Difference between revisions of "2020 AIME II Problems/Problem 5"

Revision as of 12:04, 8 June 2020

Problem

For each positive integer $n$ , left $f(n)$ be the sum of the digits in the base-four representation of $n$ and let $g(n)$ be the sum of the digits in the base-eight representation of $f(n)$ . For example, $f(2020) = f(133210_{\text{four}}) = 10 = 12_{\text{eight}}$ , and $g(2020) = \text{the digit sum of }12_{\text{eight}} = 3$ . Let $N$ be the least value of $n$ such that the base-sixteen representation of $g(n)$ cannot be expressed using only the digits $0$ through $9$ . Find the remainder when $N$ is divided by $1000$ .

Solution

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

\[N &= 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1 \\
  &\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.\] (Error compiling LaTeX. Unknown error_msg)

Video Solution

https://youtu.be/lTyiRQTtIZI ~CNCM

Video Solution 2

https://youtu.be/ZWe_99091e4

~IceMatrix

@@ Line 5: / Line 5: @@
 ==Solution 2 (Official MAA)==
-First note that if <math>h_b(s)</math> is the least positive integer whose digit sum, in some fixed base <math>b</math>, is <math>s</math>, then <math>h_b</math> is a strictly increasing function. This together with the fact that <math>g(N) \ge 10</math> shows that <math>f(N)</math> is the least positive integer whose base-eight digit sum is 10. Thus <math>f(N) = 37_\text{eight} = 31</math>, and <math>N</math> is the least positive integer whose base-four digit sum is <math>31.</math> Therefore\begin{align*}
+First note that if <math>h_b(s)</math> is the least positive integer whose digit sum, in some fixed base <math>b</math>, is <math>s</math>, then <math>h_b</math> is a strictly increasing function. This together with the fact that <math>g(N) \ge 10</math> shows that <math>f(N)</math> is the least positive integer whose base-eight digit sum is 10. Thus <math>f(N) = 37_\text{eight} = 31</math>, and <math>N</math> is the least positive integer whose base-four digit sum is <math>31.</math> Therefore <cmath>
 N &= 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1 \\
-   &\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.
+   &\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.</cmath>
-\end{align*}
 ==Video Solution==

2020 AIME II (Problems • Answer Key • Resources)
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