Difference between revisions of "2023 AMC 12A Problems/Problem 24"
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*Note: Unfortunately, <math>\binom{10}{5}</math> is not congruent to 0 mod 10, so this solution has the correct answer by coincidence. Also <math>\binom{10}{2}</math> and <math>\binom{10}{8}</math> are not 0 mod 10. Also, the negation of "there does not exist a <math>k</math> so that <math>b_k</math> is 0 or 10" is NOT "for all <math>k</math>, <math>b_k</math> is 0 or 10." | *Note: Unfortunately, <math>\binom{10}{5}</math> is not congruent to 0 mod 10, so this solution has the correct answer by coincidence. Also <math>\binom{10}{2}</math> and <math>\binom{10}{8}</math> are not 0 mod 10. Also, the negation of "there does not exist a <math>k</math> so that <math>b_k</math> is 0 or 10" is NOT "for all <math>k</math>, <math>b_k</math> is 0 or 10." | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | We observe that in each sequence, if element <math>e \in A_i</math>, then <math>e \in A_j</math> for all <math>j \geq i</math>. | ||
+ | Therefore, to determine a sequence with a fixed length <math>n</math>, we only need to determine the first set <math>A_i</math> that each element in <math>\left\{ 1, 2, \cdots , 10 \right\}</math> is inserted into, or an element is never inserted into any subset. | ||
+ | |||
+ | We have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | K & = \sum_{n = 1}^{10} \left( n + 1 \right)^{10} \\ | ||
+ | & = \sum_{m = 2}^{11} m^{10} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Modulo 10, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | K & \equiv \sum_{m = 2}^{11} m^2 \\ | ||
+ | & \equiv \sum_{m = 1}^{11} m^2 - 1^2 \\ | ||
+ | & \equiv \frac{11 \cdot \left( 11 + 1 \right) \left( 2 \cdot 11 + 1 \right)}{6} - 1\\ | ||
+ | & \equiv 505 \\ | ||
+ | & \equiv \boxed{\textbf{(C) 5}} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 18:42, 10 November 2023
Contents
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
Solution 2
Let be the number of sequences
,
,
,
such that each
is a subset of
, and
is a subset of
for
,
,
. Then
and
.
If and
, we need to get a recursive formula for
: If
, then
has
possibilities, and the subsequence
has
possibilities. Hence
By applying this formula and only considering modulo
, we get
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
Lastly, we get .
~Quantum-Phantom
Solution 3 (Cheese, answer by coincidence, incorrect logic)
Since the question only wants mod 10 of the answer, we can cheese this problem.
Let be the number of elements of the set
. Assume that
and
for any
. However, that means by symmetry, there will be
different sequences of
with the same sequence of
. Since
is 0 mod 10 for all
except for 0 and 10, we only consider sequences where each term is either the empty set or the entire set
. If
, then there are
sets. If
, then there are
sets, and so on. So the answer is
(mod 10).
- Note: Unfortunately,
is not congruent to 0 mod 10, so this solution has the correct answer by coincidence. Also
and
are not 0 mod 10. Also, the negation of "there does not exist a
so that
is 0 or 10" is NOT "for all
,
is 0 or 10."
Solution
We observe that in each sequence, if element , then
for all
.
Therefore, to determine a sequence with a fixed length
, we only need to determine the first set
that each element in
is inserted into, or an element is never inserted into any subset.
We have
Modulo 10, we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn
See also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.