Divisibility tests

Ask a Class 8 student to check whether 4,752 is divisible by 11, and a slow long division is not the way to go. There is a faster route: look only at the digits, add or subtract a few of them, and read off the answer in seconds. These shortcuts — the divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10, and 11 — are not magic tricks to be memorised blindly. Every single one falls out of writing a number in place value, $\overline{\ldots d_2d_1d_0} = \ldots + 100d_2+10d_1+d_0$ , and noticing how each power of 10 behaves when divided by the number you are testing. Once you see why the digit-sum test works for 9, you can rebuild the test for 3 in your head, and the same trick — splitting a number into "everything except the last few digits" plus "the last few digits" — cracks the tests for 4 and 8 too.

The core insight

Every divisibility test is a shortcut that follows from writing a number in place value and tracking how each power of 10 behaves under division by 2, 3, 4, 5, 6, 8, 9, 10, or 11.

The core idea

The nine tests, stated plainly:

Divisible by 2: last digit is 0, 2, 4, 6, or 8.
Divisible by 5: last digit is 0 or 5.
Divisible by 10: last digit is 0.
Divisible by 4: the number formed by the last two digits is divisible by 4.
Divisible by 8: the number formed by the last three digits is divisible by 8.
Divisible by 3: the sum of all digits is divisible by 3.
Divisible by 9: the sum of all digits is divisible by 9.
Divisible by 6: the number is divisible by both 2 and 3 at once.
Divisible by 11: the alternating sum of digits (add, subtract, add, subtract… consistently from one end) is divisible by 11, including 0.

Why place value is the engine behind all of them. Every number can be written as a sum of digits times powers of 10:

N = d_k\cdot 10^k + d_{k-1}\cdot 10^{k-1} + \cdots + d_1\cdot 10 + d_0,

where: $d_i$ is the digit sitting in the $10^i$ place (units, tens, hundreds, …) and $k+1$ is the number of digits in $N$ . The whole art of a divisibility test is choosing a divisor for which the powers of 10 have a simple, repeating remainder — either always 0 beyond a certain point (that gives the "look only at the last few digits" tests), always a multiple of the divisor after adding 1 (that gives the "add the digits" tests for 3 and 9), or an alternating $+1,-1$ pattern (that gives the "alternate the digits" test for 11). For 4 and 8, the trick is that $100=4\times25$ and $1000=8\times125$ , so any place value from the hundreds up contributes a multiple of 4 (or from the thousands up, a multiple of 8) automatically — only the tail digits can decide the remainder. For 3 and 9, the trick is that $10=9+1$ , so every power of 10 leaves remainder 1 on division by 9 (and by 3), which is exactly what turns "divide by 9" into "add up the digits and divide that." For 11, the trick is that $10=11-1$ , so the remainder alternates sign as the power of 10 climbs, which is exactly what turns "divide by 11" into "alternate-sum the digits."

Figure

Place value blocks for the number four thousand seven hundred fifty two showing the last two digits circled for the test for four and the digit sum shown below for the test for nine

Derivation AWhy the digit-sum test works for 9 and for 3

The goal is to prove, for a number of any length and not just three digits, that the digit-sum rule for 9 is not a coincidence but a direct consequence of the single fact $10 = 9+1$ , and that the identical argument works for 3 because 9 and 3 share the same key property.

Step 1. Write the number in place value as $N = d_k\cdot10^k + \cdots + d_1\cdot10 + d_0$ , and notice that every power of 10 can be rewritten as "a multiple of 9, plus 1." Indeed $10 = 9+1$ , $100 = 99+1$ , $1000=999+1$ , and in general $10^i = \underbrace{9\ldots9}_{i}+1$ for every $i \geq 1$ , since a string of $i$ nines is exactly $10^i-1$ .

Step 2. Substitute this into every term of $N$ (for $i \geq 1$ ): $d_i\cdot10^i = d_i\cdot(\underbrace{9\ldots9}_{i}+1) = d_i\cdot\underbrace{9\ldots9}_{i} + d_i$ . The first part, $d_i$ times a string of nines, is always a multiple of 9 no matter what the digit $d_i$ is, because the string of nines itself is a multiple of 9.

Step 3. Collect all the "multiple of 9" pieces together and all the leftover digits together: $N = 9\times(\text{some whole number}) + (d_k+d_{k-1}+\cdots+d_1+d_0)$ . The leftover sum is exactly the digit sum of $N$ . By the division algorithm, adding a multiple of 9 never changes the remainder on division by 9, so $N$ and its digit sum leave the same remainder when divided by 9.

Step 4. Read off the test. $N$ is divisible by 9 exactly when that remainder is 0, which happens exactly when the digit sum is divisible by 9. Since 9 and 3 are both factors of every string of nines (a string of nines is $9\times\underbrace{1\ldots1}_{i}$ , which is also $3\times3\times\underbrace{1\ldots1}_{i}$ ), the identical Step 1–3 argument shows $N$ and its digit sum also leave the same remainder on division by 3 — giving the test for 3 for free.

Worked example

Check whether 4,752 is divisible by 3, 6, and 9, then verify one result by direct division.

Step 1. Find the digit sum: $4+7+5+2 = 18$ .

Step 2. Test for 3: is 18 divisible by 3? Yes, $18 = 3\times6$ , so 4,752 is divisible by 3.

Step 3. Test for 9: is 18 divisible by 9? Yes, $18 = 9\times2$ , so 4,752 is divisible by 9 as well.

Step 4. Test for 6: divisibility by 6 needs both 2 and 3. The last digit of 4,752 is 2, which is even, so it passes the test for 2; Step 2 already showed it passes the test for 3. Both hold, so 4,752 is divisible by 6.

Step 5. Verify by direct division: $4752 \div 3 = 1584$ and $4752 \div 6 = 792$ , both exact — confirming the digit-sum reasoning. 4,752 is divisible by 3, 6, and 9.

Competitive stretch

A quicker mental check for large numbers: to test 63,180 for divisibility by 9 without adding all five digits at once, cancel digits that already sum to 9 or a multiple of 9 as you go. Here $6+3=9$ (cancel), leaving $1+8+0=9$ (cancel) — everything cancels, so 63,180 is divisible by 9 (and also by 3, and since its last digit is 0, by 2, 5, and 10 too).

Common traps

Testing divisibility by 4 using only the very last digit instead of the two-digit number formed by the tens and units places — for 4,752 you must use "52," not just "2."
Treating "divisible by 2 and by 3" as automatically the same as "divisible by 6" — divisibility by 6 genuinely needs both conditions to hold together, checked separately.
Losing track of the $+$ and $-$ signs in the alternating-sum test for 11 — always alternate consistently starting from the same end (usually the units digit).
Assuming 3 and 9 need different-looking rules — both use the plain digit sum; only the divisor you divide that sum by changes (3 versus 9).

Ask the tutor: Want to see the matching place-value proof for the last-two-digits test for 4, or try the alternating-sum test on a 6-digit number step by step?

Why does the digit-sum trick work for 9 but a totally different alternating trick is needed for 11?

Because the two divisors relate to 10 in opposite ways. Since $10=9+1$ , every power of 10 leaves the same remainder (1) on division by 9, so all digits simply add up. Since $10=11-1$ , every power of 10 leaves an alternating remainder ( $+1$ , then $-1$ , then $+1$ , …) on division by 11, so digits must be alternately added and subtracted instead of just summed.

Do I have to memorise all nine tests separately for exams?

No — memorise the three underlying ideas instead. "Look at the last digit(s)" covers 2, 5, 10, 4, and 8 (using more digits as the divisor gets bigger). "Add the digits" covers 3 and 9. "Alternate the digits" covers 11. Divisibility by 6 is not a new idea at all — it is just the tests for 2 and 3 applied together. Three ideas, not nine rules.

Can I use the digit-sum test to also find the actual remainder, not just whether it divides evenly

Yes, for 9 (and for 3). Because $N$ and its digit sum leave the same remainder on division by 9 (shown in the derivation), you can find $N \bmod 9$ just by digit-summing repeatedly until one digit remains, without ever dividing $N$ itself. This does not work for 4, 8, or 11 in the same simple additive way — those need the specific last-digits or alternating-sum construction instead.

Divisibility tests

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads