
Prove that for any positive integer n, ${n^3} - n{\text{ is divisible by 6}}$.
Answer
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Hint: The main hint for this question is that if any number is divisible by 2 and 3, then it has to be divisible by 6. So, we will try to check whether ${n^3} - n$ is divisible by 2 and 3 individually. If yes, then ${n^3} - n$ will be divisible by 6, otherwise not.
Complete step-by-step answer:
So, ${n^3} - n{\text{ can also be written as n(n - 1)(n + 1)}}$
For any positive integer n,
When we divide a number by 2, it will either have remainder 0 or 1
So clearly $n(n - 1)(n + 1){\text{ will have a factor }}$
Because for any positive integer n, 2 will be a factor of ${n^3} - n$
For example, let n=11, then assigning value of n=11 in the equation ${n^3} - n$ , we will get
$
\Rightarrow n(n - 1)(n + 1) = 11\left( {11 - 1} \right)(11 + 1) \\
= 11*10*12 \\
$
Here, firstly we will check n is divisible by 2 or not. If yes, then ${n^3} - n{\text{ is divisible by 2}}$. If not, then $(n - 1){\text{ or (}}n + 1{\text{) is divisible by 2}}$ so ${n^3} - n{\text{ is divisible by 2}}$ .
So, in any case, ${n^3} - n{\text{ is divisible by 2}}$
Similarly, when we divide a number by 3, it will have remainder of 0, 1 or 2
So either n, (n-1), (n+1) will be divisible by 3.
For example, let n=11, then assigning value of n=11 in the equation ${n^3} - n$ , we will get
$
\Rightarrow n(n - 1)(n + 1) = 11(11 - 1)(11 + 1) \\
= 11*10*12 \\
$
Here also, we will check whether n is divisible by 3 or not. If yes, then ${n^3} - n$ is divisible by 3.If not, then we will check $(n - 1)$ that is (11-1) , if that is divisible by 3 , then ${n^3} - n$ is divisible by 3. If not, then we will check $(n + 1)$ that is (11+1). Clearly it is divisible by 3.
So, ${n^3} - n$ is divisible by 3 also.
$\therefore $ ${n^3} - n$ is divisible by 2 and 3 individually.
So, ${n^3} - n$ is divisible by 6 also.
Hence, it is proved that for any positive integer n, ${n^3} - n$ is divisible by 6.
Note: Every number has some property from which we can directly conclude whether it is divisible by a number or not. For example, a number is divisible by 2 only if its last digit is even. Another example is, we can directly check whether a big number is divisible by 3 or not by just calculating the sum of its digits and if sum is divisible by 3, then the number will get divisible by 3, otherwise not. Also to check the validity of the above proof, we can put any number in place of n and check whether it is divisible by 6 or not.
Complete step-by-step answer:
So, ${n^3} - n{\text{ can also be written as n(n - 1)(n + 1)}}$
For any positive integer n,
When we divide a number by 2, it will either have remainder 0 or 1
So clearly $n(n - 1)(n + 1){\text{ will have a factor }}$
Because for any positive integer n, 2 will be a factor of ${n^3} - n$
For example, let n=11, then assigning value of n=11 in the equation ${n^3} - n$ , we will get
$
\Rightarrow n(n - 1)(n + 1) = 11\left( {11 - 1} \right)(11 + 1) \\
= 11*10*12 \\
$
Here, firstly we will check n is divisible by 2 or not. If yes, then ${n^3} - n{\text{ is divisible by 2}}$. If not, then $(n - 1){\text{ or (}}n + 1{\text{) is divisible by 2}}$ so ${n^3} - n{\text{ is divisible by 2}}$ .
So, in any case, ${n^3} - n{\text{ is divisible by 2}}$
Similarly, when we divide a number by 3, it will have remainder of 0, 1 or 2
So either n, (n-1), (n+1) will be divisible by 3.
For example, let n=11, then assigning value of n=11 in the equation ${n^3} - n$ , we will get
$
\Rightarrow n(n - 1)(n + 1) = 11(11 - 1)(11 + 1) \\
= 11*10*12 \\
$
Here also, we will check whether n is divisible by 3 or not. If yes, then ${n^3} - n$ is divisible by 3.If not, then we will check $(n - 1)$ that is (11-1) , if that is divisible by 3 , then ${n^3} - n$ is divisible by 3. If not, then we will check $(n + 1)$ that is (11+1). Clearly it is divisible by 3.
So, ${n^3} - n$ is divisible by 3 also.
$\therefore $ ${n^3} - n$ is divisible by 2 and 3 individually.
So, ${n^3} - n$ is divisible by 6 also.
Hence, it is proved that for any positive integer n, ${n^3} - n$ is divisible by 6.
Note: Every number has some property from which we can directly conclude whether it is divisible by a number or not. For example, a number is divisible by 2 only if its last digit is even. Another example is, we can directly check whether a big number is divisible by 3 or not by just calculating the sum of its digits and if sum is divisible by 3, then the number will get divisible by 3, otherwise not. Also to check the validity of the above proof, we can put any number in place of n and check whether it is divisible by 6 or not.
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