TESTS OF DIVISIBILITY 1. Divide 12, 14, 24, 26, 36, 48, and 50 each by 2. What is the ones' figure, or digit, in each number? Divide other numbers ending in 2, 4, 6, 8, or 0 by 2. A number is divisible by 2, if the ones' figure is 2, 4, 6, 8, or 0. 2. Divide 5, 15, 20, 65, 155, 180, and 240 each by 5. What is the ones' figure, or digit, in each number? Divide other numbers ending in 5 or 0 by 5. A number is divisible by 5, if the ones' figure is 5 or 0. 3. Divide 3, 33, 39, 69, 129, 339, and 369 each by 3. Is the sum of the figures, or digits, in each of these numbers divisible by 3? Divide by 3, other numbers the sum of whose digits is divisible by 3. A number is divisible by 3, if the sum of the digits is divisible by 3. 4. Divide 9, 18, 36, 81, 189, and 918 each by 9. Is the sum of the digits in each of these numbers divisible by 9? Divide by 9 other numbers the sum of whose digits is divisible by 9. A number is divisible by 9, if the sum of the digits is divisible by 9. 5. State which of the following numbers are divisible by 2; by 5; by 3; by 9: 21. Which of the foregoing numbers are divisible both by 2 and by 5? by 2, 3, and 5? by 2, 3, and 9? by 2, 3, 5, and 9? 2 120 2 60 2 30 5 Divide 120 by the prime factor 2, which gives the quotient 60. Divide 60 and the succeeding quotients by the smallest prime factors that will divide each. The last quotient, 5, is a prime number. The prime factors of 120 are 2, 2, 2, 3, 5; that is, 120 = 2 × 2 × 2 x 3 x 5. Every composite number is equal to the product of all its prime factors. 2. Find the prime factors of 700. 100)700 7 We see at a glance that 700 = 100 × 7, or 10 x 10 x 7. 100 = 10 × 10 = 2 × 5 × 2 × 5 Since the factors of 10 are 2 and 5, the factors of 700 are 2, 5, 2, 5, and 7. CANCELLATION Cancellation is the process of shortening operations in division by striking out equal factors from dividend and divisor. 48163. We may divide the dividend 48 into the factors 8 and 6, and the divisor 16 into the factors 8 and 2. Therefore, we may write this example in the following way: (8 × 6) ÷ (8 × 2) = 3. If we strike out the equal factor 8 in both dividend and divisor, the problem reads: 6 ÷ 2 = 3. Striking out equal factors from both dividend and divisor does not change the value of the quotient. Written Work 1. Divide 10 x 25 by 5 × 10. 5 10 × 25=5 3 × 10 Strike out from both dividend and divisor the factor 10. In a similar manner strike out from both dividend and divisor the factor 5, leaving 5 in the dividend and 1 (which need not be written) in the divisor. The result is 5. GREATEST COMMON DIVISOR 1. Name a number that will exactly divide 8 and 12; 10 and 15; 12 and 16. A common divisor (c. d.) of two or more numbers is a number that exactly divides each of them; thus, 4 is a common divisor of 16 and 24. The greatest common divisor (g. c. d.) of two or more numbers is the greatest number that will exactly divide each of them; thus, 8 is the g. c. d. of 16 and 24. 2. Name the g. c. d. of 12 and 16; of 20 and 30; of 18 and 27; of 10 and 15; of 22 and 33; of 63 and 81. The greatest common divisor of two or more integral numbers is the product of all their common prime factors. Written Work 1. Find the greatest common divisor of 24, 36, and 48. Divide all the numbers by a common prime factor. In the same way divide the quotients until they are prime to each other. The divisors 2, 2, and 3 are all the common prime factors. Hence the g. c. d. of 24, 36, and 48 is 2 × 2 × 3, or 12. greatest common divisor of 165 and 210. Dividing each number by 3 and the quotients by 5, we find that the only prime factors common to both are 3 and 5. Hence their product, 15, is the greatest common divisor of 165 and 210. NOTE. The chief application of greatest common divisor is in reducing fractions to their lowest terms. See pp. 148, 149. LEAST COMMON MULTIPLE 1. Name a number that contains 3 and 4 a whole number of times; 4 and 5. A common multiple of two or more numbers is a number that can be exactly divided by each of them; thus, 24 is a common multiple of 3 and 4. 2. Name the least number that can be exactly divided by 3 and 4; by 6 and 8; by 5 and 6. The least common multiple (1. c. m.) of two or more numbers is the least number that can be exactly divided by each of them; thus, 12 is the 1. c. m. of 3 and 4. 3. Name the 1.c.m. of 4 and 5; of 6 and 8; of 6 and 9. The least common multiple of two or more numbers is the product of all their prime factors each used as often as it occurs in any one of the numbers. 2 occurs 4 times as a factor in 16. It must, therefore, be used 4 times in the l. c. m. 3 occurs once as a factor in 24. It must, The 1. c. m., therefore, of 16 and therefore, be used once in the 1. c. m. 24 is 2 × 2 × 2 × 2 × 3, or 48. 2. Find the 1. c. m. of 6, 12, 16, and 24. 16 24 2)6 12 16 2) 8 12 2)4 6 2 3 Since 6 and 12 are exact divisors of 24, a multiple of 24 is also a multiple of 6 and 12, hence 6 and 12 may be rejected. Divide the remaining numbers by any prime factor that will divide two or more of them. In the same way divide the quotients until no two of them have a common divisor. 2 occurs 4 times in 16, hence it must be used 4 times in the 1. c. m. 3 occurs once in 24, hence it must be used once in the 1. c. m. Hence the 1. c. m. of 6, 12, 16, and 24 is 2 × 2 × 2 × 2 × 3 or 48. |