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/ Dividing by 7:
Dividing by 7:
February 22, 2026
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This means $ n = 7k + 3 $ for some integer $ k $.
We want $ 1 \leq 7k + 3 \leq 100 $.
Subtracting 3 from all parts:
-\frac{2}{7} \leq k \leq \frac{97}{7} \approx 13.857
Since $ k $ must be a non-negative integer, valid values are $ k = 0, 1, 2, ..., 13 $, which gives 14 values.
Thus, there are $ \boxed{14} $ integers between 1 and 100 that are congruent to 3 modulo 7.
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