Checking Divisibility Quickly

Suppose someone asks whether 4,27,536 can be shared equally among nine groups. Long division would answer the question, but the digits themselves contain a quicker clue. A useful shortcut does more than save time: it reveals how place value connects a large number with a much smaller calculation. You will begin with familiar last-digit tests and then investigate why adding digits, or adding and subtracting alternate digits, can reveal divisibility.

Keep the grouping idea from the previous lesson in mind. If part of a number is already a complete multiple of the divisor, it cannot affect the final remainder. We only need to examine what is left after removing that part. This is the reason behind every shortcut developed here.

In the digit string dcba, the letters indicate digits in the thousands, hundreds, tens and units places. The number is 1000d + 100c + 10b + a. Here dcba is a way of writing a number; it does not mean d × c × b × a. A leading digit cannot be zero if the number is to have all four digits.

For example, 4075 = 4 × 1000 + 0 × 100 + 7 × 10 + 5. Every place except the units contributes a multiple of 10. Therefore any whole number can be written 10q + a, where a is its units digit. Since 10 is also divisible by 2 and 5, only a matters for divisibility by 10, 5 or 2.

For divisibility by 4, use N = 100q + t, where t is the number formed by the last two digits. The first part is divisible by 4 because 100 = 4 × 25. So N is divisible by 4 exactly when t is. For 8, use N = 1000q + u: 1000 = 8 × 125, so only the last three digits matter. For shorter numbers, simply test the whole number.

A Shortcut for Divisibility by 9

Start with numbers made entirely from 0s and 9s, such as 999, 909, 900, 90 and 990. Each term in their place-value expansion has a factor of 9 or is zero, so the whole number is divisible by 9. But many other multiples of 9, such as 18 and 72, do not look like this.

The units digit alone cannot decide the question: 99 is divisible by 9 but 109 is not. Instead, inspect the place values. We can write 10 = 9 + 1, 100 = 99 + 1 and 1000 = 999 + 1. Each decimal place value is one more than a multiple of 9.

In 427, the four hundreds contribute 4 × 99 plus four leftovers. The two tens contribute 2 × 9 plus two leftovers. Add the seven units: 427 = (4 × 99 + 2 × 9) + (4 + 2 + 7). The bracketed large part is divisible by 9. Thus 427 and 13 have the same remainder when divided by 9.

Thirteen leftovers can be regrouped as 9 + 4. So the final remainder is 4, not 13. A digit of 9 contributes nine units to the leftover sum, and those units form another complete group. Do not call 9 a remainder on division by 9.

The same reasoning works with any number of digits. The number and its digit sum differ by a multiple of 9, so either both are divisible by 9 or neither is. This gives both directions of the test: a divisible number has a divisible digit sum, and a divisible digit sum identifies a divisible number.

It also gives the two negative statements: if the number is not divisible by 9, its digit sum is not; if the digit sum is not divisible by 9, the number is not. These statements all follow from the same shared remainder, rather than from four separate rules.

A Shortcut for Divisibility by 3

Every multiple of 9 is a multiple of 3, because 9k = 3(3k). The converse fails: 15, 33 and 87 are multiples of 3 but not of 9. So a divisibility test for 3 must recognise more numbers.

Use the same place-value decomposition. The numbers 9, 99 and 999 are multiples of 3 as well as 9. Removing these parts leaves the digit sum. Therefore the original number and its digit sum also have the same remainder on division by 3.

You must test the digit sum against 3 when your divisor is 3. A digit sum of 12 passes the test for 3 even though it fails the test for 9. The remainder on division by 3 must be 0, 1 or 2.

A Shortcut for Divisibility by 11

For 11 the place values behave differently. One is one more than a multiple of 11: 1 = 0 + 1. Ten is one less: 10 = 11 − 1. One hundred is one more: 100 = 99 + 1. One thousand is one less: 1000 = 1001 − 1.

The pattern continues, alternating between an extra one and a missing one. Consequently the units, hundreds and ten-thousands digits contribute with a plus sign, while the tens, thousands and hundred-thousands digits contribute with a minus sign.

Start at the units digit with a plus sign and alternate as you move left. The resulting signed sum differs from the original number by a multiple of 11. If the sum is 0 or any positive or negative multiple of 11, the original number is divisible by 11.

For divisibility alone, reversing all signs makes no difference: a number is a multiple of 11 exactly when its negative is. For the actual remainder, however, the sign convention matters. Use a plus on the units digit, then add or subtract multiples of 11 until the result lies from 0 to 10.

For example, a signed sum of 22 means remainder 0, and a signed sum of −14 becomes −14 + 22 = 8. Do not stop after adding 11 if the value is still negative.

More on Divisibility Shortcuts

Choose the test that uses the least information. First inspect the final digit for 2, 5 and 10. Next inspect the last two or three digits for 4 and 8. Compute the digit sum once and reuse it for both 3 and 9. Keep the alternating sum separate for 11.

A number can pass several tests at once. For example, 128 is divisible by 2, 4 and 8: it ends in 8, its last two digits form 28, and 128 = 8 × 16. But its digit sum 11 rules out both 3 and 9. A failed test for one divisor says nothing by itself about an unrelated divisor.

To check a row rather than trust a list, take 3060. The ending 0 passes 2, 5 and 10; 60 passes 4; 060 = 60 fails 8. The digit sum 9 passes 3 and 9, and the combination of 2 and 3 passes 6. The alternating sum 0 − 6 + 0 − 3 = −9 fails 11. Each entry has its own reason.

Divisibility Shortcuts for Other Numbers

A divisibility test for a composite number can sometimes be built from tests you already know. For 6, test 2 and 3: their LCM is 6. Among 38, 225, 186 and 64, only 186 is both even and has a digit sum divisible by 3. So only 186 is divisible by 6.

Why must both tests pass? The factorisation 6 = 2 × 3 requires both prime factors. Being even supplies the factor 2; passing the test for 3 supplies the factor 3. Neither condition on its own is enough.

For 24, checking 4 and 6 is insufficient. The factorisations 4 = 2² and 6 = 2 × 3 together guarantee only 2² × 3 = 12. The same factor 2 can satisfy both tests. Instead use 8 and 3: these require 2³ and 3, giving 24.

Digital Roots

You have already shortened digit sums more than once. For 489710, the first sum is 29, then 2 + 9 = 11, then 1 + 1 = 2. Once you reach one digit, stop. The final digit is called the digital root.

For a positive integer, the digital root is one of 1 through 9. The digital root of 0 is 0. Every digit-sum step preserves the remainder on division by 9, which explains both the usefulness and the limitation of this idea.

For a positive number with digital root 1 through 8, the digital root equals its remainder on division by 9. For digital root 9, the remainder is 0. Thus 18 has digital root 9 but leaves no remainder when divided by 9. A remainder of 9 is impossible for this divisor.

The digital root also determines the remainder on division by 3. Roots 1, 4 and 7 leave remainder 1; roots 2, 5 and 8 leave remainder 2; roots 3, 6 and 9 leave remainder 0. Zero itself also leaves remainder 0.

For consecutive positive integers, the digital roots repeat 1, 2, 3, 4, 5, 6, 7, 8, 9. For example, 17 through 28 have roots 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1. When adding 1 takes the remainder past 8, the number becomes a multiple of 9, represented by digital root 9, before returning to 1.

A sequence formed by repeatedly adding a fixed amount follows the same remainder pattern. Adding 11 adds one complete group of 9 and two extra units, so each digital root advances by two positions around the repeating cycle.