Divisibility Rules

Read the common base-ten tests

Apply the tests to the digits of the non-negative magnitude of an integer. A negative sign does not change whether a positive integer divides it.

Shorter numbers can be padded with leading zeros when a test asks for two or three final digits. The tests are statements about base-ten notation; a different base changes which digit shortcuts work.

Explain the last-digit and last-block rules

Write a decimal integer as $n=10q+r$, where $r$ is the final digit. The term $10q$ is divisible by 2 and 5, so divisibility by either depends only on $r$.

For 4, use $n=100q+r$ with $0\le r<100$. Since 100 is divisible by 4, the last two digits determine the remainder. Similarly, 1000 is divisible by 8, so the final three digits determine divisibility by 8.

For example, 73 128 ends in 28 and 128. Both are divisible by 4, and 128 is divisible by 8. Thus 73 128 passes both tests. Testing only the final two digits for 8 would fail as a general method because 100 is not divisible by 8.

Derive the digit-sum tests

For a three-digit number,

The first two terms are divisible by 9 and therefore by 3. The original number and its digit sum have the same remainder modulo either divisor. The same reasoning extends to longer numerals because $10^k-1$ is divisible by 9 for every positive integer $k$.

For 73 128, the digit sum is $7+3+1+2+8=21$. The number is divisible by 3, but not by 9. Since it is also even, it is divisible by 6. Since it is divisible by 4 and 3, it is divisible by 12:

The general remainder explanation and additional tests are developed in Wolfram MathWorld’s divisibility reference.

Use alternating signs for eleven

Powers of ten alternate between remainders 1 and $-1$ modulo 11. That produces an alternating digit sum. For 2728, working from the right gives

Since 11 is divisible by 11, so is 2728; indeed, $2728=11\times248$. Starting from the left changes only the overall sign for an even-length numeral, so divisibility is unaffected.

An alternating sum of zero also passes. The test is not “the sum must equal 11”; any integer multiple of 11, positive, negative or zero, is suitable.

Derive a useful test for seven

Write the number as $10q+r$. Subtract twice its final digit from the remaining prefix: $q-2r$. The identity

shows that the two sides have the same remainder modulo 7. Since 10 and 7 are coprime, 7 divides the original number exactly when it divides $q-2r$.

For 203, calculate $20-2\times3=14$. Since 14 is divisible by 7, so is 203, with quotient 29. A negative result is acceptable; test its magnitude. Repeating the procedure may make a large number easier to handle, but direct division can be simpler once the result is small.

Combine tests without overclaiming

Divisibility by coprime numbers implies divisibility by their product. That is why the tests for 2 and 3 combine into a test for 6, and those for 3 and 4 combine into a test for 12.

When the divisors share a factor, the guaranteed combined divisor is their LCM, not necessarily their product. A number divisible by both 4 and 6 is divisible by 12, but need not be divisible by 24. The number 12 itself is a counterexample.

Recover a missing digit and recognise a weak check

Suppose the three-digit number $4d2$ is divisible by 9. Its digit sum is $6+d$. With $d\in\{0,1,\ldots,9\}$, the only multiple of 9 in the possible range is 9. Thus $d=3$ and the number is 432. Its last two digits, 32, also make it divisible by 4.

A digit check cannot certify an arbitrary calculation. The correct product $23\times17=391$ and the incorrect value 319 have the same digit sum. A remainder-nine check cannot distinguish them. Passing a necessary test is weaker than proving an answer correct.

Related topics

Remainders formalises what these tests detect. Modular Arithmetic extends the same idea from divisibility to general arithmetic with remainders.