Why Do the Divisibility Tests Work?

Shuffle and subtract

Here are four tries at the trick. Look at the last column.

Number	Shuffled	Bigger − smaller	Its digits added
482	248	234	2 + 3 + 4 = 9
9,051	1,590	7,461	7 + 4 + 6 + 1 = 18
63	36	27	2 + 7 = 9
52,701	10,257	42,444	4 + 2 + 4 + 4 + 4 = 18

The sum of a number’s digits is its digit sum: the digit sum of 7,315 is 7 + 3 + 1 + 5 = 16. Sabina’s trick, and the digit-sum tests for 9 and for 3, are all about digit sums.

Remainders and digit sums

Most numbers don’t divide by 9 exactly. Divide one by 9, and something is left over: the remainder, a number from 0 to 8. Here are some numbers, their remainders, and their digit sums.

Number	Divided by 9	Its digit sum	The digit sum divided by 9
352	9 × 39 + 1	10	9 × 1 + 1
7,315	9 × 812 + 7	16	9 × 1 + 7
4,096	9 × 455 + 1	19	9 × 2 + 1
12,345	9 × 1,371 + 6	15	9 × 1 + 6
80,008	9 × 8,889 + 7	16	9 × 1 + 7

So here is a guess: a number and its digit sum always leave the same remainder when divided by 9. If it is true, the test for 9 follows at once: a number divides by 9 exactly (remainder 0) when its digit sum does. But five rows of a table don’t prove anything. A pattern can hold for five numbers, or five thousand, and then fail (topic 16, Is a Pattern That Works Ten Times Always True?, has some that do). To be sure, we need to see why.

Sharing money among nine friends

You have Rs 352, as three Rs 100 notes, five Rs 10 notes and two Rs 1 coins, and nine friends to share it among, as fairly as you can.

Take a Rs 100 note and change it into small money. Rs 99 of it shares out perfectly among nine friends, Rs 11 each, and Re 1 is left over. Each of the three Rs 100 notes leaves Re 1 over like this. Now a Rs 10 note: Rs 9 of it gives each friend Re 1, and Re 1 is left over again. Each of the five Rs 10 notes leaves Re 1. And the two coins are left over as they are.

Count what is left over: Re 1 from each Rs 100 note (3), Re 1 from each Rs 10 note (5), and the coins (2): 3 + 5 + 2 = Rs 10, the digit sum of 352. Every digit counts how many notes of its place you have, and each note leaves exactly Re 1. Then you share those Rs 10 too: each friend gets Re 1 more, and Re 1 is left. So 352 divided by 9 leaves 1, the same as its digit sum, 10.

That is the whole secret. It works for every place, however far left, because every place value is one more than a number made only of nines:

• 10 = 9 + 1
• 100 = 99 + 1
• 1,000 = 999 + 1
• 1,00,000 (a lakh) = 99,999 + 1

and every number made only of nines is a multiple of 9: 9 = 9 × 1, 99 = 9 × 11, 999 = 9 × 111, 99,999 = 9 × 11,111. So each digit, whatever its place, leaves itself over, and the leftovers of the whole number add up to its digit sum.

This is a fact about place value: the value of a digit depends on its place, ones, tens, hundreds, and each place is ten times the one to its right. It is because of place value that 352 is 3 hundreds and 5 tens and 2 ones, and each kind leaves its own leftover.

The same thing with letters

Grade 7’s algebra says it for every three-digit number at once. Call the digits a, b and c: the number is 100a + 10b + c. Split each place the way the notes were changed:

$$100a + 10b + c = (99a + 9b) + (a + b + c) = 9 \times (11a + b) + (a + b + c).$$

The first part is 9 times a whole number, so it shares out among nine with nothing over. What is left, a + b + c, is the digit sum. Every number is a multiple of 9 plus its digit sum. A number with more digits works the same way, with 999d for a thousands digit d, 9,999e for a ten-thousands digit, and so on. So the guess from the table is true for every number there is.

When the digit sum is itself a big number, do it again: 7,315 has digit sum 16, and 16 has digit sum 7. Keep going until one digit is left and you have the number’s digital root. It is the remainder on dividing by 9, except that a multiple of 9 has digital root 9 instead of 0.

Back to the trick

Now Sabina’s trick explains itself.

1. Your number and its shuffle have the same digits, so the same digit sum, so they leave the same remainder when divided by 9.
2. Take one from the other, and the two equal leftovers cancel. The answer leaves 0: it is a multiple of 9.
3. A multiple of 9 has a digit sum that is a multiple of 9 too (that is the test for 9).
4. So when you read out all the digits but one, Sabina adds them up, and the circled digit is whatever takes the sum up to the next multiple of 9.

For 7,315: 7,315 − 1,537 = 5,778, and 5 + 7 + 7 + 8 = 27. Circle a 7, read out 5, 7 and 8 (which make 20), and the circled digit must be 27 − 20 = 7.

The notes gave the secret away: what matters is how much each place leaves over. For 9, every place leaves 1, so the leftovers add up to the digit sum. Ask the same question for another divisor, and you get its test. This is how the French mathematician Blaise Pascal made a divisibility test for any number, nearly four hundred years ago (the story is below).

Here is the rule, for any divisor:

1. Find what each place (1, 10, 100, 1,000…) leaves over when divided by the divisor.
2. Multiply each digit by what its place leaves over, and add.
3. The total leaves the same remainder as the number. So the number divides exactly when the total does.

It works for the same reason as the notes: each note of a place leaves the same leftover, and the leftovers add up. Mathematicians say that remainders can be added and multiplied: to find the remainder of a sum, you can add the remainders and take the remainder of that. This arithmetic of remainders is called modular arithmetic, or clock arithmetic. Topic 6, What Day of the Week Will Your 100th Birthday Be?, uses it on the calendar.

A leftover can also be written as how far a place is short of a multiple: 10 is 1 short of 11, so for 11 we write that 10 leaves −1. The widget does this whenever the shortfall is the smaller number.

2, 5 and 10: only the last digit

10 = 2 × 5, so 10 shares out evenly among two or among five, and so do 100, 1,000 and every place to the left. Every place except the ones leaves 0. So only the ones digit counts: a number divides by 2 when its last digit is even, by 5 when it ends in 0 or 5, and by 10 when it ends in 0. That is why these tests look only at the end of the number.

4 and 8: the last two or three digits

10 doesn’t divide by 4 (it leaves 2), but 100 = 4 × 25 does, and so does every place from the hundreds to the left. So only the last two digits count: 7,316 divides by 4 because 16 does. For 8, it is 1,000 = 8 × 125 that shares out evenly, so the last three digits decide. And 6 = 2 × 3: a number divides by 6 when it passes the tests for 2 and for 3 together.

11: plus, minus, plus, minus

For 11, the places take turns.

The ones leave +1, the tens −1, the hundreds +1, the thousands −1, and so on. So: starting from the ones digit, add and subtract the digits in turn. If the total is a multiple of 11 (0, 11, −11, 22…), so is the number. For 4,829: 9 − 2 + 8 − 4 = 11, so 4,829 divides by 11 (it is 11 × 439). For 2,431: 1 − 3 + 4 − 2 = 0, and 0 is a multiple of 11 (11 × 0), so 2,431 divides by 11 too.

Chop off the last digit

Some books, Nepal’s Grade 6 books among them, teach a different test for 11: take the ones digit off the end and subtract it from the number that is left; repeat until you know. For 2,431: 243 − 1 = 242, then 24 − 2 = 22, a multiple of 11. Why does that work?

Call the number left after chopping x, and the chopped digit y. The number is 10x + y: 2,431 is 10 × 243 + 1. Now look at ten times the test’s number:

$$10 \times (x - y) = 10x - 10y = (10x + y) - 11y.$$

So ten times the test’s number differs from the real number by 11y, a multiple of 11. If the test’s number is a multiple of 11, then the real number is too. And if the test’s number leaves a remainder of 1, 2, … or 10, then ten times it leaves 10 × 1, 10 × 2, … or 10 × 10 over a multiple of 11: 10, 20, 30, 40, 50, 60, 70, 80, 90 or 100, and you can check that none of those is a multiple of 11. So then the real number isn’t either. The test is right both ways.

Try 7 in the widget above. Its places leave 1, 3, 2, −1, −3, −2, and then the same again: 1, 3, 2… No pattern as simple as “all 1” or “all 0 after the first”, which is why 7 is so often said to have no easy test. You can still use the leftovers. For 7,315, from the ones digit: 5 × 1 + 1 × 3 + 3 × 2 + 7 × (−1) = 5 + 3 + 6 − 7 = 7. That is a multiple of 7, so 7,315 is too (7 × 1,045).

The Grade 6 test for 7 uses the chopping trick instead: subtract twice the last digit from the rest. For 651: 65 − 2 × 1 = 63 = 7 × 9, so 651 divides by 7. It works for the same reason as the chopping test for 11:

$$10 \times (x - 2y) = 10x - 20y = (10x + y) - 21y,$$

and 21 = 3 × 7 is a multiple of 7. (And 10 × 1, 10 × 2, … 10 × 6, that is 10, 20, 30, 40, 50 and 60, are none of them multiples of 7, so the test is right both ways again.)

In 2019 a 12-year-old pupil at Westminster Under School in London, Chika Ofili, found a test for 7 of his own: add five times the last digit to the rest. For 532: 53 + 5 × 2 = 63, a multiple of 7, so 532 is one too (7 × 76).

Three tests at once: 1,001

Here is a surprise in a number you might never look at twice:

$$1{,}001 = 7 \times 11 \times 13.$$

1,000 is one short of 1,001, so for 7, 11 and 13 the thousands place leaves −1, the millions +1, and so on. So cut the number into groups of three digits from the right, and add and subtract the groups in turn. Try it on 287,542,178, the number Pascal used as his example: 178 − 542 + 287 = −77 = −7 × 11. So 287,542,178 divides by 7 and by 11, but not by 13, since 77 isn’t a multiple of 13.

Does the digit sum work for 27?

999 = 27 × 37. So does the digit-sum test work for 27 too? Try 27 itself: its digit sum is 9, which isn’t a multiple of 27, yet 27 certainly is. And 1,899 has digit sum 1 + 8 + 9 + 9 = 27, but 1,899 = 27 × 70 + 9. The test fails both ways.

The leftovers say why. For 27, the places leave 1, 10, 19, 1, 10, 19… (10 ÷ 27 leaves 10 itself), not 1 every time. A digit sum works only when 10 leaves 1, and that happens only for 9 and the numbers that divide 9, which are 3 and 1. For 27 the pattern repeats every three places, because 1,000 = 999 + 1 and 999 is a multiple of 27: add the groups of three. For 1,899: 1 + 899 = 900 = 27 × 33 + 9, the same 9 left over as 1,899.

And 6?

Ram says: “A number divides by 6 when its digit sum does.” He is wrong both ways. 15 has digit sum 6, but 15 isn’t a multiple of 6; and 12 is a multiple of 6, but its digit sum, 3, isn’t. For 6, the places leave 1, 4, 4, 4… (10 = 6 + 4), not all 1.

In another base

The digit sum works for 9 because we count in tens and 9 is one less than 10. In Grade 8 you write numbers in base 5 (the quinary system), where the places are 1, 5, 25, 125 instead of 1, 10, 100, 1,000. Each of those is one more than a multiple of 4: 5 = 4 + 1, 25 = 24 + 1, 125 = 124 + 1. So in base 5 it is the test for 4 that adds up the digits. The base-5 number 1234 is 1 × 125 + 2 × 25 + 3 × 5 + 4 = 194, its digit sum is 1 + 2 + 3 + 4 = 10, and both leave 2 when divided by 4 (194 = 4 × 48 + 2).

• Why Does 1089 Always Come Back? Reverse, subtract, reverse, add: a mind-reading trick like Sabina’s, with the same place-value algebra behind it.
• Is There a Biggest Prime Number? Numbers that only 1 and themselves divide, and a proof from 2,300 years ago that they never run out.
• What Day of the Week Will Your 100th Birthday Be? Remainders on dividing by 7, used on the calendar.
• Is the Answer Roughly Right? Checking sums by casting out nines, and the mistakes it can’t see.
• How Can Two Strangers Agree on a Secret in Public? (Grades 9–12) The arithmetic of remainders, keeping messages safe on the internet.
• How Does a Barcode Catch a Typing Mistake? (Computing, Going Further) Check digits built from remainders, and a program that tries every mistake.