Divisibility tests

How can you tell, in one second, whether $87654321$ is a multiple of $9$? Long division would take a minute. A divisibility test takes about three seconds , sometimes one. These tricks were known to Indian and Greek mathematicians thousands of years ago and they are still useful today.

Concept

A divisibility test is a quick rule that tells you whether one number divides another, using only the digits. Here are the standard tests for the small divisors.

By $2$. Last digit is even: $0, 2, 4, 6, 8$. Example: $5876$ ends in $6$, so divisible by $2$.

By $5$. Last digit is $0$ or $5$. Example: $980$ , yes. $987$ , no.

By $10$. Last digit is $0$.

By $4$. Look at the last two digits. If they form a number divisible by $4$, so is the whole. Example: $7316$ , last two digits $16 = 4 \times 4$, so $7316$ is divisible by $4$.

Why? $7316 = 7300 + 16$, and $7300 = 100 \times 73$ is a multiple of $4$ (since $100$ is). So divisibility of $7316$ by $4$ is identical to divisibility of $16$ by $4$.

By $8$. Look at the last three digits. Same logic: $1000 = 8 \times 125$, so we only need to test the last three. Example: $52168$ , last three digits $168 = 21 \times 8$, divisible.

By $3$. Add the digits. If the sum is divisible by $3$, so is the whole. Example: $4521$ , digit sum $4+5+2+1 = 12 = 3 \times 4$. Divisible.

By $9$. Same rule but the digit sum must be divisible by $9$. Example: $4527$ , digit sum $4+5+2+7 = 18 = 9 \times 2$. Divisible.

Why does the digit-sum trick work? Because $10 \equiv 1 \pmod{9}$, $100 \equiv 1 \pmod{9}$, etc. So $4527 \equiv 4 + 5 + 2 + 7 \pmod{9}$. We will see this idea formally in the next subtopic.

By $6$. Divisible by both $2$ and $3$. Example: $4521$ is divisible by $3$ but ends in $1$, so not by $6$.

By $11$. Compute the alternating sum: take digits from right to left, alternately adding and subtracting. If the result is a multiple of $11$ (including $0$), so is the original. Example: $7129$ , alternating from right: $9 - 2 + 1 - 7 = 1$. Not divisible. Now $7128$: $8 - 2 + 1 - 7 = 0$. Yes, $7128 = 11 \times 648$.

Combining tests. A number is divisible by $12$ iff it is divisible by both $3$ and $4$. By $15$ iff by both $3$ and $5$. By $36$ iff by both $4$ and $9$. (The two divisors must be coprime for this combining trick to work.)

Worked examples

Example 1. Is $63{,}480$ divisible by $8$?

• Last three digits: $480$.
• $480 \div 8 = 60$, no remainder.
• So yes, $63{,}480$ is divisible by $8$.

Example 2. Is $87{,}654{,}321$ divisible by $9$?

• Digit sum: $8+7+6+5+4+3+2+1 = 36 = 9 \times 4$.
• Yes , divisible by $9$.

Example 3. Test $94{,}864$ for divisibility by $11$.

• Alternating sum from right: $4 - 6 + 8 - 4 + 9 = 11$.
• $11$ is a multiple of $11$. So yes.

Example 4. Find the smallest digit $d$ to put in the blank so that $2348\,d$ is divisible by $9$.

• Digit sum $= 2+3+4+8+d = 17+d$.
• For divisibility by $9$ we need $17+d \equiv 0 \pmod 9$, i.e., $d \equiv 1 \pmod 9$.
• Smallest such single digit: $d = 1$. So $23481$ is divisible by $9$.

Try it yourself

1. Test $738$ for divisibility by $3$ and by $9$.
2. Test $4536$ for divisibility by $4$ and by $8$.
3. Test $7392$ for divisibility by $11$.
4. Find the smallest digit $d$ that makes $4521d$ divisible by $9$.
5. Is $123{,}456$ divisible by $6$? Justify using two simpler tests.
6. Find the smallest $5$-digit number divisible by $11$.
7. Without dividing, find which of $13{,}572$ and $13{,}575$ is divisible by $4$.
8. Show that $1024$ is divisible by $8$ but not by $16$. (Check the last three digits and explain.)

Activity

Test the test. Write down ten random $4$-digit numbers. Apply the divisibility-by-$11$ test to each. Then check by actual division. Every one of your predictions should be right. Repeat with the $9$ test. Once you trust the tests, you have a powerful new shortcut for life.