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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:
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|
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 |
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Blue: Divisible by 3, but not 9
Red: Divisible by 9 and 3
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 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:
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 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:
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