Difference between revisions of "2020 AIME II Problems/Problem 5"
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==Solution 2 (Official MAA)== | ==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 | + | 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 \ | 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> |
− | |||
==Video Solution== | ==Video Solution== |
Revision as of 12:04, 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
\[N &= 13333333333_\text{four} = 2\cdot4^{10} - 1 = 2\cdot1024^2 - 1 \\ &\equiv 2\cdot24^2 - 1 \equiv 151 \pmod{1000}.\] (Error compiling LaTeX. Unknown error_msg)
Video Solution
https://youtu.be/lTyiRQTtIZI ~CNCM
Video Solution 2
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.