Difference between revisions of "2023 AMC 12A Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
− | Consider any sequence with <math>n</math> terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the <math>n</math>th spot, which means every number has <math>(n+1)</math> choices to show up in the sequence. Consequently, for each sequence with length <math>n</math>, there are <math>(n+1)^10</math> possible ways. | + | Consider any sequence with <math>n</math> terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the <math>n</math>th spot, which means every number has <math>(n+1)</math> choices to show up in the sequence. Consequently, for each sequence with length <math>n</math>, there are <math>(n+1)^{10}</math> possible ways. |
Thus, the desired value is <math>\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}</math> | Thus, the desired value is <math>\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}</math> | ||
~bluesoul | ~bluesoul |
Revision as of 23:21, 9 November 2023
Problem
Let be the number of sequences
,
,
,
such that
is a positive integer less than or equal to
, each
is a subset of
, and
is a subset of
for each
between
and
, inclusive. For example,
,
,
,
,
is one such sequence, with
.What is the remainder when
is divided by
?
Solution 1
Consider any sequence with terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the
th spot, which means every number has
choices to show up in the sequence. Consequently, for each sequence with length
, there are
possible ways.
Thus, the desired value is
~bluesoul