Difference between revisions of "2020 AIME II Problems/Problem 5"
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==Solution== | ==Solution== | ||
Let's work backwards. The minimum base-sixteen representation of <math>g(n)</math> that cannot be expressed using only the digits <math>0</math> through <math>9</math> is <math>A_{16}</math>, which is equal to <math>10</math> in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of <math>f(n)</math> is <math>10</math>. The minimum value for which this is achieved is <math>37_8</math>. We have that <math>37_8 = 31</math>. Thus, the sum of the digits of the base-four representation of <math>n</math> is <math>31</math>. The minimum value for which this is achieved is <math>13,333,333,333_4</math>. We just need this value in base 10 modulo 1000. We get <math>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</math>. Taking this value modulo <math>1000</math>, we get the final answer of <math>\boxed{151}</math>. (If you are having trouble with this step, note that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>) ~ TopNotchMath | Let's work backwards. The minimum base-sixteen representation of <math>g(n)</math> that cannot be expressed using only the digits <math>0</math> through <math>9</math> is <math>A_{16}</math>, which is equal to <math>10</math> in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of <math>f(n)</math> is <math>10</math>. The minimum value for which this is achieved is <math>37_8</math>. We have that <math>37_8 = 31</math>. Thus, the sum of the digits of the base-four representation of <math>n</math> is <math>31</math>. The minimum value for which this is achieved is <math>13,333,333,333_4</math>. We just need this value in base 10 modulo 1000. We get <math>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</math>. Taking this value modulo <math>1000</math>, we get the final answer of <math>\boxed{151}</math>. (If you are having trouble with this step, note that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>) ~ TopNotchMath | ||
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+ | ==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*} | ||
+ | N &= 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1 \ | ||
+ | &\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}. | ||
+ | \end{align*} | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/lTyiRQTtIZI ~CNCM | https://youtu.be/lTyiRQTtIZI ~CNCM |
Revision as of 12:01, 8 June 2020
Contents
[hide]Problem
For each positive integer , left be the sum of the digits in the base-four representation of and let be the sum of the digits in the base-eight representation of . For example, , and . Let be the least value of such that the base-sixteen representation of cannot be expressed using only the digits through . Find the remainder when is divided by .
Solution
Let's work backwards. The minimum base-sixteen representation of that cannot be expressed using only the digits through is , which is equal to in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of is . The minimum value for which this is achieved is . We have that . Thus, the sum of the digits of the base-four representation of is . The minimum value for which this is achieved is . We just need this value in base 10 modulo 1000. We get . Taking this value modulo , we get the final answer of . (If you are having trouble with this step, note that ) ~ TopNotchMath
Solution 2 (Official MAA)
First note that if is the least positive integer whose digit sum, in some fixed base , is , then is a strictly increasing function. This together with the fact that shows that is the least positive integer whose base-eight digit sum is 10. Thus , and is the least positive integer whose base-four digit sum is Therefore\begin{align*} N &= 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1 \
&\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.
\end{align*}
Video Solution
https://youtu.be/lTyiRQTtIZI ~CNCM
Video Solution 2
~IceMatrix
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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