(a) 3 consecutive integers are divisible by 2,3. (b) $1^{2015}+6^{2015}=1^{2015}-1^{2015}=0(mod 7)$. Similarly for$(2,5),(3,4)=(2,-2),(3,-3)(mod7)$ Q.E.D.

(a) $(n-1)\times n\times(n+1)=(n^2-n)(n+1)$ [According to fermat's little theorem given expression is divided by $2$ ] $(n-1)\times n\times (n+1)=(n^2-n)(n+1)=n^3-n$ [According to fermat's little theorem given expression is divided by $3$ ] So, $(n-1)\times n\times (n+1)$ is divided by $2\times 3$ or $6$. Proved.

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Solution

Do these types of problem really appear in NMO?

We can also use induction for the first question.

a) We know that every second integer is divisible by 2, and every third is divisible by 3, so there is at least one factor of 2 and 3, so we are done.

Show that $n(n + 1)(n + 2)$ is divisible by $6$. (I'm assuming for positive numbers, can be extended similarly to negative numbers and so on) $\text{Proof:}$ $\textbf{Base Case:}$ let the least number, $n$, be $1$ then, we have $n(n+1)(n+2)=1(2)(3)$, which is clearly divisible by $3$ $\textbf{Inductive Step:}$ assume this is true for some $n=k, k\in\mathbb{Z^+}$ we have $k(k+1)(k+2)=3m, m\in\mathbb{Z^+}$ $\textbf{Proof for } n=k+1\textbf{:}$ if our argument is to hold true, then $(k+1)(k+2)(k+3)=3p, p\in\mathbb{Z^+}$ simplifying the $\textbf{LHS},$ we see it is equal to $$k^3+3k^2+3k^2+9k+2k+6$$but we also know that $k(k+1)(k+2)=3m=k^3+3k^2+2k$; substituting this in, we see $$k^3+3k^2+3k^2+9k+2k+6=3m+3k^2+9k+6=3m+3(k^2+9k+6)$$now since, $k\in\mathbb{Z^+},$ we know that $k^2+9k+6$ is divisible by $3,$ meaning $3|$$k^3+3k^2+3k^2+9k+2k+6\Rightarrow3|(k+1)(k+2)(k+3)$ and satisfying $p\in\mathbb{Z^+}.$ By the $\textbf{PMI}$, this result holds true for all positive $n.$ Similar arguments can be made to extend this argument to other sets of numbers.

Prabh1512 wrote:

Solution

If I'm not wrong isn't this circular logic... ?? Or does such a theorem exist already because I don't know about it (a) By definition, there is a multiple of 3 every three consecutive numbers, there is a multiple of 2 every two consecutive numbers. Since $n(n + 1)(n + 2)$ is a set of three consecutive numbers. Thus there would be one multiple of 3 and 1 or 2 multiples of 2, so it would divide 6 (as $3\cdot 2 = 6$). (b) Notice that $1^{2015} + 6^{2015} \equiv 0\pmod7$ (using pattern bash), as well as $2^{2015} + 5^{2015} \equiv 0\pmod7$, $3^{2015} + 4^{2015} \equiv 0\pmod7$, thus the sum would be divisble by $7$. $\blacksquare$

For first it is direct induction and for second we can use the identity that if n is odd then $a^n+b^n$=(a+b) ($a^{n-1}....... b^{n-1}$) then proof of second is done

We assume $n$ is a positive integer.

Part (a)

Part (b)