30, and to subtract its equivalent 30 times 2; but, as the result will not be affected by these operations, they may be omitted. Therefore, we have 28 times 32 = = 30 times 30- 2 times 2 302-22 = 900 4 896. As these principles will apply to the multiplication of all numbers similar to the above, we infer that the product of two factors, one of which is the sum of two numbers, and the other their difference, is equal to the difference of their squares. Show the truth of the following equations: K. We give the following rules for squaring numbers ending in 25 and 75, leaving the pupil to exercise his ingenuity in finding their demonstration. To square a number ending in 25, multiply the number representing the hundreds by the same number increased by, call the result ten-thousands, and add to it 625. 680625. 2. 7252 = 625525625. 8 times 8 ten-thousands + 625 = 680000 + 1 73 times 7 ten-thousands + 625 = 525000+ 3. What is the square of 225? of 925? of 325? of 525? of 425? of 125? of 1125? of 1225? of 1425? of 625? of 1325? To square a number ending in 75, multiply the number representing the hundreds by the same number increased by 11, call the result ten-thousands, and add to it 5625. 8 times 7 ten-thousands +5625 = 595000 + 4. 7752 5625600625. 5. 4752 = 5625 = 5 times 4 ten-thousands +5625 = 220000 + 225625. 6. What is the square of 975? of 275? of 875? of 1275? of 175? of 575? of 375? of 1175? of 675? of 1075? of 1375? L. The contractions before mentioned may frequently be applied in the method illustrated in the following equations and examples: 1. 126 times a number= 125 times the number+the number. 2. 124 times a number 125 times the number the number. 3. 33 times a number the number. = 33 times the number + of 4. 87 times a number 100 times the number 12 times the number. 5. 64 times 57 = 64 times 56 + 64, or it equals 63 times 57 +57. 47, or it equals 48 times 78 times 2, or it equals 7. 78 times 84 78 times 82 76 times 84+ 2 times 84. 8. How many are 124 times 6737? 9. How many are 126 times 54797? 10. How many are 333 times 97848? 11. How many are 873 times 64734 ? 12. How many are 166 times 8352? 13. How many are 933 times 6832? 14. How many are 17625 times 987439? Suggestion. 176 = 7 × 25 + 1. = 15. How many are 37253 times 784362? Suggestion. 372 7 × 53+ 1. 16. How many are 64772 times 5943786? Suggestion. 647 = 9 × 72 — 1. 17. How many are 93 times 86? 18. How many are 24 times 38? 19. How many are 96 times 83? 20. How many are 73 times 68? 44 times 37? M. When several factors are to be multiplied together, care should be taken to arrange them in such order as to render the work as simple as possible. A habit of examining each question carefully before attempting its numerical solution is one of the most valuable a pupil can acquire. A distinguished teacher once said, "If I were required, on peril of my life, to perform a complicated problem in two minutes, I would spend the first minute in considering how to do it." The following equations and examples furnish illustrations of principles that can sometimes be applied : 1. 4 X 7 X 5 X 3 X 5 X 29 = (4 × 5 × 5) X (3 X 7) X 29 100 x 21 x 2960900. Explanation. In this example, as we have among the factors 4, 5, and 5, the product of which is 100, it is most convenient to consider those factors by themselves. We then have for the remaining factors, 7, 3, and 29, the product of which equals 21 times 29 609, which, multiplied by 100,60900. = 2. 167 16 × 5 × 3 × 12 = (163 × 3 × 2) X (12 X 8) X 7 X 5 100 x 100 x 35 350000. = = Explanation. In solving this example, we observe that 163= of 100, and that 121 of 100. By dividing 16 into the factors 8 and 2, and considering the factors 163, 3, and 2 together, and the factors 8 and 12 together, we have the product equal to 100 X 100 X 35 = 350000. NOTE. The contractions in such examples as the two immediately preceding this note, are made by so arranging the factors as to get for a product a number by which we can easily multiply; as, for instance, 100, 1000, 1001, &c. 1001=7× 11 × 13, a fact which the pupil should remember, and apply in every instance that admits of it. Persons unaccustomed to look for contractions would be surprised to see how many there are, and how frequently they can be applied to advantage. Scarcely an example presents itself, that does not admit of some contraction or other. Practice will make them familiar and easy to all. 3. 11 x 23 x 7 × 6 × 13 (11 X 7 X 13) X (6 X x = 138138. 23) = 1001 × 138 × × 4. 39 X 55 X 28 X 235 = 3 × 13 × 5 × 11 × 7 X 4 |