AIME 2020 II · 第 5 题

Solution 1

Let's work backwards. The minimum base-sixteen representation of $g(n)$ that cannot be expressed using only the digits $0$ through $9$ is $A_{16}$, which is equal to $10$ in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of $f(n)$ is $10$. The minimum value for which this is achieved is $37_8$. We have that $37_8 = 31$. Thus, the sum of the digits of the base-four representation of $n$ is $31$. The minimum value for which this is achieved is $13,333,333,333_4$. We just need this value in base 10 modulo 1000. We get $13,333,333,333_4 = 3(1 + 4 + 4^2 + \dots + 4^8 + 4^9) + 4^{10} = 3\left(\dfrac{4^{10} - 1}{3}\right) + 4^{10} = 2*4^{10} - 1$. Taking this value modulo $1000$, we get the final answer of $\boxed{151}$. (If you are having trouble with this step, note that $2^{10} = 1024 \equiv 24 \pmod{1000}$) ~ TopNotchMath

Solution 2 (Official MAA)

First note that if $h_b(s)$ is the least positive integer whose digit sum, in some fixed base $b$, is $s$, then $h_b$ is a strictly increasing function. This together with the fact that $g(N) \ge 10$ shows that $f(N)$ is the least positive integer whose base-eight digit sum is 10. Thus $f(N) = 37_\text{eight} = 31$, and $N$ is the least positive integer whose base-four digit sum is $31.$ Therefore

$$N = 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1$$ $$\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.$$

Video Solutions

https://youtu.be/lTyiRQTtIZI

https://youtu.be/ZWe_99091e4

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=5032

~ pi_is_3.14