2020 AIME II Problems/Problem 5
Contents
Problem
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
.
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
Video Solution
https://youtu.be/lTyiRQTtIZI ~CNCM
Video Solution 2
~IceMatrix
Video Solution 3
https://youtu.be/ZhAZ1oPe5Ds?t=5032
~ pi_is_3.14
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 6 | |
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