Difference between revisions of "2023 AMC 12A Problems/Problem 24"
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Let <math>f(x,\ell)</math> be the number of sequences <math>A_1</math>, <math>A_2</math>, <math>\dots</math>, <math>A_\ell</math> such that each <math>A_i</math> is a subset of <math>\{1, 2,\dots,x\}</math>, and <math>A_i</math> is a subset of <math>A_{i+1}</math> for <math>i=1</math>, <math>2\dots</math>, <math>\ell-1</math>. Then <math>f(x,1)=2^x</math> and <math>f(0,\ell)=1</math>. | Let <math>f(x,\ell)</math> be the number of sequences <math>A_1</math>, <math>A_2</math>, <math>\dots</math>, <math>A_\ell</math> such that each <math>A_i</math> is a subset of <math>\{1, 2,\dots,x\}</math>, and <math>A_i</math> is a subset of <math>A_{i+1}</math> for <math>i=1</math>, <math>2\dots</math>, <math>\ell-1</math>. Then <math>f(x,1)=2^x</math> and <math>f(0,\ell)=1</math>. | ||
− | If <math>\ell\ge2</math> and <math>x\ge1</math>, we need to get a recursive formula for <math>f(x,\ell)</math>: If <math>|A_1|=i</math>, then <math>A_1</math> has <math>\text C_x^i</math> possibilities, and the subsequence <math>\{A_i\}_{2\le i\le\ell}</math> has <math>f(x-i,\ell-1)</math> possibilities. Hence | + | If <math>\ell\ge2</math> and <math>x\ge1</math>, we need to get a recursive formula for <math>f(x,\ell)</math>: If <math>|A_1|=i</math>, then <math>A_1</math> has <math>\text C_x^i</math> possibilities, and the subsequence <math>\{A_i\}_{2\le i\le\ell}</math> has <math>f(x-i,\ell-1)</math> possibilities. Hence |
<cmath>f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).</cmath> | <cmath>f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).</cmath> | ||
By applying this formula and only considering modulo <math>10</math>, we get <math>f(1,2)=3</math>, <math>f(1,3)=4</math>, <math>f(1,4)=5</math>, <math>f(1,5)=6</math>, <math>f(1,6)=7</math>, <math>f(1,7)=8</math>, <math>f(1,8)=9</math>, <math>f(1,9)=0</math>, <math>f(1,10)=1</math>, <math>f(2,2)=9</math>, <math>f(2,3)=6</math>, <math>f(2,4)=5</math>, <math>f(2,5)=6</math>, <math>f(2,6)=9</math>, <math>f(2,7)=4</math>, <math>f(2,8)=1</math>, <math>f(2,9)=0</math>, <math>f(2,10)=1</math>, <math>f(3,2)=7</math>, <math>f(3,3)=4</math>, <math>f(3,4)=5</math>, <math>f(3,5)=6</math>, <math>f(3,6)=3</math>, <math>f(3,7)=2</math>, <math>f(3,8)=9</math>, <math>f(3,9)=0</math>, <math>f(3,10)=1</math>, <math>f(4,2)=1</math>, <math>f(4,3)=6</math>, <math>f(4,4)=5</math>, <math>f(4,5)=6</math>, <math>f(4,6)=1</math>, <math>f(4,7)=6</math>, <math>f(4,8)=1</math>, <math>f(4,9)=0</math>, <math>f(4,10)=1</math>, <math>f(5,2)=3</math>, <math>f(5,3)=4</math>, <math>f(5,4)=5</math>, <math>f(5,5)=6</math>, <math>f(5,6)=7</math>, <math>f(5,7)=8</math>, <math>f(5,8)=9</math>, <math>f(5,9)=0</math>, <math>f(5,10)=1</math>, <math>f(6,2)=9</math>, <math>f(6,3)=6</math>, <math>f(6,4)=5</math>, <math>f(6,5)=6</math>, <math>f(6,6)=9</math>, <math>f(6,7)=4</math>, <math>f(6,8)=1</math>, <math>f(6,9)=0</math>, <math>f(6,10)=1</math>, <math>f(7,2)=7</math>, <math>f(7,3)=4</math>, <math>f(7,4)=5</math>, <math>f(7,5)=6</math>, <math>f(7,6)=3</math>, <math>f(7,7)=2</math>, <math>f(7,8)=9</math>, <math>f(7,9)=0</math>, <math>f(7,10)=1</math>, <math>f(8,2)=1</math>, <math>f(8,3)=6</math>, <math>f(8,4)=5</math>, <math>f(8,5)=6</math>, <math>f(8,6)=1</math>, <math>f(8,7)=6</math>, <math>f(8,8)=1</math>, <math>f(8,9)=0</math>, <math>f(8,10)=1</math>, <math>f(9,2)=3</math>, <math>f(9,3)=4</math>, <math>f(9,4)=5</math>, <math>f(9,5)=6</math>, <math>f(9,6)=7</math>, <math>f(9,7)=8</math>, <math>f(9,8)=9</math>, <math>f(9,9)=0</math>, <math>f(9,10)=1</math>, <math>f(10,2)=9</math>, <math>f(10,3)=6</math>, <math>f(10,4)=5</math>, <math>f(10,5)=6</math>, <math>f(10,6)=9</math>, <math>f(10,7)=4</math>, <math>f(10,8)=1</math>, <math>f(10,9)=0</math>, <math>f(10,10)=1</math>. | By applying this formula and only considering modulo <math>10</math>, we get <math>f(1,2)=3</math>, <math>f(1,3)=4</math>, <math>f(1,4)=5</math>, <math>f(1,5)=6</math>, <math>f(1,6)=7</math>, <math>f(1,7)=8</math>, <math>f(1,8)=9</math>, <math>f(1,9)=0</math>, <math>f(1,10)=1</math>, <math>f(2,2)=9</math>, <math>f(2,3)=6</math>, <math>f(2,4)=5</math>, <math>f(2,5)=6</math>, <math>f(2,6)=9</math>, <math>f(2,7)=4</math>, <math>f(2,8)=1</math>, <math>f(2,9)=0</math>, <math>f(2,10)=1</math>, <math>f(3,2)=7</math>, <math>f(3,3)=4</math>, <math>f(3,4)=5</math>, <math>f(3,5)=6</math>, <math>f(3,6)=3</math>, <math>f(3,7)=2</math>, <math>f(3,8)=9</math>, <math>f(3,9)=0</math>, <math>f(3,10)=1</math>, <math>f(4,2)=1</math>, <math>f(4,3)=6</math>, <math>f(4,4)=5</math>, <math>f(4,5)=6</math>, <math>f(4,6)=1</math>, <math>f(4,7)=6</math>, <math>f(4,8)=1</math>, <math>f(4,9)=0</math>, <math>f(4,10)=1</math>, <math>f(5,2)=3</math>, <math>f(5,3)=4</math>, <math>f(5,4)=5</math>, <math>f(5,5)=6</math>, <math>f(5,6)=7</math>, <math>f(5,7)=8</math>, <math>f(5,8)=9</math>, <math>f(5,9)=0</math>, <math>f(5,10)=1</math>, <math>f(6,2)=9</math>, <math>f(6,3)=6</math>, <math>f(6,4)=5</math>, <math>f(6,5)=6</math>, <math>f(6,6)=9</math>, <math>f(6,7)=4</math>, <math>f(6,8)=1</math>, <math>f(6,9)=0</math>, <math>f(6,10)=1</math>, <math>f(7,2)=7</math>, <math>f(7,3)=4</math>, <math>f(7,4)=5</math>, <math>f(7,5)=6</math>, <math>f(7,6)=3</math>, <math>f(7,7)=2</math>, <math>f(7,8)=9</math>, <math>f(7,9)=0</math>, <math>f(7,10)=1</math>, <math>f(8,2)=1</math>, <math>f(8,3)=6</math>, <math>f(8,4)=5</math>, <math>f(8,5)=6</math>, <math>f(8,6)=1</math>, <math>f(8,7)=6</math>, <math>f(8,8)=1</math>, <math>f(8,9)=0</math>, <math>f(8,10)=1</math>, <math>f(9,2)=3</math>, <math>f(9,3)=4</math>, <math>f(9,4)=5</math>, <math>f(9,5)=6</math>, <math>f(9,6)=7</math>, <math>f(9,7)=8</math>, <math>f(9,8)=9</math>, <math>f(9,9)=0</math>, <math>f(9,10)=1</math>, <math>f(10,2)=9</math>, <math>f(10,3)=6</math>, <math>f(10,4)=5</math>, <math>f(10,5)=6</math>, <math>f(10,6)=9</math>, <math>f(10,7)=4</math>, <math>f(10,8)=1</math>, <math>f(10,9)=0</math>, <math>f(10,10)=1</math>. | ||
Line 22: | Line 22: | ||
Lastly, we get <math>K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{\textbf{(C) } 5}\pmod{10}</math>. | Lastly, we get <math>K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{\textbf{(C) } 5}\pmod{10}</math>. | ||
~Quantum-Phantom | ~Quantum-Phantom | ||
+ | ==Solution 3 (Cheese, but this time it actually works)== | ||
+ | Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem. | ||
− | + | Let <math>A_1, A_2, \cdots, A_n</math> be one sequence satisfying the constraints of the problem. Let <math>b_1, b_2, \cdots, b_n, b_{n+1}</math> be the sequence of nonnegative integers such that <math>A_k</math> has <math>b_k</math> elements for all <math>1\leq{}k\leq{}n</math>, and <math>b_{n+1}=10</math>. Note that we can generate the number of valid sequences of <math>A</math> by first generating all sequences of <math>b</math> such that <math>b_i\leq{}b_{i+1}</math> for all <math>1\leq{}i<n</math>, then choosing the elements from <math>A_{k+1}</math> that we keep in <math>A_k</math>, given the sequence of <math>b</math> as the restraint for the number of elements. For each sequence <math>b_1, b_2, \cdots{}, b_{n+1}</math>, there are <math>\prod_{i=2}^{n+1}\binom{b_i}{b_{i-1}}</math> corresponding sequences for <math>A</math>. Now, consider two cases - either all terms in <math>b</math> are either 0, 5, or 10, or there is at least one term in <math>b</math> that is neither 0, 5 nor 10. In the second case, consider the last term in <math>b</math> that is not 10 or 5, say <math>b_m</math>. However, that implies <math>b_{m+1}=10</math>, and so the number of corresponding sequences of <math>A</math> is <math>\binom{10}{b_m}\cdot{}</math> something or <math>\binom{5}{b_m}\cdot</math> something, which is always a multiple of <math>5</math>. Therefore, we only need to consider sequences of <math>b</math> where each term is <math>0</math> <math>5</math> or <math>10</math>. If all terms in <math>b</math> are 0 or 10, then for each <math>1\leq{}n\leq{}10</math> there are <math>n+1</math> sequences of <math>b</math> (since there are <math>n+1</math> places to turn from <math>0</math> to <math>10</math>), for a total of <math>2+3+\cdots+11=65\equiv0</math> (mod 5). If there exists at least one term <math>b_k=5</math>, then we use stars and bars to count the number of sequences of <math>b</math>, and each sequence of <math>b</math> corresponds to <math>\binom{10}{5}</math> sequences of <math>A</math>. For each <math>n</math>, we must have at least one term of <math>5</math>. After that, there are <math>n-1</math> stars and <math>2</math> bars (separating <math>0</math> to <math>5</math> and <math>5</math> to <math>10</math>), so that is <math>\binom{n+1}{2}</math> sequences of <math>b</math>. So the sum is <math>\binom{2}{2}+\binom{3}{2}+\cdots+\binom{11}{2}=\binom{12}{3}\equiv0</math> (mod 5). Therefore, the answer is 0 mod 5, and it must be <math>\boxed{\textbf{(C) } 5}</math> | |
− | + | ||
− | Let <math>b_n</math> be the number of elements of | + | ==Solution 4 == |
+ | |||
+ | 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> | ||
+ | |||
+ | Recalling or noticing that <math>x^n \equiv x^{n | ||
+ | \mod 4} \pmod {10}</math>, then, | ||
+ | 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) | ||
+ | |||
+ | Slightly easier, observe that <math>11^{10} \equiv 1^{10 }\pmod {10}</math>, so, working <math>{\mod 10}</math>, we have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | K & \equiv \sum_{m = 2}^{11} m^2 \\ | ||
+ | & \equiv \sum_{m = 1}^{10} m^2 \\ | ||
+ | & \equiv \frac{10 \cdot \left( 10 + 1 \right) \left( 2 \cdot 10 + 1 \right)}{6} \\ | ||
+ | & \equiv \frac{10 \cdot 11 \cdot 21}{6} \\ | ||
+ | & \equiv 5 \cdot 11 \cdot 7 \\ | ||
+ | & \equiv \boxed{ 5} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~oinava | ||
+ | |||
+ | ==Solution 5 (Constructive Counting)== | ||
+ | |||
+ | (See 2021 aime ii p6 for a better understanding. It covers similar material, albeit a bit more manipulation). | ||
+ | |||
+ | |||
+ | We try this for <math>n=2</math>. Clearly, this is just the number of subsets, giving <math>2^{10}</math> sequences. | ||
+ | |||
+ | |||
+ | For <math>n=3</math>, we have <math>3</math> choices for each element in the list: | ||
+ | |||
+ | <math>1</math>. The element is in both <math>A_1</math> and <math>A_2</math>. | ||
+ | |||
+ | <math>2</math>. The element is in only <math>A_1</math> and NOT in <math>A_2</math>. | ||
+ | |||
+ | <math>3</math>. The element is in neither <math>A_1</math> nor <math>A_2</math>. | ||
+ | |||
+ | Hence we have <math>3^{10}</math> sequences. | ||
+ | |||
+ | |||
+ | For <math>n=4</math>, we follow a similar logic; we have <math>4</math> choices for each element in the list: | ||
+ | |||
+ | <math>1</math>. The element is in all of <math>A_1</math>, <math>A_2</math>, and <math>A_3</math>. | ||
+ | |||
+ | <math>2</math>. The element is in only <math>A_1</math> and <math>A_2</math>. | ||
+ | |||
+ | <math>3</math>. The element is in only <math>A_1</math>. | ||
+ | |||
+ | <math>4</math>. The element is in none of the above. | ||
+ | |||
+ | Hence we have <math>4^{10}</math> sequences. | ||
+ | |||
+ | |||
+ | Continuing so, we just need to find the units digit of <math>2^{10} + 3^{10} + 4^{10} + ... + 11^{10}</math>. By using mod patterns, we easily arrive upon <math>\boxed{\textbf{(C) } 5}</math> as our answer. | ||
+ | |||
+ | |||
+ | ~xHypotenuse | ||
− | |||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
https://youtu.be/0LLQW0XCKsQ | https://youtu.be/0LLQW0XCKsQ | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/FpagLq1uzBI | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
{{AMC12 box|ab=A|year=2023|num-b=23|num-a=25}} | {{AMC12 box|ab=A|year=2023|num-b=23|num-a=25}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:55, 15 August 2024
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, but this time it actually works)
Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let be one sequence satisfying the constraints of the problem. Let
be the sequence of nonnegative integers such that
has
elements for all
, and
. Note that we can generate the number of valid sequences of
by first generating all sequences of
such that
for all
, then choosing the elements from
that we keep in
, given the sequence of
as the restraint for the number of elements. For each sequence
, there are
corresponding sequences for
. Now, consider two cases - either all terms in
are either 0, 5, or 10, or there is at least one term in
that is neither 0, 5 nor 10. In the second case, consider the last term in
that is not 10 or 5, say
. However, that implies
, and so the number of corresponding sequences of
is
something or
something, which is always a multiple of
. Therefore, we only need to consider sequences of
where each term is
or
. If all terms in
are 0 or 10, then for each
there are
sequences of
(since there are
places to turn from
to
), for a total of
(mod 5). If there exists at least one term
, then we use stars and bars to count the number of sequences of
, and each sequence of
corresponds to
sequences of
. For each
, we must have at least one term of
. After that, there are
stars and
bars (separating
to
and
to
), so that is
sequences of
. So the sum is
(mod 5). Therefore, the answer is 0 mod 5, and it must be
Solution 4
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
Recalling or noticing that , then,
Modulo 10, we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Slightly easier, observe that , so, working
, we have
~oinava
Solution 5 (Constructive Counting)
(See 2021 aime ii p6 for a better understanding. It covers similar material, albeit a bit more manipulation).
We try this for . Clearly, this is just the number of subsets, giving
sequences.
For , we have
choices for each element in the list:
. The element is in both
and
.
. The element is in only
and NOT in
.
. The element is in neither
nor
.
Hence we have sequences.
For , we follow a similar logic; we have
choices for each element in the list:
. The element is in all of
,
, and
.
. The element is in only
and
.
. The element is in only
.
. The element is in none of the above.
Hence we have sequences.
Continuing so, we just need to find the units digit of . By using mod patterns, we easily arrive upon
as our answer.
~xHypotenuse
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.