# Problem Of The Week # 293 - Dec 14, 2017

• MHB
• Ackbach
In summary, the conversation focused on the importance of effective communication in relationships. The speakers discussed the need for active listening, understanding and empathy, and setting clear boundaries. They also emphasized the role of effective communication in conflict resolution and maintaining healthy relationships. Overall, the conversation highlighted the crucial role of communication in building strong and fulfilling relationships.
Ackbach
Gold Member
MHB
Here is this week's POTW:

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What is the greatest common divisor of the set of numbers $\left\{16^n+10n-1 \;|\; n=1, 2, 3, \dots\right\}?$

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Congratulations to kaliprasad, kiwi, johng, and castor28 for their correct solutions to this week's POTW, which was the MAA Challenges Problem 174. castor28's solution follows:

[sp]
Let us write $f(n)=16^n+10n-1$ and $d$ for the GCD of $\{f(n)\mid n \ge 1\}$.

Since $f(1)=25$, $d$ can only be 1, 5, or 25. To prove that $d=25$, we only need to prove that $25\mid f(n)$ for all $n\ge 1$.

We prove this by induction on $n$.

We obviously have $25\mid f(1)=25$. Assume now that $n>1$. We have:

\begin{align*} f(n+1) - f(n) &= 15\cdot16^{n-1} + 10\\ &= 5(3\cdot16^{n-1} + 2) \end{align*}

Since $3\cdot16^{n-1} + 2\equiv 3\cdot1^{n-1}+2\equiv0\pmod5$, we conclude that $f(n+1)\equiv f(n)\equiv0\pmod{25}$ (using the induction hypothesis), and $d=25$.
[/sp]

## 1. What is the "Problem of the Week #293"?

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## 2. When was "Problem of the Week #293" posted?

"Problem of the Week #293" was posted on December 14, 2017.

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Some websites or publications may offer rewards or prizes for individuals who successfully solve "Problem of the Week #293". However, the main reward is the satisfaction and sense of accomplishment that comes from solving a challenging problem.

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