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# Base numbers

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147
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What is the sum of all positive integers that have twice as many digits when written in base 2 as they have when written in base 4? Express your answer in base 10.

Apr 16, 2022

#1
+9461
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Suppose the positive integer $$x$$ satisfies the condition.

Then, for some positive integer $$n$$, we have $$4^{n - 1} \leq x < 4^n$$ and $$2^{2n - 1} \leq x < 2^{2n}$$. i.e., $$\dfrac12 \cdot 4^n \leq x < 4^n$$.

In fact, if there exists a positive integer $$n$$ such that $$\dfrac12 \cdot 4^n \leq x < 4^n$$, then the positive integer $$x$$ satisfies the condition.

It turns out that there are infinitely many numbers satisfying the condition.

Apr 17, 2022