Divisibility Rules (2,3,5,7,11,13,17,19,...) | Brilliant Math & Science Wiki

A positive integer \(N\) is divisible by

• \(\color{green}{\boxed{\mathbf{2}}}\) if the last digit of \(N\) is 2, 4, 6, 8, or 0;
• \(\color{green}{\boxed{\mathbf{3}}}\) if the sum of digits of \(N\) is a multiple of 3;
• \(\color{green}{\boxed{\mathbf{4}}}\) if the last 2 digits of \(N\) are a multiple of 4;
• \(\color{green}{\boxed{\mathbf{5}}}\) if the last digit of \(N\) is either 0 or 5;
• \(\color{green}{\boxed{\mathbf{6}}}\) if \(N\) is divisible by both 2 and 3;
• \(\color{green}{\boxed{\mathbf{7}}}\) if subtracting twice the last digit of \(N\) from the remaining digits gives a multiple of 7 (e.g. 658 is divisible by 7 because 65 - 2 x 8 = 49, which is a multiple of 7);
• \(\color{green}{\boxed{\mathbf{8}}}\) if the last 3 digits of \(N\) are a multiple of 8;
• \(\color{green}{\boxed{\mathbf{9}}}\) if the sum of digits of \(N\) is a multiple of 9;
• \(\color{green}{\boxed{\mathbf{10}}}\) if the last digit of \(N\) is 0;
• \(\color{green}{\boxed{\mathbf{11}}}\) if the difference of the alternating sum of digits of \(N\) is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 - 3 + 4 - 3 = 0, which is a multiple of 11);
• \(\color{green}{\boxed{\mathbf{12}}}\) if \(N\) is divisible by both 3 and 4.

Here are some example questions that can be solved using some of the divisibility rules above.

Without performing actual division, show that the number below is an integer:

\[\dfrac{1,481,481,468}{12}.\]

From the divisibility rules, we know that a number is divisible by 12 if it is divisible by both 3 and 4. Therefore, we just need to check that 1,481,481,468 is divisible by 3 and 4.

Applying the divisibility test for 3, we get that \(1+4+8+1+4+8+1+4+6+8=45,\) which is divisible by 3. Hence 1,481,481,468 is divisible by 3.

Applying the divisibility test for 4, we get that the last two digits, 68, is divisible by 4. Hence 1,481,481,468 is also divisible by 4.

Now, since we know that 1,481,481,468 is divisible by both 3 and 4, it is divisible by 12. Therefore, \(\frac{1,481,481,468}{12}\) will be an integer. \(_\square\)

Find all the possible values of \( a\) such that the number \( \overline{98a6}\) is a multiple of \(3.\)

From the rules of divisibility, the number \( \overline{98a6}\) is a multiple of \(3\) if and only if the sum of its digits \( 9 + 8 + a + 6 = 23+a\) is a multiple of \(3.\) Since \( 0 \leq a \leq 9\), this implies that \( a = 1, 4, 7\) are all the possible values. \(_\square\)

Without performing division, explain why the number \(987654321\) is a multiple of \(9\).

By the rule of divisibility of \(9,\) since \(9+8+7+6+5+4+3+2+1 = 45 \) is a multiple of \(9,\) \(987654321\) is a multiple of \(9. \ _\square\)

Without performing actual division, show that \(87456399\) is not divisible by \(11\).

Applying the divisibility rule of \(11,\) the difference between the sum of digits at the odd places \((8+4+6+9 = 27 )\) and the sum of digits at even places \((7+5+3+9 = 24)\) is \(27-24=3,\) which is not divisible by \(11\). Hence \(87456399\) is not divisible by \(11\). \(_\square\)

For what values of \(a\) and \(b\) is \( \overline{12ab} \) a multiple of \(99?\)

Since \( 99 = 9 \times 11 \), the number must be a multiple of \(9\) and \(11.\)

The divisibility rule of \(9\) tells us that \( 1 + 2 + a + b \) is a multiple of \(9.\) Since it is a number from \(3\) to \(21,\) it must be either \(9\) or \(18.\)

Now, the divisibility rule of \(11\) tells us that \( 1 - 2 + a - b \) is a multiple of \(11.\) Since it is a number from \(-10\) to \(8,\) it must be \(0.\)

Solving \( \begin{cases} 1 + 2 + a + b = 9 \\ 1 - 2 + a - b = 0, \end{cases} \) there are no integer solutions.

Solving \( \begin{cases} 1 + 2 + a + b = 18 \\ 1 - 2 + a - b = 0, \end{cases} \) we get \( a = 8, b = 7 \).

Hence, the only solution is \( 1287 \) with \( a = 8\) and \(b = 7. \) \(_\square\)

Is \(65973390\) divisible by \(210?\)

We do not know the divisibility rule of 210. However, we can easily see that \(210=2\times 3\times 5\times7\), so if 65973390 is divisible by 2, 3, 5, 7, then it is divisible by 210.

• Since the last digit of 65973390 is 0, it is divisible by 2.
• Since \(6+5+9+7+3+3+9+0=42\), which is divisible by 3, it follows that 65973390 is divisible by 3.
• Since the last digit of 65973390 is 0, hence it is divisible by 5.
• To check divisibility by 7, as the initial step, we calculate \(6597339-2(0)=6597339\). However, this number is still a little too big for us to tell whether it's divisible by 7. In such cases, we keep applying the divisibility rule again and again until we have a small enough number to work with: \[\begin{align}659733-2(9)&=659715\\65971-2(5)&=65961\\6596-2(1)&=6594\\659-2(4)&=651\\65-2(1)&=63.\end{align}\] Now we can see that we are left with \(63,\) which we can easily identify as a multiple of 7. Hence 65973390 is a multiple of 7 also.

Since 65973390 is divisible by all of 2, 3, 5, 7, it is divisible by \(2\times3\times5\times7=210. \ _\square\)

Try some problems for yourself to see if you understand this topic:

Reveal the answer

If we know an integer is a multiple of 5, how many possibilities are there for the last two digits of the integer?

The correct answer is: 20

By the rule of divisibility of 5, if the integer is a multiple of 5, then the last digit can either be 0 or 5. The next to last digit can be anything. Hence, there are 20 possibilities.

Yes No Reveal the answer

Is 87985 divisible by 7?

The correct answer is: No

Let's just divide! If we begin to do long-division on 87985, we get \[87985 = 12569 \times 7 + 2.\] Thus, 87985 is not divisible by 7 as there is a remainder of 2.