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Problem 686

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Powers of Two

$2^7=128$ is the first power of two whose leading digits are “$12$”.
The next power of two whose leading digits are “$12$” is $2^{80}$.

Define $p(L,n)$ to be the $n$th-smallest value of $j$ such that the base $10$ representation of $2^j$ begins with the digits of $L$.
So $p(12,1)=7$ and $p(12,2)=80$.

You are also given that $p(123,45)=12710$.

Find $p(123,678910)$.

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二的幂

在二的幂中，第一个拥有前导数字“$12$”的是$2^7=128$。下一个拥有前导数字“$12$”的二的幂是$2^{80}$。

记$p(L,n)$为第$n$个使得$10$进制表示的$2^j$拥有前导数字$L$的数$j$。
因此$p(12,1)=7$，$p(12,2)=80$。

已知$p(123,45)=12710$。

求$p(123,678910)$。

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