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Session 5 Divisibility Tests and Factors Session 5 Divisibility Tests and Factors Key Terms in This Session Previously Introduced • base New in This Session • divisibility test • factor • factor tree • figurate number • prime number • relatively prime numbers Introduction This session introduces some topics related to number theory. Number theory allows us to consider why mathe- matics works the way it does. You will work with “Alpha math”problems to explore relationships among numbers, which is an important part of thinking about mathematics. You’ll look at divisibility tests and why they work, and then move on to examine factors. Learning Objectives In this session, you will do the following: • Use number theory to build your problem-solving skills • Use “Alpha math”problems to deepen your understanding of relationships among numbers and build your problem-solving skills • Find and use divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10, and 11 and understand why these tests work • Begin to explore and understand factors Number and Operations - 95 - Session 5 Part A: Alpha Math (35 min.) “Alpha math” problems, where each letter stands for one digit of a number, can help you identify some of the things you know about the behaviors of particular base ten digits under various operations. Your task is to decode each of the following problems, figuring out what digit each letter represents. [See Note 1] Problem A1. In the following sums, one letter always represents the same digit in each problem, and no digit is represented by more than one letter. Replace the letters with digits: Problem A2. Each letter represents a different digit of a number: Decode the problem to determine the following value: Problem A3. In these problems, the asterisks represent missing digits (though they do not all represent the same digit, as do the letters in the previous problems). Identify the missing digits in the following multiplication problems: Problem A4. Each letter represents a different digit of a three-digit number. Decode the problem: [See Tip A4, page 109] Take It Further Problem A5. Jen has found a special five-digit number that she calls abcde. If you enter the number 1 and then her number on a calculator and multiply it by 3, the result is the same number with a 1 on the other end: 1abcde • 3 = abcde1 What is her number? Note 1. Doing Alpha math is like decoding a cipher—it helps the solver think about how different number/letter combinations relate to one another. For example, ab + b = cdd suggests that a must be 9, because no other digit in the tens place would give a three-digit sum. The most that could be added to 9 in the tens column is 1, because two one-digit numbers cannot add to more than 18. That means that d must be 0. Since the sum has equal ones and tens digits, b must be 5. This type of reasoning is an important step toward a deep understanding of the operations involved in the Alpha math problems. Session 5 - 96 - Number and Operations Part B: Divisibility Tests (50 min.) Developing Testing Rules Another way to examine the characteristics of base ten numbers is to look at the patterns that emerge when we try to determine whether a particular number is evenly divisible by another number. [See Note 2] Use the following numbers for Problems B1 and B2. Remember, we are only looking for patterns here. The actual divisibility rules will be explored in the following sections. Row Numbers 1 1 11 21 31 151 2461 2 2 12 22 32 152 2462 3 3 13 23 33 153 2463 4 4 14 24 34 154 2464 5 5 15 25 35 155 2465 6 6 16 26 36 156 2466 7 7 17 27 37 157 2467 8 8 18 28 38 158 2468 9 9 19 29 39 159 2469 10 10 20 30 40 160 2470 Problem B1. a. Which of the above numbers are divisible by 2? [See Tip B1(a), page 109] b. Which of the above numbers are divisible by 5? c. Which of the above numbers are divisible by 10? d. What is the test for divisibility by 2, 5, or 10? [See Tip B1(d), page 109] Problem B2. a. Which of the above numbers are divisible by 9? [See Tip B2, page 109] b. Which of the above numbers are divisible by 3? c. What is the test for divisibility by 9 or 3? Note 2. Knowing and understanding divisibility tests is an important mathematical skill. It is not enough to know how to do the tests— knowing why the tests work is even more important. When you understand why a test works, you can recreate the test even if you have for- gotten the details. Number and Operations - 97 - Session 5 Part B, cont’d. Divisibility Tests for 2, 5, and 10 As you’ve seen, it is easy to tell if a counting number is divisible by 2, 5, or 10—just look at the units digit: Row Numbers 1 1 11 21 31 151 2461 2 2 12 22 32 152 2462 3 3 13 23 33 153 2463 4 4 14 24 34 154 2464 5 5 15 25 35 155 2465 6 6 16 26 36 156 2466 7 7 17 27 37 157 2467 8 8 18 28 38 158 2468 9 9 19 29 39 159 2469 10 [10] [20] [30] [40] [160] [2470] Key: Bold: Divisible by 2, but not 5 or 10 Italic: Divisible by 5, but not 2 or 10 [Brackets]: Divisible by 2, 5, and 10 Since 2, 5, and 10 all divide 10 evenly, the divisibility tests for 2, 5 and 10 are similar in that you only have to examine the units digit. If the units digit is 0, then 10 divides the number. If the units digit is 0 or 5, then 5 divides the number. If the units digit is even (0, 2, 4, 6, or 8), then 2 divides the number. Why does this work? Any multi-digit number can be written as a sum by replacing the units digit with a 0 and adding the original units digit. For example, the five-digit number 12,345 can be written as 12,340 + 5, where 5 is the units digit of the number. The idea that abcde = abcd0 + e can be extended to any number of digits. The number abcd0 is 10 • abcd, so 2, 5, and 10 all divide the number abcd0. So if the units digit is 0, then 10 divides the number. If the units digit is not 0, then 10 does not divide the number. Similarly, you only need to check the units digit for divisibility by 2 or 5. Session 5 - 98 - Number and Operations Part B, cont’d. Divisibility Tests for 3 and 9 Knowing how divisibility tests work helps us think carefully about factors and multiples. All the tests for divisibility so far are for factors of 10. What about other numbers, like 3 or 9? Let’s start with divisibility by 9, because 9 is one less than 10. First, let’s look at the numbers that are divisible by 3 and/or by 9: Row Numbers 1 1 11 21 31 151 2461 2 2 12 22 32 152 2462 3 3 13 23 33 [153] 2463 4 4 14 24 34 154 2464 5 5 15 25 35 155 2465 6 6 16 26 [36] 156 [2466] 7 7 17 [27] 37 157 2467 8 8 [18] 28 38 158 2468 9 [9] 19 29 39 159 2469 10 10 20 30 40 160 2470 Key: Bold: Divisible by 3, but not 9 [Brackets]: Divisible by 9 and 3 Notice the sum of the digits of each number in brackets. If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. Why does this work? Consider the four-digit number 1,116, which in expanded form is (1 • 1,000) + (1 • 100) + (1 • 10) + (6 • 1). Below, the number is modeled with base ten blocks: Number and Operations - 99 - Session 5 Part B, cont’d. The sum 1,000 + 100 + 10 + 6 could be expressed as (999 + 1) + (99 + 1) + (9 + 1) + 6, which can be re-expressed as (999 + 99 + 9) + (1 + 1 + 1 + 6) or 9(111 + 11 + 1) + 9. This shows that the number 1,116 is divisible by 9, because it can be expressed as 9 more than a multiple of 9, which in itself is a multiple of 9. The distributive law also allows us to pull a 9 out of each piece of the sum 999 + 99 + 9 + 9 = 9 • (111 + 11 + 1 + 1), thus showing that the sum is divisible by 9. Let’s apply a similar analysis to the number 261: 261 = (2 • 100) + (6 • 10) + (1 • 1) = (2 • [99 + 1]) + (6 • [9 + 1]) + (1 • 1) = ([2 • 99] + [6 • 9]) + ([2 • 1] + [6 • 1] + [1 • 1]) = 9 • ([2 • 11] + 6) + 9 Thus, we’ve shown that the above sum is divisible by 9, and, as a result, that the number 261 is divisible by 9. Let’s test another number, for example, 3,455: 3,455 = (3 • 1,000) + (4 • 100) + (5 • 10) + (5 • 1) = (3 • [999 + 1]) + (4 • [99 + 1]) + (5 • [9 + 1]) + (5 • 1) = ([3 • 999] + [4 • 99] + [5 • 9]) + ([3 • 1] + [4 • 1] + [5 • 1] + [5 • 1]) Factoring out 9 from the first parentheses, we get: 3,455 = 9 • (333 + 44 + 5) + (3 + 4 + 5 + 5) Since we know that 9 • (333 + 44 + 5) is divisible by 9, we only need to examine the second parentheses.
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