## Thursday, August 10, 2006

### Keep Looking. Topic: Number Theory. Level: AIME.

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.