Did someone say 'division'?

Number Theory Level 2

The largest number $N$ , by which the expression ${ n }^{ 3 }-n$ is divisible by $N$ for all possible integral values of $n$ , is:

6 3 5 2 4

3 solutions

Michael Ng
Jan 23, 2015

Nice problem! We can factorise: $n^3-n = n(n^2-1) = n(n+1)(n-1)$

Now in any set of three consecutive integers there must be one divisible by $3$ and at least one divisible by $2$ . Therefore their product must be divisible by $\mathrm{LCM} (2,3) = 6$ .

But why is it the largest (apart from the fact that it is the largest given answer)? It's because if we consider $n=2$ we get $n^3 - n = 6$ . But then any divisor must be $\leq 6$ so the maximum is indeed $\boxed{6}$ .

Since there are $3$ consecutive numbers, one must be divisible by $3$ because every multiple of $3$ appears every 3 times. . Now every multiple of $2$ also appear every 2 times. Therefore the possible divisors are $1,2,3$ . So the largest would be the $LCM(1,2,3)$ which is $\boxed{6}$

William Isoroku - 6 years, 4 months ago

Same solution.

Roman Frago - 6 years, 4 months ago

Saunak Bhattacharjee
Jul 24, 2015

n^{3} - n = n(n+1)(n-1) =(n-1)n(n+1), for all integral values of n , this expression is the product of three consecutive integers and we know that the product of any three consecutive integers is divisible by 6. Hence for all possible integral values of n, n^{3} - n is divisible by 6. Thus N = 6.

Shivam Pal
Jul 29, 2016

Rule-any consecutive number is divisible by that number's factorial Ex-5. 6. 7. 8. Is divisible by 4! And. 56. 57. 58. 59. 60. 61 is divisible by 6!

Did someone say 'division'?

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