Back to Lesson

Teacher Guide: Combined Divisibility Rules

Combine simple rules to test for 12, 15, 18 and 36, learn the alternating sum rule for 11, and find every divisor from 2 to 12 in seconds.

Present LessonShare to ClassroomView Student Page

Use this lesson with your class

Free, no student accounts needed.

Share with students

Students open the lesson and practise with instant feedback.

Printable worksheet

All practice problems on paper, with a separate answer key.

Worksheet Answer key

Class quiz

10 questions on Divisibility. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives
  • Combine divisibility rules for coprime factors to test for 12, 15, 18 and 36
  • Explain with a counterexample why rules for factors that are not coprime, such as 2 and 6, cannot be combined to test their product
  • Apply the alternating sum rule to test divisibility by 11
  • Find all divisors from 2 to 12 of a number quickly and solve missing digit problems under two rules
Prerequisites
  • • Divisibility rules for 2, 5 and 10 (lesson Divisibility by 2, 5, and 10)
  • • Digit sum rules for 3 and 9 (lesson Divisibility by 3 and 9)
  • • Rules for 4, 8 and 6 = 2 × 3 (lesson Divisibility by 4, 6, and 8)
  • • Factors, multiples and remainders
Discussion Starters
  • 1. 18 is divisible by 2 and by 6. Why is it not divisible by 12?
  • 2. Which pairs of numbers could you use to test for 24? Which pairs would give wrong answers?
  • 3. Why does 11 need an alternating sum while 9 only needs the digit sum?
  • 4. A number passes the tests for 4 and 9. Which other numbers must divide it?
  • 5. Where in daily life would you want to know if a number divides by 12 or 15?
Common Misconceptions

Any two factors whose product is the divisor give a valid test (12 = 2 × 6)

Remediation: Let students test 18, 30 and 42 against 2 and 6, then divide by 12. Link the failure to the shared factor 2 and to lcm(2, 6) = 6.

An alternating sum of 0 means the number is not divisible by 11

Remediation: Ask what 0 ÷ 11 is. Show 1386: 6 - 8 + 3 - 1 = 0 and 1386 = 11 × 126.

If a number is divisible by 12, it must also be divisible by 24

Remediation: Contrast 36 (divisible by 12, not 24) with 48. Divisibility goes from the product to its factors, not the other way.

Differentiation Ideas

For Struggling Students:

  • • Give a two-column table: test 1, test 2, both passed?
  • • Start with 15 = 3 × 5 and 18 = 2 × 9, where both tests are easy
  • • Use three- and four-digit numbers only

For On-Level Students:

  • • Run full factor searches from 2 to 12 on four-digit numbers
  • • Solve missing digit problems for 12, 18 and 36
  • • Find and explain counterexamples for pairs that are not coprime

For Advanced Students:

  • • Build tests for 24, 45, 72 and 99 from coprime pairs
  • • Prove the rule for 11 with remainders of powers of 10
  • • Test 12-digit numbers for 11 and 12 and check with modular arithmetic
Standards Alignment
  • 4.OA.B.4 (CCSS.MATH.CONTENT.4.OA.B.4)

    Find all factor pairs for a whole number and determine whether it is a multiple of a given one-digit number

  • 6.NS.B.4 (CCSS.MATH.CONTENT.6.NS.B.4)

    Find the greatest common factor and the least common multiple of two whole numbers

Lesson Resources
  • visualDivisibility Practice Board

    Students guess yes or no for every divisor from 2 to 12, then reveal each rule and its verdict

  • activityFind the Counterexample

    Pairs hunt for numbers that pass two tests but fail the product (2 and 6, 4 and 6, 3 and 12) and explain why

  • gameMissing Digit Race

    Teams race to find the digit that makes a number divisible by 12, 18 or 36

  • worksheetFactor Search Grid

    A grid of numbers to test against 2 to 12, including 11 with the alternating sum

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

Some divisors have no rule of their own, but you can combine two simple rules.
Combined rule: if aa and bb are coprime (their only common factor is 11), then a number divisible by both aa and bb is divisible by a×ba \times b.
DivisorTestWhy it works
12=3×412 = 3 \times 4digit sum divisible by 3 and last two digits divisible by 43 and 4 are coprime
15=3×515 = 3 \times 5digit sum divisible by 3 and last digit 0 or 53 and 5 are coprime
18=2×918 = 2 \times 9last digit even and digit sum divisible by 92 and 9 are coprime
36=4×936 = 4 \times 9last two digits divisible by 4 and digit sum divisible by 94 and 9 are coprime
Careful: 12=2×612 = 2 \times 6 is NOT a valid test, because 2 and 6 share the factor 2. The number 1818 is divisible by 2 and by 6, but 18÷12=1.518 \div 12 = 1.5. In general, passing the tests for aa and bb proves divisibility by lcm(a,b)\text{lcm}(a, b), which equals a×ba \times b only when aa and bb are coprime.
Divisibility by 11 (alternating sum): start at the ones digit and add and subtract the digits in turn: +−+−…+ - + - \dots If the result is divisible by 11 (this includes 00, 1111, −11-11, 2222), the number is divisible by 11.
82,918:8−1+9−2+8=22⇒82,918 is divisible by 1182{,}918: \quad 8 - 1 + 9 - 2 + 8 = 22 \Rightarrow 82{,}918 \text{ is divisible by } 11

Worked Examples

Is 71287128 divisible by 1212?
1

Split 12 into coprime factors

12=3×412 = 3 \times 4, and 3 and 4 have no common factor except 1 → Test 3 and 4
2

Test 3 with the digit sum

7+1+2+8=187 + 1 + 2 + 8 = 18, and 1818 is divisible by 3
3

Test 4 with the last two digits

The last two digits form 2828, and 28÷4=728 \div 4 = 7 → Divisible by 4
4

Combine and check

Both tests pass, so 12 divides it: 7128÷12=5947128 \div 12 = 594 → Divisible by 12

Common Mistakes

Testing 12 with the rules for 2 and 6

Why it's wrong: 2 and 6 share the factor 2, so passing both only proves divisibility by 6. Counterexample: 1818 (and also 3030) is divisible by 2 and 6 but not by 12.

Correct: Use coprime factors: 12=3×412 = 3 \times 4. Check the digit sum for 3 and the last two digits for 4.

Testing 36 with 3 and 12

Why it's wrong: 3 and 12 share the factor 3. 2424 is divisible by 3 and by 12, but 24÷3624 \div 36 is not a whole number.

Correct: Use 36=4×936 = 4 \times 9: last two digits divisible by 4 and digit sum divisible by 9.

Thinking one passed test is enough

Why it's wrong: A combined rule needs BOTH tests. 2525 ends in 5 but its digit sum is 7, so it is not divisible by 15.

Correct: Check every factor of the pair. If one test fails, the number is not divisible by the product.

Reading an alternating sum of 00 or −11-11 as not divisible by 11

Why it's wrong: 00, −11-11 and −22-22 are multiples of 11 too. For 91749174: 4−7+1−9=−114 - 7 + 1 - 9 = -11, and indeed 9174=11×8349174 = 11 \times 834.

Correct: The number is divisible by 11 whenever the alternating sum is 00 or any multiple of 11, positive or negative.

Why It Matters

Combined rules turn big questions into two quick checks:
  • Sharing equally: can 1350 stickers go to 15 classes with none left over? Two checks say yes.
  • Timetables: a day has 1440 minutes, so a tram every 36 minutes fits the day exactly if 1440 is divisible by 36.
  • Fractions: spotting that 12 or 18 divides the top and the bottom simplifies a fraction in one step.
  • Huge numbers: a 12-digit number can be tested for 11 or 12 in seconds, even where a calculator shows too few digits.
The idea behind every combined rule, coprime factors, is also the key to the least common multiple and to many puzzles with missing digits.

Real World Applications

Sharing Equally

Teachers, coaches and organizers often need to know whether items can be shared out with nothing left over.

Example:

A school has 1350 stickers for 15 classes. 15=3×515 = 3 \times 5: the last digit is 0 (divisible by 5) and 1+3+5+0=91 + 3 + 5 + 0 = 9 (divisible by 3). So the stickers can be shared equally.

1Try It Yourself

A school wants to give 1350 stickers to 15 classes, the same number to each class.

How many stickers does each class get?

Step 1: Write the mathematical expression

Write the division that gives the stickers per class:

Timetables

Transport planners want departures that repeat in the same pattern every day. A day has $24 \times 60 = 1440$ minutes.

Example:

A bus every 12 minutes fits a day exactly: 1+4+4+0=91 + 4 + 4 + 0 = 9 (divisible by 3) and 40÷4=1040 \div 4 = 10, so 1440 is divisible by 12, giving 120 departures.

2Try It Yourself

A tram leaves every 36 minutes, all day and all night, starting at midnight.

How many trams leave in one day of 1440 minutes?

Step 1: Write the mathematical expression

Write the division that gives the number of departures:

The 12-Digit Check

Many calculators show only 8 to 10 digits, so they cannot divide a 12-digit number exactly. Divisibility rules work on numbers of any length.

Example:

For 123,456,789,012123{,}456{,}789{,}012 the digit sum is 4848 (divisible by 3) and the last two digits form 1212 (divisible by 4), so the number is divisible by 12.

3Try It Yourself

A 12-digit code is 123,456,789,012123{,}456{,}789{,}012. You want to know if it is divisible by 12 and by 11, without a calculator.

What is the remainder when 123,456,789,012123{,}456{,}789{,}012 is divided by 12?

Step 1: Write the mathematical expression

Enter the remainder:

Key Takeaways

  • 1If aa and bb are coprime, a number divisible by both aa and bb is divisible by a×ba \times b
  • 2Test 12 with 3 and 4, 15 with 3 and 5, 18 with 2 and 9, 36 with 4 and 9
  • 3Never test a product with two factors that share a factor: 18 is divisible by 2 and 6 but not by 12
  • 4Divisible by 11: the alternating sum of the digits (+−+−…+ - + - \dots from the ones digit) is a multiple of 11, including 00 and negative multiples
  • 5For a quick factor search, use last-digit tests, digit sums and the alternating sum, then combine coprime pairs
  • 6For a missing digit under two rules, list the digits that pass each rule and keep the ones in both lists

Frequently Asked Questions

Why must the two factors be coprime?

A number divisible by aa and by bb is a common multiple of aa and bb, so it is divisible by lcm(a,b)\text{lcm}(a, b). When aa and bb are coprime, lcm(a,b)=a×b\text{lcm}(a, b) = a \times b. When they share a factor, the lcm is smaller: lcm(2,6)=6\text{lcm}(2, 6) = 6, not 12.

Why does the alternating sum work for 11?

10=11−110 = 11 - 1, so 10 leaves the remainder −1-1 when divided by 11, 100=99+1100 = 99 + 1 leaves +1+1, 1000=1001−11000 = 1001 - 1 leaves −1-1, and so on. So each digit counts with the sign +,−,+,−,…+, -, +, -, \dots from the ones digit, and the number has the same remainder as its alternating sum.

Does it matter whether I start the alternating sum from the left or the right?

For the yes or no answer it does not: the two results are equal or opposite numbers, and −22-22 is a multiple of 11 exactly when 2222 is. Starting at the ones digit with ++ also keeps the remainder of the number (add or subtract 11 until you land between 0 and 10), so it is the safer habit.

Can I test for 24 with 4 and 6?

No, 4 and 6 share the factor 2. 1212 is divisible by 4 and 6 but not by 24. Use the coprime pair 24=3×824 = 3 \times 8: digit sum divisible by 3 and last three digits divisible by 8.

Glossary

Coprime
Two numbers are coprime if their only common factor is 1. Example: 4 and 9 are coprime, 2 and 6 are not.
Combined divisibility rule
A test for a product a×ba \times b of coprime numbers: the number must pass the test for aa and the test for bb.
Alternating sum
The digits added and subtracted in turn from the ones digit: for 2816 it is 6−1+8−2=116 - 1 + 8 - 2 = 11.
Digit sum
The sum of all digits of a number. The digit sum of 1980 is 1+9+8+0=181 + 9 + 8 + 0 = 18.
Remainder
What is left over after dividing. 18÷12=118 \div 12 = 1 remainder 66.
Least common multiple (lcm)
The smallest number that is a multiple of both numbers. lcm(3,4)=12\text{lcm}(3, 4) = 12, lcm(2,6)=6\text{lcm}(2, 6) = 6.

Formula Card

Combined rule

a,ba, b coprime: a∣na \mid n and b∣n⇒a×b∣nb \mid n \Rightarrow a \times b \mid n

Works only when the two factors share no factor except 1

Divisible by 12

12=3×412 = 3 \times 4

Digit sum divisible by 3 and last two digits divisible by 4

Divisible by 15

15=3×515 = 3 \times 5

Digit sum divisible by 3 and last digit 0 or 5

Divisible by 18

18=2×918 = 2 \times 9

Last digit even and digit sum divisible by 9

Divisible by 36

36=4×936 = 4 \times 9

Last two digits divisible by 4 and digit sum divisible by 9

Divisible by 11

d0−d1+d2−d3+…d_0 - d_1 + d_2 - d_3 + \dots

Alternating sum from the ones digit is a multiple of 11 (including 0)

More in This Topic

Divisibility Rules

Elementary25 min

Divisibility by 2, 5, and 10

Elementary15 min

Divisibility by 3 and 9

Elementary20 min

Divisibility by 4, 6, and 8

Intermediate20 min