Divisibility by 3 and 9

Learn the digit sum rule: add the digits of a number to see at once whether it is divisible by 3 or by 9, find missing digits and check your sums.

Elementary20 min18 problems
Practice Problems

Definition

To test whether a number is divisible by 3 or divisible by 9, the last digit does not help. Instead, you add up all of its digits. This total is called the digit sum.

The rule for 3

A number is divisible by 33 when its digit sum is divisible by 33.
516:5+1+6=12,12÷3=4✓(516÷3=172)516: \quad 5 + 1 + 6 = 12, \quad 12 \div 3 = 4 \quad \checkmark \quad (516 \div 3 = 172)

The rule for 9

A number is divisible by 99 when its digit sum is divisible by 99.
4527:4+5+2+7=18,18÷9=2✓(4527÷9=503)4527: \quad 4 + 5 + 2 + 7 = 18, \quad 18 \div 9 = 2 \quad \checkmark \quad (4527 \div 9 = 503)

Big numbers: add again

If the digit sum is still big, add its digits again until one digit is left:
987,654→39→12→3987{,}654 \rightarrow 39 \rightarrow 12 \rightarrow 3
The last digit you reach tells you everything: 33, 66 or 99 means divisible by 33, and only 99 means divisible by 99.

9 is stricter than 3

Every number divisible by 99 is also divisible by 33 (because 9=3×39 = 3 \times 3). The reverse is not true: 516516 is divisible by 33, but its digit sum 1212 is not divisible by 99, so 516516 is not divisible by 99.

Try it now

Which number is divisible by 33?

Worked Examples

Is 516516 divisible by 33? Is it divisible by 99?
1

Add all the digits

5+1+6=125 + 1 + 6 = 12 → Digit sum: 1212
2

Test the digit sum for 3

12÷3=412 \div 3 = 4, no remainder → 516516 is divisible by 33: 516÷3=172516 \div 3 = 172
3

Test the digit sum for 9

12÷9=112 \div 9 = 1 remainder 33 → 516516 is not divisible by 99

Common Mistakes

Looking at the last digit, as for 2, 5 and 10

Why it's wrong: The last digit says nothing about 3 or 9. 2323 ends in 33 but is not divisible by 33, while 2727 ends in 77 and is divisible by 33 and by 99.

Correct: Always add all the digits: 2+7=92 + 7 = 9, so 2727 is divisible by 33 and by 99.

Thinking every number divisible by 3 is also divisible by 9

Why it's wrong: The rule for 99 is stricter. A digit sum of 1212 or 1515 is a multiple of 33 but not of 99.

Correct: 516516 (digit sum 1212) is divisible by 33 but not by 99. The other way round always works: every multiple of 99 is a multiple of 33.

Stopping after the first digit that works in a missing digit puzzle

Why it's wrong: For 33 there are usually three or four digits that work, because the multiples of 33 are only 33 apart.

Correct: For 52□452\square4 and 33, check 1212, 1515 and 1818: the box can be 11, 44 or 77.

Believing that a matching digit sum check proves the answer is right

Why it's wrong: Casting out nines only finds some mistakes. Swapping two digits does not change the digit sum, so 40324032 passes the same check as 40234023.

Correct: If the check does not match, the answer is surely wrong. If it matches, the answer is probably right, but not certainly.

Interactive Visual

Your turn: is it divisible by 3? By 9? Guess first, then check the digit sum

0 of 0 right
372

For 3: add all the digits.

Pick a divisor, guess yes or no, then see the rule.

Is 372 divisible by 3?

For 3: add all the digits.

Why does the digit sum work? Split a long number into 999..., 99..., 9 and its digit sum

123456789012

For 3: add all the digits.

Tap a divisor to see its rule on this number.

Is 123,456,789,012 divisible by 3?

Add all the digits:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 + 1 + 2 = 48

Still more than one digit, so add again:

4 + 8 = 12

Still more than one digit, so add again:

1 + 2 = 3

3 is a multiple of 3.

123,456,789,012 is divisible by 3.

Check with division: 123,456,789,012 ÷ 3 = 41,152,263,004

123,456,789,012 = 1 × 100,000,000,000 + 2 × 10,000,000,000 + 3 × 1,000,000,000 + 4 × 100,000,000 + 5 × 10,000,000 + 6 × 1,000,000 + 7 × 100,000 + 8 × 10,000 + 9 × 1000 + 1 × 10 + 2

= 1 × 99,999,999,999 + 2 × 9,999,999,999 + 3 × 999,999,999 + 4 × 99,999,999 + 5 × 9,999,999 + 6 × 999,999 + 7 × 99,999 + 8 × 9999 + 9 × 999 + 1 × 9 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 2)

10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1. The parts with 9, 99, 999 are multiples of 3, so only the digit sum is left over: it decides.

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which number is divisible by 33?

Why It Matters

The digit sum rule is one of the most useful shortcuts in arithmetic:
  • Sharing fairly: Can 378 stickers be split among 9 friends? 3+7+8=183 + 7 + 8 = 18, so yes, without dividing first.
  • Simplifying fractions: In 5187\frac{51}{87} both digit sums (66 and 1515) are divisible by 33, so you can divide both by 33: 1729\frac{17}{29}.
  • Checking your work: The digit sum keeps the remainder after dividing by 99, so you can check a long addition in a few seconds (this is called casting out nines).
  • Number puzzles: Missing digit puzzles and codes often hide a number divisible by 33 or 99.
After the rules for 2, 5 and 10 (which look at the last digit), this is the first rule that looks at every digit, and it prepares you for the rules for 6 and 12.

Real World Applications

Making Equal Teams

Coaches and teachers often need to split a group into equal teams. The digit sum tells them at once whether it works.

Example:

Can 135135 runners start in 99 equal groups? 1+3+5=91 + 3 + 5 = 9, so yes: 135÷9=15135 \div 9 = 15 runners per group.

1Try It Yourself

A school sports day has 378378 students. The teachers want 99 teams of the same size.

How many students will be in each team?

Step 1: Write the mathematical expression

Divide the students into 9 teams:

Checking a Bill

Before calculators, shopkeepers and bookkeepers checked long additions with digit sums. It is still a quick way to catch a slip in your own calculations.

Example:

1236+2458=36941236 + 2458 = 3694: the remainders are 33 and 11, which add to 44, and 36943694 also gives 3+6+9+4=22→43 + 6 + 9 + 4 = 22 \rightarrow 4. The check passes.

2Try It Yourself

A cashier adds two delivery notes: 1236+2458=36941236 + 2458 = 3694. You want to check the total with remainders after dividing by 99.

What remainder does the total 36943694 leave when divided by 99?

Step 1: Write the mathematical expression

Find the remainder of 3694 divided by 9:

Cracking a Code

Puzzle books and escape rooms like codes with a missing digit. A divisibility clue often gives the answer in one step.

Example:

A code 7□27\square2 is divisible by 99: 7+2=97 + 2 = 9, so the box must be 00 or 99 (702702 or 792792).

3Try It Yourself

The code of a bike lock is 4□714\square71. A note says: the code is divisible by 99.

Which digit is missing?

Step 1: Write the mathematical expression

Find the digit that brings the digit sum to the next multiple of 9:

Key Takeaways

  • 1The digit sum is the total of all digits of a number: for 45274527 it is 4+5+2+7=184 + 5 + 2 + 7 = 18
  • 2A number is divisible by 33 when its digit sum is divisible by 33, and by 99 when its digit sum is divisible by 99
  • 3For big numbers, add the digits again and again until one digit is left: 33, 66, 99 mean divisible by 33, only 99 means divisible by 99
  • 4Every multiple of 99 is a multiple of 33, but not every multiple of 33 is a multiple of 99
  • 5It works because 10=9+110 = 9 + 1, 100=99+1100 = 99 + 1, 1000=999+11000 = 999 + 1: the nines drop out and only the digit sum is left
  • 6A number and its digit sum leave the same remainder when divided by 33 or 99, which lets you find missing digits and check sums (casting out nines)

Frequently Asked Questions

No. 45274527, 72547254 and 27452745 all have digit sum 1818, so all three are divisible by 99. Moving digits around never changes the digit sum.
No. 45274527, 72547254 and 27452745 all have digit sum 1818, so all three are divisible by 99. Moving digits around never changes the digit sum.
A 00 adds nothing to the digit sum, so you can skip it. 30513051 and 351351 have the same digit sum 99, and both are divisible by 99.
It works for 33 and 99 because 1010 is exactly 11 more than a multiple of 99. For 77 that is not true (10=7+310 = 7 + 3), so the place values do not drop out. Seven needs a different test.
Reduce the number to one digit. For 99: that digit is the remainder, except that 99 means remainder 00. For example 7654→22→47654 \rightarrow 22 \rightarrow 4, and indeed 7654=9×850+47654 = 9 \times 850 + 4.

Glossary

Digit sum
The total you get by adding all the digits of a number, e.g. 372→3+7+2=12372 \rightarrow 3 + 7 + 2 = 12
Repeated digit sum
Adding the digits again and again until only one digit is left, e.g. 987,654→39→12→3987{,}654 \rightarrow 39 \rightarrow 12 \rightarrow 3
Divisible by
A number is divisible by 33 if dividing it by 33 leaves no remainder, e.g. 372÷3=124372 \div 3 = 124
Divisibility rule
A shortcut that tells whether a number is divisible by another without doing the division
Remainder
What is left over after dividing, e.g. 23÷9=223 \div 9 = 2 remainder 55
Casting out nines
Checking a calculation by replacing every number with its remainder after dividing by 99 (its repeated digit sum)

Formula Card

Rule for 3

3∣n  ⟺  3∣digit sum of n3 \mid n \iff 3 \mid \text{digit sum of } n

A number is divisible by 3 when its digit sum is divisible by 3

Rule for 9

9∣n  ⟺  9∣digit sum of n9 \mid n \iff 9 \mid \text{digit sum of } n

A number is divisible by 9 when its digit sum is divisible by 9

Why it works

4527=4×999+5×99+2×9+(4+5+2+7)4527 = 4 \times 999 + 5 \times 99 + 2 \times 9 + (4 + 5 + 2 + 7)

The parts with 999, 99 and 9 are always multiples of 9, so only the digit sum decides

Same remainder

7654→22→4,7654=9×850+47654 \rightarrow 22 \rightarrow 4, \quad 7654 = 9 \times 850 + 4

A number leaves the same remainder as its digit sum when divided by 9 (or 3)

Today's challengeOne problem, three tries

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