**Problem**: Find a positive integer $ n $ such that if you put a $ 2 $ on the left and a $ 1 $ on the right the new number is $ 33n $.

**Solution**: First, we mathize the problem statement. Adding a $ 2 $ on the left is equivalent to adding $ 2 \cdot 10^k $ for some $ k $. Adding a $ 1 $ to the right is equivalent to multiplying $ n $ by $ 10 $ and adding $ 1 $. So the new number should be

$ 2 \cdot 10^k+10n+1 = 33n $

$ 2 \cdot 10^k+1 = 23n $.

Taking this modulo $ 23 $, we have

$ 2 \cdot 10^k+1 \equiv 0 \pmod{23} $.

We easily check to find $ k = 3 $ works. So, plugging in, we solve

$ 2 \cdot 1000 + 1 = 23n \Rightarrow n = 87 $.

To make sure we didn't do anything stupid, we can see that

$ 33 \cdot 87 = 2871 $,

as desired. QED.

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Comment: A pretty elementary number theory problem; the hardest part was probably converting the problem statement into its mathematical counterpart. Straightforward calculations with mods gave us the answer quickly.

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Practice Problem: Characterize all positive integers with the property in the above problem.

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