I am working on an old math contest and came across this problem:

What is the largest integer less than $2013$ that can be obtained by repeatedly doubling a positive integer less than $100$?

Here's what I've done: $$x<100$$ $$2^nx<2013$$ $$x<\frac{2013}{2^n}$$

The first time $\frac{2013}{2^n}$ is less than $100$ occurs when $n=5$. Therefore $x < 62.90625$. The largest integer that satisfies the inequality is $x=62$.

As it turns out $62$ is the correct answer, but I feel like my solution is missing something. All I have done is narrow down the possibilities, going from $x<100$ to $x\leq 62$. Can anyone point me in the direction of what I am missing? I don't want the solution just a hint/nudge in the right direction. Any tips would be appreciated!

  • $\begingroup$ You got a multiple of $64$. Since $2012\lt 1984+32$, we can't do better by dividing by $32$. $\endgroup$ Commented Aug 7, 2016 at 16:43

2 Answers 2


If you start with a number less than $100$, you can double it at least $4$ times without exceeding $2012$, so the answer must be a multiple of $2^4=16$. The largest multiple of $16$ less than $2013$ is $2000=16\cdot 125$, which doesn't work: $125\not<100$. The next largest is $1984=31\cdot 2^6$, which works and must therefore be the solution.


I might have approached it using logarithms, i.e. $$\log_2[n]\leq\log_2[2013]-\log_2[x]$$


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