Difference between revisions of "2020 AIME II Problems/Problem 5"
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==Video Solution== | ==Video Solution== | ||
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+ | https://youtu.be/lTyiRQTtIZI ~CNCM | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=4|num-a=6}} | {{AIME box|year=2020|n=II|num-b=4|num-a=6}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:23, 8 June 2020
Contents
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://www.youtube.com/watch?v=ZWe_99091e4
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.