n_k = 3 + (k-1)\cdot 7 = 7k - 4.
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- Solution: We seek the number of positive integers $n \leq 100$ such that $n \equiv 3 \pmod{7}$.
- Such numbers form an arithmetic sequence: $3, 10, 17, \dots$, with first term $a = 3$, common difference $d = 7$.
- The general term is:
- We require $n_k \leq 100$:
- 7k - 4 \leq 100 \Rightarrow 7k \leq 104 \Rightarrow k \leq rac{104}{7} pprox 14.857.
- So the largest integer $k$ is $14$.
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