Let $$a_n$$ and $$b_n$$ be two sequences such that $$a_n = 13 + 6(n-1)$$ and $$b_n = 15 + 7(n-1)$$ for all natural numbers $$n$$. Then, the largest three-digit integer that is common to both these sequences is

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Accepted Answer

1. **General Form of Sequences:**  
   - For $$a_n$$: $$a_n = 13 + 6(n-1) = 7 + 6n$$.  
   - For $$b_n$$: $$b_n = 15 + 7(n-1) = 8 + 7m$$.

2. **Finding Common Terms:**  
   To find common terms, equate $$a_n$$ and $$b_m$$:  
   $$
   7 + 6n = 8 + 7m.
   $$  
   Simplify:  
   $$
   6n - 7m = 1.
   $$

3. **Solving for $$n$$ and $$m$$:**  
   Using integer values for $$n$$ and $$m$$:  
   - For $$n = 6$$ and $$m = 5$$, we get:  
     $$
     a_6 = b_5 = 43.
     $$

4. **Common Terms Formula:**  
   The least common multiple (LCM) of the common differences (6 and 7) is:  
   $$
   	ext{LCM}(6,7) = 42.
   $$  
   Thus, the common terms are of the form:  
   $$
   a_n = b_m = 43 + 42k \quad 	ext{for integer } k.
   $$

5. **Finding the Largest Three-Digit Common Term:**  
   To find the largest three-digit number in this form:  
   $$
   	ext{Largest } k 	ext{ such that } 43 + 42k < 1000.
   $$  
   Solve:  
   $$
   k < \frac{1000 - 43}{42} \approx \frac{957}{42} \approx 22.79.
   $$  
   Thus, the largest integer value of $$k$$ is $$22$$. Substituting:  
   $$
   a_{22} = b_{22} = 43 + 42 	imes 22 = 967.
   $$

Thus, the largest three-digit integer common to both sequences is **967**.

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