2020 AIME II Problems/Problem 5
For each positive integer , let 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 .
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
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