The Smallest $k$

$f(x)$ can be written as

$7x^{11}-7x+11x^{7}-11x+9kx +18x$

$= 7x(x^{10}-1) +11x(x^{6}-1)+x(9k+18)$

With the help of FLT it could be seen that Both the terms leaving $x(9k+18)$ are divisible by 77

Now we need to find the value of $k$ for which $77|9k+18$

The values will come out to be 75,152,229.........so on it will be an AP with $a=75$ and $d=77$

These values will not satisfy the condition for $g(x)$

Now $g(x)$ could be written as $7x(x^{12}-1) +13x(x^{6}-1)+x(10k+20)$

With the help of FLT it could be seen that Both the terms leaving $x(10k+20)$ are divisible by 91.

Similarly the values k for which $91|10k+20$ will be

89,180,271........so on It will be an AP with $a=89 \ and \ d=91$

Now we have 2AP's for the value of $k$

• $75 ,\ 152, \ 229.................$ .
• $89 , \ 180, \ 271..............$

The value of $k$ that will satisfy both the conditions will be the common term of both the AP's.

So the smallest common term of the AP's will be 999

So the smallest $k$ will be $\boxed{999}$

If we stay in the realm of modular arithmetic and apply the Chinese Remainder Theorem, we get a slightly different perspective on the nice solutions by shubhendra singh and Aditya Raut.

Since $77 = 7 \cdot 11$ and $91 = 7 \cdot 13$ , it follows from Fermat's Little Theorem that $\begin{aligned} x^6 - 1 & \equiv 0 \pmod{7} \\ x^{10} - 1 & \equiv 0 \pmod{11} \\ x^{12} - 1 & \equiv 0 \pmod{13} \end{aligned}$ Observe that $\begin{aligned} f(x) & = 7 (x^{10} - 1) x + 11 (x^6 - 1) x + (9 k + 18) x \\ & \equiv (9 k + 18) x \pmod{7, 11} \end{aligned}$ and $\begin{aligned} g(x) & = 7 (x^{12} - 1) x + 13 (x^6 - 1) x + (10 k + 20) x \\ & \equiv (10 k + 20) x \pmod{7, 13} \end{aligned}$ We would like that $9k + 18 \equiv 10 k + 20 \equiv 0 \pmod{7}$ . So we conclude that the integer $k$ should satisfy the three linear congruences $\begin{aligned} 9k + 18 & \equiv 0 \pmod{7} & \Rightarrow & k \equiv 5 \pmod{7} \\ 9k + 18 & \equiv 0 \pmod{11} & \Rightarrow & k \equiv 9 \pmod{11} \\ 10k + 20 & \equiv 0 \pmod{13} & \Rightarrow & k \equiv 11 \pmod{13} \end{aligned}$ By the Chinese Remainder Theorem, the system has a solution uniquely determined modulo $7 \cdot 11 \cdot 13 = 1001$ , namely $k \equiv 999 \pmod{1001}$ Hence, the solution to the original problem is $\boxed{k = 999}$ .