Applications of Bases: Part II - 2014 AIME II #15

First, before I begin, I'd like to say WOW, was I surprised by the reaction. I haven't been posting anything new (part II) because I've been looking/drained of good problems to add. I was totally in for a surprise on this year's AIME II.

First, the problem.

For any integer \(k\ge 1\), let \(p(k)\) be the smallest prime which does not divide k. Define the integer function \(X(k)\) to be the product of all primes less than \(p(k)\) if \(p(k)>2\), and \(X(k)=1\) if \(p(k)=2\). Let \(\{x_n\}\) be the sequence defined by \(x_0=1\), and \(x_{n+1}X(x_n)=x_n p(x_n)\) for \(n\ge 0\). Find the smallest positive integer \(t\) such that \(x_t=2090\).

So, let's go through my thinking process seeing this on the test. At first, I did not see the trick, but it seemed like the kind of problem with a trick. So, what better than compute a few values. Around 10 minutes later (and with a good deal of errors having been made), I had finally obtained the following:

\[x_0=1,~ x_1=2, ~ x_2=3, ~ x_3=6, ~ x_4=5\]

Just looking at that, I wasn't totally sure I saw any pattern. But I did see that there was one, just not an immediately obvious one.

So, I started factorizing. I knew that this was a question about prime factorization simply because we were multiplying and dividing by products of primes constantly.

Here's the important part, though, which saved me quite a bit of time and is why I am posting this to Applications of Bases. I was kind of lazy. I wrote 1 instead of \(2^1\), 31 instead of \(3^3\cdot 2^1\), 1000 instead of \(7^1\), etc. In other words, I wrote the number \( n=(p_k^{e_k})\cdot \ldots \cdot (p_1^{e_1}) \) as \(e_k e_{k-1}\ldots e_1\) where \(p_k\) is the \(k\) - th prime.

Now, I went back to our sequence.

\[x_0=0, ~ x_1=1, ~ x_2=10, ~x_3=11, ~ x_4=100\]

By that point, I needed little confirmation that this was a strange kind of binary, so I rushed off, found that 2090 was 10010101, and converted to get \(\boxed{t=149}\).

Yup, 149 was right. But let's prove it, shall we? (I actually did prove it in my head on the test. It wasn't too hard to see.)

First, we get \(x_{n+1}=\frac{x_n p(x_n)}{X(x_n)}\). Let \(x_n\) be something ending in 011....1 (there could be 0 1s). Then, when we multiply by \(p(x_n)\), we turn this ending into 111...1. Finally, when we divide by \(X(x_n)\), we turn this into 100...0. But, since this is just how we count up in binary, we conclude that \(x_n=n+k\) for some k. Since \(x_0=0\), we see that \(k=0\), and the proof is complete.

Note by Matthew Lipman
4 years, 8 months ago

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Hi, I read your notes and the problem sets but I don't quite understand the third problems wording, can you give some more information on that?

Christopher Boo - 4 years, 8 months ago

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@Matthew Lipman Can you add this to the Brilliant Wiki under a suitable skill in Number Bases? Thanks!

Calvin Lin Staff - 4 years, 2 months ago

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