If a 10-digit number **M30348462N** is divisible by both 8 and 11, then what is the value of $$ M^2 + N^2 - 18 $$?

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Accepted Answer

### Step-by-Step Solution:

1. **Divisibility by 8:**  
   A number is divisible by 8 if its last three digits are divisible by 8.  
   - The last three digits of the number are **62N**. For divisibility by 8, substitute $$ N = 4 $$:  
     $$
     624 \div 8 = 78 \, (	ext{exactly divisible}).
     $$  
   Thus, $$ N = 4 $$.

2. **Divisibility by 11:**  
   A number is divisible by 11 if the difference between the sum of its digits in odd positions and even positions is either zero or a multiple of 11.  
   - Digits of the number: $$ M, 3, 0, 3, 4, 8, 4, 6, 2, N $$.  
   - Odd-position digits: $$ M, 0, 4, 4, 2 $$. Sum = $$ M + 0 + 4 + 4 + 2 = M + 10 $$.  
   - Even-position digits: $$ 3, 3, 8, 6, N $$. Sum = $$ 3 + 3 + 8 + 6 + N = 20 + N = 20 + 4 = 24 $$.

For divisibility by $$11$$:  
   $$
   |(M + 10) - (24)| = 	ext{multiple of }11.
   $$  
   Simplify:  
   $$
   |M + 10 - 24| = |M - 14| = 	ext{multiple of }11.
   $$  
   Solve for $$ M $$: $$ M - 14 = -11 \implies M = 3 $$.

3. **Calculate $$ M^2 + N^2 -18 $$:**  
   Substitute $$ M = 3 $$ and $$ N = 4 $$:  
   $$
   M^2 + N^2 -18 = (3)^2 + (4)^2 -18 = 9 +16 -18 =7.
   $$

### Final Answer:  
The value of $$ M^2 + N^2 -18 $$ is **7**, corresponding to option **(c)**.

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