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In This Part: 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 below.
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  


Problem B1
a. Which of the above numbers are divisible by 2?
Look across the rows of numbers. Do you see any patterns?
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?
Which digits of the number must you examine to test for divisibility by 2, 5, or 10? Why?
Problem B2
a. Which of the above numbers are divisible by 9?
Look across the rows or columns of numbers. Do you see any patterns?
b. Which of the above numbers are divisible by 3?
c. What is the test for divisibility by 9 or 3?
In This Part: 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:
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 multidigit number can be written as a sum by replacing the units digit with a 0 and adding the original units digit. For example, the fivedigit 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.
In This Part: 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:
Notice the sum of the digits of each number in red. 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 fourdigit 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:
The sum 1,000 + 100 + 10 + 6 could be expressed as (999 + 1) + (99 + 1) + (9 + 1) + 6, which can be reexpressed 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 parenthesis, 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 parenthesis. Note that this is the same as the sum of the digits of the original number! The sum is 17, which is not divisible by 9. Thus, 3,455 is not divisible by 9.
Because 3 is a factor of 9, the divisibility test for 3 is related to the test for 9. If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
The number 3,455 is not divisible by 3, since 3 + 4 + 5 + 5 = 17, which is not divisible by 3. Why does this work?
Applying the same analysis we’ve been using, let’s determine why 3,455 is not divisible by 3:
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 parenthesis, we get:
Problem B3
Since you know the tests for 2 and 3, can you devise a divisibility test for 6? Use the following chart to check your rule:
Row  Numbers  


Problem B4
What do you notice about the factors you used to check for divisibility by 6? What is the divisibility test for 15?
Video Segment
In this video segment, Vicky and Nancy explore different properties of numbers that are multiples of 6 in order to figure out the rule for divisibility by 6. Watch this segment after you’ve completed Problems B3 and B4.
Did you use similar methods to come up with a rule?
You can find this segment on the session video approximately 7 minutes and 17 seconds after the Annenberg Media logo.
Video Segment
In this video segment, Donna explains her group’s thinking in finding the divisibility test for 6. With Professor Findell, she further explores the relationship between the numbers 2 and 3, which together guarantee divisibility by 6.
You can find this segment on the session video approximately 12 minutes and 1 second after the Annenberg Media logo.
In This Part: Divisibility Tests for 4 and 8
You probably used the test for 2 and the test for 3 to check divisibility by 6. We cannot, however, use the test for 2 twice to check for divisibility by 4. Using 2 and 3 works because they are relatively prime; that is, the only factor they have in common is 1.
The test for divisibility by 2 can be modified for testing divisibility by 4 and 8. Here’s how it works.
Since 4 does not divide 10, but 4 does divide 100, rewrite the number, such as the fivedigit number abcde, into two parts, abc00 + de; this is 100 times the threedigit number abc plus the twodigit number de. Since 4 divides 100, then 4 divides abc00, and all that needs to be checked is the twodigit number de. If de is divisible by 4, then the entire number is divisible by 4.
For example, to check whether the number 23,456 is divisible by 4, rewrite the number as 23,400 + 56. We know that 4 divides 23,400. Since 4 also divides 56, then 4 divides 23,456.
The test for divisibility by 8 continues this pattern. Because 8 does not evenly divide either 10 or 100, but it does divide 1,000, separate the number — for example, separate abcde into two parts, ab000 + cde. To test 23,456, write the numbers 23,000 + 456. Since 8 divides 23,000 and 8 divides 456, then 8 evenly divides 23,456.
You can try these tests with the numbers in the chart below:
Problem B5
Use the divisibility test to determine whether the following numbers are divisible by 4 and 8:
a. 32,464
b. 82,426
Video Segment
In this video segment, Rhonda and Andrea explain how they found the test for divisibility by 4. They then extend the same reasoning to discover the divisibility test for 8 and other powers of 2.
You can find this segment on the session video approximately 9 minutes and 31 seconds after the Annenberg Media logo.
In This Part: Divisibility Test for 11
Because 11 is one more than 10, the divisibility test for 11 is related to the test for 9. Remember that each power of 10 is one more than a multiple of 9. Some powers of 10 are also one more than a multiple of 11. For example, 1 is (0 • 11) + 1, and 100 is (9 • 11) + 1.
Moreover, although 10 and 1,000 are not one more than a multiple of 11, they are one less than a multiple of 11. That is, 1,000 = (91 • 11) – 1, and 10 = (1 • 11) – 1. So what powers of 10 are one more than a multiple of 11? And what powers of 10 are one less than a multiple of 11?
The base ten blocks below represent the number 1,111:
To determine if 1,111 is divisible by 11, we express 1,111 as a sum:
1,000 + 100 + 10 + 1 = (1,001 – 1) + (99 + 1) + (11 – 1) + 1
This can be rewritten as:
(1,001 + 99 + 11) + (1 + 1 – 1 + 1), or 11 • (91 + 9 + 1) + 0.
Thus, 1,111 is divisible by 11.
Knowing this leads to the divisibility rule for 11. Here’s the rule: Find the sum of the digits indicating odd powers of 10 (e.g., 10^{1}, 10^{3}, 10^{5}, etc.) and the sum of the digits indicating even powers of 10 (e.g., 10^{0}, 10^{2}, 10^{4}, etc.). If the difference between these two sums is divisible by 11, then the number is divisible by 11. In our example, we have (1 + 1 – 1 + 1) which yields 0, and 0 is divisible by 11.
There are divisibility tests for 7 as well, but the calculations involved take longer than dividing by 7! See Note 3 below.
TAKE IT FURTHER
Problem B6
Use the divisibility test to determine if 11 divides 3,456.
Problem B7
Apply divisibility tests to find the missing digits so that
a. 124,73_ is divisible by 9
b. 364,12_ is divisible by 33
Problem B8
If the tests for both 2 and 6 work, can you assume that a number is divisible by 12? Explain.
Problem B9
Devise a general rule for a divisibility test for 16.
Problem B10
Devise divisibility tests for 12, 18, and 72.
TAKE IT FURTHER
Problem B11
a. Devise a divisibility test for 3 in base four.
b. Devise a divisibility test for 2 in base five.
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 forgotten the details.
Note 3
Here’s one way to test if a number — let’s say, 55,762 — is divisible by 7:
1. Remove the last digit, multiply it by 2, and subtract that from the revised number:
2. Repeat until you’re left with a number that is either a multiple of 7 or not a multiple of 7:
Fortynine is a multiple of 7. Therefore, the number 55,762 is divisible by 7. Notice that this test looks at least as long, if not longer, than actually dividing the number by 7. Still, it is interesting to do these tests, because it tells us more about how numbers relate to one another.
Problem B1
a. All numbers ending in 2, 4, 6, 8, or 0 are multiples of 2.
b. All numbers ending in 5 or 0 are multiples of 5.
c. All numbers ending in 0 are multiples of 10.
d. The test for divisibility by any of these numbers tests whether the units digit of the given number is divisible by 2, 5, or 10. You only need to look at the units digit, since any tens digit is automatically a multiple of 2, 5, and 10. For this same reason, all digits higher than the units digit can be ignored. We’ll explore why this works later in this part of the session.
Problem B2
a. The following number are divisible by 9: 9, 18, 27, 36, 153, 2,466.
b. The following numbers are divisible by three: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 153, 156, 159, 2,463, 2,466, and 2,469.
c. One test for divisibility by 9 or 3 is to add the digits of the number being used and then test that for divisibility by 9 or 3. So, for example, 2 + 4 + 6 + 6 = 18, so 2,466 is divisible by 9. We’ll explore why later in this part.
Problem B3
Since a number is divisible by 6 only when it is divisible by both 2 AND 3, we can combine the rules for 2 and 3: A number is divisible by 6 when its units digit is divisible by 2 (i.e., it is 2, 4, 6, 8, or 0) and when the sum of its digits is divisible by 3.
Problem B4
Using 2 and 3 to check the divisibility by 6 is possible because 2 and 3 are relatively prime; that is, the only factor they have in common is 1.
Similarly, the divisibility test for 15 is constructed from the divisibility tests for 3 and 5 (they are also relatively prime): A number is divisible by 15 when its units digit is divisible by 5 (i.e., it is 5 or 0) and when the sum of its digits is divisible by 3.
Solution B5
a. To test for divisibility by 4, we need to check the number that comprise the units and tens digit of the number. So to test whether 32,464 is divisible by 4, we test 64 to see if it’s divisible by 4. It is, so 32,464 is divisible by 4.
To test for divisibility by 8, you need to check the numbers that comprise the last three digits of the number. So, to test whether 32,464 is divisible by 8, we test 464 to see if it’s divisible by 8. It is, so 32,464 is divisible by 8.
b. Likewise, to test 82,426, we see if 26 is divisible by 4; it isn’t, so 82,426 is not divisible by 4.Since 82,426 is not divisible by 4, it can’t possibly be divisible by 8 either, so we don’t need to apply the test.
Problem B6
In the number 3,456, the digits for even powers of 10 are 4 and 6. The digits for odd powers of 10 are 3 and 5. The sum for the even powers is 10, and the sum for the odd powers is 8. The difference between the sums is 2. Since 2 is not divisible by 11, 3,456 is not a multiple of 11.
Problem B7
a. For this number to be divisible by 9, the sum of its digits must be a multiple of 9. If we call the missing digit x, then 1 + 2 + 4 + 7 + 3 + x = (17 + x) must be a multiple of 9. The only possible value of x that satisfies this is 1, so the number is 124,731, a multiple of 9.
b. For this number to be divisible by 33, it must satisfy the tests for 3 and 11. Let’s look at 11 first; we need the sum of the odd and even powers of 10:
Even powers: 6 + 1 + x = 7 + x
Odd powers: 3 + 4 + 2 = 9
The difference between these must be a multiple of 11, and therefore the only possible value for x is 2. The number 364,122 is a multiple of 11.
To check whether it is a multiple of 3, add its digits: 3 + 6 + 4 + 1 + 2 + 2 = 18, a multiple of 3. Therefore 364,122 is a multiple of 33, since it is a multiple of both 3 and 11.
Solution B8
No! If a number is a multiple of 6, then we already know it is a multiple of 2. This makes the test for 2 irrelevant and unhelpful. In general, if we know that a number is divisible by both a and b, then it must be divisible by the least common multiple of a and b. For a = 6 and b = 2, the least common multiple is 6.
Solution B9
Refer to the divisibility tests for 2, 4, and 8. Since 10,000 is divisible by 16, we can test the last four digits of our given number. So, for example, to test whether the number 3,450,128 is divisible by 16, test the last four digits (0128) for divisibility by 16.
Solution B10
a. The test for divisibility by 12 encompasses the tests for 3 and 4. The sum of the digits must be a multiple of 3, and the last two digits must be a multiple of 4.
b. The test for divisibility by 18 encompasses the tests for 2 and 9. The units digit must be a multiple of 2, and the sum of the digits must be a multiple of 9.
c. The test for divisibility by 72 encompasses the tests for 8 and 9. The last three digits must be a multiple of 8, and the sum of the digits must be a multiple of 9.
Solution B11
a. The test for divisibility by 3 in base four is equivalent to the test for 9 in base ten: Add the digits of the number and check whether the sum is a multiple of 3. For example, the base four number 2313_{four} is a multiple of 3, since the sum is 2 + 3 + 1 + 3 = 21_{four}, which is a multiple of 3, since 3_{four} • 3_{four} = 21_{four}.
You could also check this by converting 21_{four} to base ten. In base ten, 21_{four}equals 9, which is also a multiple of 3:
21_{four} = (2 • 4^{1}) + (1 • 4^{0}) = 9
b. The test for divisibility by 2 in base five is equivalent to the test for 3 in base ten. The test is to add the digits of the number and then check whether the sum is a multiple of 2. For example, to check whether 42 _{five} is divisible by 2, we add the digits, 4 + 2 = 11_{five}. This number is divisible by 2, since 11_{five} = 2_{five} • 3_{five}.