What is the smallest five-digit number which, when divided by 7, 11, and 21, leaves a remainder of 3 in each case?

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Accepted Answer

To solve this:  
1. The least common multiple (LCM) of 7, 11, and 21 is calculated:  
   $$ 	ext{LCM}(7, 11, 21) = 231 $$.

2. Let the required number be $$N$$. It satisfies:  
   $$ N \mod 231 = 3 \implies N = 231k + 3 $$ for some integer $$k$$.

3. The smallest five-digit number is **10000**. Solve for $$k$$:  
   $$ N \geq 10000 \implies 231k + 3 \geq 10000 \implies k = \lceil \frac{9997}{231} ceil = 44 $$.

4. Substitute $$k = 44$$:  
   $$ N = 231 	imes 44 + 3 = 10019 $$.

Thus, the smallest five-digit number is **10019**.

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