Understand Divisibility and Build a Test Toolkit
理解整除，建立整除判断工具箱
Learn what “divides” really means, then use the last digit, digit sum, last two or three digits, and alternating sums to test numbers efficiently.
Divisible means “remainder zero”
Use a nonnegative whole-number dividend and a positive whole-number divisor. A quick rule predicts whether ordinary division leaves remainder 0. The divisor can never be 0; a dividend of 0 is allowed because 0 divided by any positive divisor is 0 with remainder 0.
Division machine
Check the language
Look at the last digit for 2, 5, and 10
Tens, hundreds, and larger place values are already multiples of 2, 5, and 10. The ones digit decides what remains.
Divisible by 2
Last digit is one of:
Divisible by 5
Last digit is:
Divisible by 10
Last digit is:
Last-digit explorer
Choose a last digit. Imagine any number ending in that digit.
Test 6,735
Add the digits for 3 and 9
A number and its digit sum leave the same remainder when divided by 3 or 9. Replace a long number with one small sum.
Test for 3
A number is divisible by 3 exactly when its digit sum is divisible by 3.
Test for 9
A number is divisible by 9 exactly when its digit sum is divisible by 9.
Digit-sum scanner
Test 4,572
Inspect the last two digits for 4 and the last three for 8
Every complete hundred is divisible by 4, and every complete thousand is divisible by 8. The tail means the number formed by the last two or three digits. If the number has fewer digits, use the whole number. Leading zeros in a tail do not change its value: 008 means 8.
Test for 4
Example: 45,312 has tail 12, and 12 is divisible by 4.
Test for 8
Example: 45,312 has tail 312, and 312 = 8 × 39.
Tail scanner
Read the right tail
Alternate the digit positions for 11
Add every other digit into two groups. Subtract the smaller group sum from the larger. If this difference is 0 or a multiple of 11, the number is divisible by 11.
Split alternating positions
Add each group
Find their difference
Check 0, 11, 22, …
Alternating-sum scanner
Use the two groups
Combine tests with the least common multiple
To test a composite divisor (a whole number greater than 1 that is not prime), combine simpler tests. The least common multiple, or LCM, is the smallest positive number divisible by both component divisors. Coprime divisors share no factor greater than 1; only for these does their product equal their LCM.
| Target divisor | Quick component tests | Why |
|---|---|---|
| 6 | 2 and 3 | LCM(2,3)=6 |
| 12 | 3 and 4 | LCM(3,4)=12 |
| 15 | 3 and 5 | LCM(3,5)=15 |
| 18 | 2 and 9 | LCM(2,9)=18 |
| 24 | 3 and 8 | LCM(3,8)=24 |
| 33 | 3 and 11 | LCM(3,11)=33 |
| 72 | 8 and 9 | LCM(8,9)=72 |
Divisibility toolkit scanner
Combine the right conditions
Worked example: make 9a8a divisible by 15
This is a complete worked investigation from Chapter 15. The same digit a, chosen from 0 through 9, appears twice, so both occurrences must match. The notation 9a8a names a four-digit number, not a multiplication.
Problem
The four-digit number 9a8a is divisible by 15. Find the digit a.
First: test for 5
The last digit is a, so:
Then: test for 3
The digit sum is:
Test the two surviving candidates
| a | Number | Pass 5? | Digit sum | Pass 3? | Valid for 15? |
|---|
Complete the worked example reasoning
Extra practice: make A1999311B divisible by 72
A and B are digits: A is from 1 through 9 because the first digit cannot be 0; B is from 0 through 9. A1999311B names a nine-digit number, not a product. Use the last three digits to find B, then the whole digit sum to find A.
Problem
The number A1999311B is divisible by 72. Find the difference between digits A and B, taking the larger minus the smaller.
Step 1: find B
The last three digits are 11B.
| B | Last three digits | Divisible by 8? |
|---|
Step 2: find A
After B is fixed, the digit sum is A + 35.
| A | Digit sum | Divisible by 9? |
|---|
Complete the worked example practice
Divisibility toolkit workshop
Choose the smallest useful test. Correct all eight answers to complete the workshop.
Worked explanation for question 1
56 = 8 × 7, so nothing remains.
Worked explanation for question 2
6 + 3 + 7 + 2 = 18.
Worked explanation for question 3
Hundreds are multiples of 4; test the tail 24.
Worked explanation for question 4
Thousands are multiples of 8; 416 = 8 × 52.
Worked explanation for question 5
(4 + 6) − (7 + 3) = 0; zero passes the 11 test.
Worked explanation for question 6
Multiples of 8: 8,16,24. Multiples of 12: 12,24.
Worked explanation for question 7
4 and 9 share no factor greater than 1, so use 4 × 9.
Worked explanation for question 8
30 is the first shared multiple, not the product 150.
Exit ticket
Complete all ten missions and earn 5 out of 5 here to earn your certificate. All questions are objectively checked; the optional reflection is not graded.
Worked explanation for question 1
75 = 6 × 12 + 3.
Worked explanation for question 2
7 + 1 + 2 + 8 = 18; it passes both 3 and 9.
Worked explanation for question 3
(5 + 4) − (8 + 1) = 0.
Worked explanation for question 4
The common multiples begin 36,72,…
Worked explanation for question 5
128 = 8 × 16, so the whole number passes.
Optional reflection — not automatically graded
Certificate of completion
Divisibility Toolkit Builder
has completed Lesson 15.1 and demonstrated accurate use of divisibility tests and combined conditions.
About this lesson
This lesson teaches the divisibility vocabulary and basic tests that Chapter 15 assumes. The complete 9a8a divisible by 15 investigation follows Example 3, and the complete A1999311B divisible by 72 investigation follows Guided Practice 3 on. The scanners, visual evidence cards, extra examples, workshop, and exit ticket are added instructional scaffolds.