Divisibility
There are several equivalent ways of stating the above definition. The most helpful way is “If I attempt to divide the number \(n\) by \(m\) and I get an integer, then \(m\) (evenly) divides \(n\). Otherwise, \(m\) does not divide \(n\).
Putting the above definition another, more verbal, way: “\(m\) divides \(n\) if and only if \(n\) can be written as a product of \(m\) and some other integer.”
Testing for Divisibility Tricks
Each of the boxes below explain a trick for determining whether a number \(n\) is divisible by the given number.
A number \(n\) is divisible by \(2\) if the last digit (i.e. the 1’s digit) of \(n\) is divisible by \(2\).
If \(n=12436\), \(n\) is divisible by \(2\) because the 1’s digit is \(6\) which is clearly divisible by \(2\).
Because integers with more than one digit can be written as a sum of some multiple of \(10\) plus a number less than \(10\). Since \(2\vert 10\), \(2\) divides any multiple of \(10\), so we only need to check that \(2\) divides the 1’s digit.
To see this reasoning through example, let \(n=1378\). Notice that \(n=1378=1370+8\). Notice also that \(2\vert 1370\) (because \(1370\) is a multiple of \(10\) and \(2\vert 10\)), and \(2\vert 8\). Since each of these two added numbers is divisible by 2, so is the sum; i.e. so is \(n\).
A number \(n\) is divisible by \(3\) if the sum of the digits of \(n\) is divisible by \(3\).
Note that if even after adding the digits the number is still too big, you can repeat the addition process until you have a number that is easy to test for divisibility by 3.
Let \(n=5754\). Then since the sum of the digits of \(n\) is \(21\), and \(21\) is divisible by \(3\), \(n\) is also divisible by \(3\).
(See Modular Arithmetic lesson before reading this)
Note that \(10\cong 1 (mod\ 3)\). So, powers of \(10\) such as \(10,100,1000,\)etc. all have remainder \(1\) when divided by \(3\).
So, if you have an integer with decimal representation \(n=a_ka_{k-1}a_{k-2}\dots a_3a_2a_1a_0\) (i.e. the 1’s digit is \(a_0\) the 10’s digit is \(a_1\) etc.\) notice that
$$a_k\cdot 10^k+a_{k-1}\cdot 10^{k-1}+\dots+a_2\cdot 100+a_1\cdot 10+a_0\equiv a_k+a_{k-1}+a_{k-2}+\dots+a_2+a_1+a_0\ (mod\ 3).$$
Therefore, our number \(n\) is divisible by \(3\) if and only if the sum of our digits \(a_k+a_{k-1}+a_{k-2}+\dots+a_2+a_1+a_0\equiv 0\ (mod\ 3)\), i.e. divisible by 3.
Illustrative example: let \(n=6123\). Then
$$\begin{align} 6123&=6\cdot 1000+1\cdot 100+2\cdot 10+3\\ & \equiv 6\cdot 1\ (mod\ 3)+ 1\cdot 1\ (mod\ 3)+2\cdot 1\ (mod\ 3)+3\cdot 1\ (mod\ 3) &\equiv 6+1+2+3\ (mod 3)\\ &= 12\ (mod\ 3)\\& \equiv 0\ (mod 3)\end{align}$$
Thus, \(6123\equiv 0 (mod\ 3)\) which implies, by definition of congruence, that \(6123\) is divisible by \(3\).
A number \(n\) is divisible by \(4\) if the number formed by the last two digits (i.e. the 1’s and 10’s digits) of \(n\) is divisible by \(4\).
Let \(n=5756\). Since the number formed by the last two digits, i.e. 56, is divisible by 4, it follows that \(n=5756\) is also divisible by 4.
Because integers with more than two digits can be written as a sum of some multiple of \(100\) plus a number less than \(100\). Since \(4\vert 100\), \(4\) divides any multiple of \(100\), so we only need to check that \(4\) divides the number formed by the last two digits.
To see this reasoning through example, let \(n=2328\). Notice that \(n=2328=2300+28\). Notice also that \(4\vert 2300\) (because \(2300\) is a multiple of \(100\) and \(4\vert 100\)), and \(4\vert 28\). Since each of these two added numbers is divisible by 4, so is the sum; i.e. so is \(n\).
A number \(n\) is divisible by \(5\) if the last digit (i.e. the 1’s digit) of \(n\) is divisible by \(5\) (i.e. the \(1\)’s digit is a \(0\) or \(5\)).
If \(n=12430\), \(n\) is divisible by \(5\) because the 1’s digit is \(0\) which is clearly divisible by \(5\). (technically, \(0\) is divisible by everything because anything times \(0\) is \(0\)).
Because integers with more than one digit can be written as a sum of some multiple of \(10\) plus a number less than \(10\). Since \(5\vert 10\), \(5\) divides any multiple of \(10\), so we only need to check that \(5\) divides the 1’s digit.
To see this reasoning through example, let \(n=1375\). Notice that \(n=1375=1370+5\). Notice also that \(5\vert 1370\) (because \(1370\) is a multiple of \(10\) and \(5\vert 10\)), and \(2\vert 5\). Since each of these two added numbers is divisible by 5, so is the sum; i.e. so is \(n\).
A number \(n\) is divisible by \(6\) if and only if it is divisible by \(2\) AND \(3\).
Let \(n=5754\). Then since the sum of the digits of \(n\) is \(21\), and \(21\) is divisible by \(3\), \(n\) is also divisible by \(3\). Since the last digit of \(n\) is even, \(n\) is also divisible by \(2\). Hence, \(n=5754\) is divisible by \(6\).
Proof of Test:
Since both \(2\) and \(3\) divide \(6\), if \(6\) divides a number \(n\), then it follows that \(2\) and \(3\) must also divide this number \(n\) as well.
If \(6\) divides a number \(n\), then since \(2\) and \(3\) divide \(6\), \(2\) and \(3\) will divide any multiple of \(6\); in this case \(n\).
A number \(n\) is divisible by \(8\) if the number formed by the last three digits (i.e. the 1’s, 10’s, and 100’s digits) of \(n\) is divisible by \(8\).
Let \(n=5856\). Since the number formed by the last three digits, i.e. 856, is divisible by 8, it follows that \(n=5856\) is also divisible by 8.
Because integers with more than three digits can be written as a sum of some multiple of \(1000\) plus a number less than \(1000\). Since \(8\vert 1000\), \(8\) divides any multiple of \(1000\), so we only need to check that \(8\) divides the number formed by the last three digits.
To see this reasoning through example, let \(n=2128\). Notice that \(n=2128=2100+28\). Notice also that \(8\vert 2100\) (because \(2100\) is a multiple of \(1000\) and \(8\vert 1000\)), and \(8\vert 128\). Since each of these two added numbers is divisible by 8, so is the sum; i.e. so is \(n\).
A number \(n\) is divisible by \(9\) if the sum of the digits of \(n\) is divisible by \(9\).
Note that if even after adding the digits the number is still too big, you can repeat the addition process until you have a number that is easy to test for divisibility by 9.
Let \(n=5778\). Then since the sum of the digits of \(n\) is \(27\), and \(27\) is divisible by \(9\), \(n\) is also divisible by \(9\).
(See Modular Arithmetic lesson before reading this if you care)
(essentially the same reason as the test for divisibility by \(3\))
Note that \(10\cong 1 (mod\ 9)\). So, powers of \(10\) such as \(10,100,1000,\)etc. all have remainder \(1\) when divided by \(9\).
So, if you have an integer with decimal representation \(n=a_ka_{k-1}a_{k-2}\dots a_3a_2a_1a_0\) (i.e. the 1’s digit is \(a_0\) the 10’s digit is \(a_1\) etc.\) notice that
$$a_k\cdot 10^k+a_{k-1}\cdot 10^{k-1}+\dots+a_2\cdot 100+a_1\cdot 10+a_0\equiv a_k+a_{k-1}+a_{k-2}+\dots+a_2+a_1+a_0\ (mod\ 9).$$
Therefore, our number \(n\) is divisible by \(9\) if and only if the sum of our digits \(a_k+a_{k-1}+a_{k-2}+\dots+a_2+a_1+a_0\equiv 0\ (mod\ 9)\), i.e. divisible by 9.
Illustrative example: let \(n=6345\). Then
$$\begin{align} 6345&=6\cdot 1000+3\cdot 100+4\cdot 10+5\\ & \equiv 6\cdot 1\ (mod\ 9)+ 3\cdot 1\ (mod\ 9)+4\cdot 1\ (mod\ 9)+5\cdot 1\ (mod\ 9) &\equiv 6+3+4+5\ (mod 9)\\ &= 18\ (mod\ 9)\\& \equiv 0\ (mod 9)\end{align}$$
Thus, \(6345\equiv 0 (mod\ 9)\) which implies, by definition of congruence, that \(6345\) is divisible by \(9\).
True
\(512346\) is even; i.e. the 1’s digit is divisible by \(2\).
True
The sum of the digits of \(621245\) is \(21\) which is divisible by \(3\).
False
\(2\vert 85616\) because the number given is even, but the digits sum to \(26\) and \(3\not| 26\). Need both \(2\) AND \(3\) to divide the number in question.
True.
Note that the last three digits give us the number \(432\) which is divisible by \(8\) (\(54\cdot 8=432\)). Thus it follows that \(8\vert 172432\) (because \(8\vert 172000\) and \(8\vert 432\), so \(8\) divides the sum).
Find an integer \(c\) so that \(11\cdot c= 352\).
Note that \(11\cdot 32 =352\), and so by the definition of “divides,” there is an integer \(c\) such that \(11\cdot c =352\), namely \(c=32\).
Find an integer \(c\) so that \(13\cdot c= -8151\).
Note that \(13\cdot -627 =-8151\), and so by the definition of “divides,” there is an integer \(c\) such that \(13\cdot c =-8151\), namely \(c=-627\).
Don’t overthink this. We are looking for an explanation as to why \(17\) doesn’t (evenly) divide \(300\).
What happens when you try to divide \(300\) by \(17\)? What type of number do you get (i.e. integer, ugly decimal that isn’t a whole number, etc.)? What type of number does \(c\) need to be in the definition of “divides” ?
Note that \(300/17=17.64705882353…\), which is not an integer. (Stated slightly differently, 17\cdot 17.64705882352…=300\) and \(17.64705882352\notin \mathbb{Z}\)). Therefore, there is no integer \(c\) such that \(18\cdot c=300\).