Checking 3 and 9 with the Digit Sum

Is $516$ divisible by $3$? Is it divisible by $9$?

Add all the digits: $5 + 1 + 6 = 12$ = Digit sum: $12$

Test the digit sum for 3: $12 \div 3 = 4$, no remainder = $516$ is divisible by $3$: $516 \div 3 = 172$

Test the digit sum for 9: $12 \div 9 = 1$ remainder $3$ = $516$ is not divisible by $9$

Answer: $516$ is divisible by $3$ ($516 \div 3 = 172$) but not by $9$, because its digit sum $12$ is a multiple of $3$ but not of $9$.

A Big Number: Repeated Digit Sum

Is $987{,}654$ divisible by $3$? Is it divisible by $9$?

Add all six digits: $9 + 8 + 7 + 6 + 5 + 4 = 39$ = First digit sum: $39$

The sum is still big, so add its digits: $3 + 9 = 12$ = Second digit sum: $12$

Add once more to get one digit: $1 + 2 = 3$ = Final digit: $3$

Read the result: $3$ is divisible by $3$ but is not $9$ = Divisible by $3$, not by $9$

Answer: $987{,}654$ is divisible by $3$ ($987{,}654 \div 3 = 329{,}218$) but not by $9$.

Why the Rule Works

Show why $4527$ is divisible by $9$ exactly when $4 + 5 + 2 + 7$ is.

Write the number with place values: $4527 = 4 \times 1000 + 5 \times 100 + 2 \times 10 + 7$ = Thousands, hundreds, tens, ones

Split each place value as nines plus 1: $1000 = 999 + 1, \; 100 = 99 + 1, \; 10 = 9 + 1$ = $4527 = 4 \times 999 + 5 \times 99 + 2 \times 9 + (4 + 5 + 2 + 7)$

Look at the first part: $4 \times 999 + 5 \times 99 + 2 \times 9 = 3996 + 495 + 18 = 4509$ = $4509 = 9 \times 501$, always a multiple of $9$

Only the digit sum is left to check: $4527 = 4509 + 18$ and $18 = 9 \times 2$ = $4527 = 9 \times 503$

Answer: The parts $999$, $99$ and $9$ are always multiples of $9$ (and of $3$), so only the digit sum decides. Since $18$ is divisible by $9$, so is $4527$.

Finding a Missing Digit

Which digits can go in the box so that $52\square4$ is divisible by $3$? Which one makes it divisible by $9$?

Add the digits you know: $5 + 2 + 4 = 11$ = Known sum: $11$

List the multiples of 3 you can reach with one digit (0 to 9): $11 + \square$ can be $12$, $15$ or $18$ = $\square = 1$, $4$ or $7$

Now look for a multiple of 9: Between $11$ and $20$ the only multiple of $9$ is $18$ = $\square = 7$

Check with a division: $5274 \div 9 = 586$ = No remainder

Answer: For $3$ the box can be $1$, $4$ or $7$ ($5214$, $5244$, $5274$). Only $7$ also makes it divisible by $9$: $5274 = 9 \times 586$.

Casting Out Nines: Checking a Sum

Mia added $2345 + 1678$ and got $4023$. Use digit sums to check her answer.

Reduce the first number to one digit: $2 + 3 + 4 + 5 = 14 \rightarrow 1 + 4 = 5$ = $2345$ leaves remainder $5$ when divided by $9$

Reduce the second number: $1 + 6 + 7 + 8 = 22 \rightarrow 2 + 2 = 4$ = $1678$ leaves remainder $4$

Add the two remainders: $5 + 4 = 9$, and $9$ counts as remainder $0$ = The sum should leave remainder $0$

Reduce Mia's answer: $4 + 0 + 2 + 3 = 9$, remainder $0$ = Both sides match

Answer: Both checks give remainder $0$, so the answer $4023$ passes the test. If she had written $4013$, the check would give $8$ and show at once that something is wrong.

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

Why: The last digit says nothing about 3 or 9. $23$ ends in $3$ but is not divisible by $3$, while $27$ ends in $7$ and is divisible by $3$ and by $9$.

Correct: Always add all the digits: $2 + 7 = 9$, so $27$ is divisible by $3$ and by $9$.

Mistake: Thinking every number divisible by 3 is also divisible by 9

Why: The rule for $9$ is stricter. A digit sum of $12$ or $15$ is a multiple of $3$ but not of $9$.

Correct: $516$ (digit sum $12$) is divisible by $3$ but not by $9$. The other way round always works: every multiple of $9$ is a multiple of $3$.

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

Why: For $3$ there are usually three or four digits that work, because the multiples of $3$ are only $3$ apart.

Correct: For $52\square4$ and $3$, check $12$, $15$ and $18$: the box can be $1$, $4$ or $7$.

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

Why: Casting out nines only finds some mistakes. Swapping two digits does not change the digit sum, so $4032$ passes the same check as $4023$.

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

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.

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

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.

$1236 + 2458 = 3694$: the remainders are $3$ and $1$, which add to $4$, and $3694$ also gives $3 + 6 + 9 + 4 = 22 \rightarrow 4$. The check passes.

Cracking a Code

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

A code $7\square2$ is divisible by $9$: $7 + 2 = 9$, so the box must be $0$ or $9$ ($702$ or $792$).

Q: Does the order of the digits matter?

A: No. $4527$, $7254$ and $2745$ all have digit sum $18$, so all three are divisible by $9$. Moving digits around never changes the digit sum.

Q: What about the digit 0?

A: A $0$ adds nothing to the digit sum, so you can skip it. $3051$ and $351$ have the same digit sum $9$, and both are divisible by $9$.

Q: Why does this trick not work for 7?

A: It works for $3$ and $9$ because $10$ is exactly $1$ more than a multiple of $9$. For $7$ that is not true ($10 = 7 + 3$), so the place values do not drop out. Seven needs a different test.

Q: How do I get the remainder from the digit sum?

A: Reduce the number to one digit. For $9$: that digit is the remainder, except that $9$ means remainder $0$. For example $7654 \rightarrow 22 \rightarrow 4$, and indeed $7654 = 9 \times 850 + 4$.