Math Core

Contest problems love hidden digits: "The number $7A38B$ is divisible by $72$ . Find $A$ ." You can't try all hundred digit pairs in the time you have. Divisibility rules let you test a number without dividing, and they turn missing-digit puzzles into two short equations.

The rules you should know cold

Divisor	Test
$2$	Last digit is even.
$3$	Digit sum is divisible by $3$ .
$4$	Last two digits form a multiple of $4$ .
$5$	Last digit is $0$ or $5$ .
$6$	Divisible by both $2$ and $3$ .
$8$	Last three digits form a multiple of $8$ .
$9$	Digit sum is divisible by $9$ .
$10$	Last digit is $0$ .
$11$	Alternating sum of digits (add, subtract, add, …) is divisible by $11$ .

Why they work

The rules come from how powers of $10$ behave.

Last-digit rules (2, 4, 5, 8, 10). $10$ is divisible by $2$ and $5$ , $100$ is divisible by $4$ , and $1000$ is divisible by $8$ . So in $75{,}384 = 75{,}000 + 384$ , the $75{,}000$ part is automatically a multiple of $8$ . Only the last three digits, $384$ , can spoil divisibility by $8$ .
Digit-sum rules (3, 9). Every power of $10$ is one more than a multiple of $9$ : $10 = 9 + 1$ , $100 = 99 + 1$ , and so on. So $4 \cdot 100 + 5 \cdot 10 + 6$ is $(4 + 5 + 6)$ plus a multiple of $9$ . The number and its digit sum leave the same remainder when divided by $9$ (and by $3$ ).
Alternating rule (11). $10 = 11 - 1$ , $100 = 99 + 1$ , $1000 = 1001 - 1$ , … The powers of $10$ are alternately one less and one more than a multiple of $11$ . That is why the digits get alternating signs.

Combine rules for composite divisors

To test divisibility by a composite number, split it into factors that share no common factor and test each one: $12 = 3 \cdot 4$ , $\;18 = 2 \cdot 9$ , $\;36 = 4 \cdot 9$ , $\;45 = 5 \cdot 9$ , $\;72 = 8 \cdot 9$ , $\;88 = 8 \cdot 11$ .

Common mistake

The factors must share no common factor. Being divisible by $4$ and by $6$ does not mean divisible by $24$ . For example, $12$ is divisible by both, but not by $24$ . Use $24 = 3 \cdot 8$ instead.

Worked examples

Worked example: The 11 test

Is $918{,}082$ divisible by $11$ ?

Alternating sum from the left: $9 - 1 + 8 - 0 + 8 - 2 = 22$ . Since $22$ is a multiple of $11$ , so is $918{,}082$ . (Indeed, $918{,}082 = 11 \times 83{,}462$ .)

Worked example: Two unknown digits, divisor 45

The four-digit number $5A6B$ is divisible by $45$ . How many such numbers are there?

$45 = 5 \cdot 9$ . For divisibility by $5$ , $B = 0$ or $B = 5$ .

If $B = 0$ : digit sum $5 + A + 6 + 0 = 11 + A$ must be a multiple of $9$ , so $A = 7$ . The number is $5760$ .
If $B = 5$ : digit sum $16 + A$ must be a multiple of $9$ , so $A = 2$ . The number is $5265$ .

There are 2 such numbers.

Worked example: Divisor 72: do the last digits first

The five-digit number $7A38B$ is divisible by $72$ . Find $A + B$ .

$72 = 8 \cdot 9$ . Start with $8$ , because it involves only the last three digits, $38B$ . Checking $380$ through $389$ , only $384 = 8 \cdot 48$ works, so $B = 4$ .

Now the digit sum: $7 + A + 3 + 8 + 4 = 22 + A$ . The next multiple of $9$ is $27$ , so $A = 5$ . (The one after, $36$ , would need $A = 14$ .)

So $A + B = 9$ . Check: $75{,}384 = 72 \times 1047$ .

Tip

When a problem has two unknown digits, handle the rule that pins down one digit first (the last-digit rules). Then the digit-sum or alternating-sum rule finishes the job.

Practice

Practice 1

Which of the following numbers is divisible by $9$ ?

Practice 2

The four-digit number $3A52$ is divisible by $3$ . How many different digits could $A$ be?