Difference between revisions of "2020 AIME II Problems/Problem 5"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/lTyiRQTtIZI ~CNCM | https://youtu.be/lTyiRQTtIZI ~CNCM | ||
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==See Also== | ==See Also== |
Revision as of 01:30, 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
Video Solution
https://youtu.be/lTyiRQTtIZI ~CNCM
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.