IMO ShortList 2001, combinatorics problem 5

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IMO ShortList 2001, combinatorics problem 5

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Source: IMO ShortList 2001, combinatorics problem 5

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orl

#1 h Sep 30, 2004, 5:29 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$ , $0 \leq j \leq n$ , $x_j$ equals the number of times $j$ appears in the sequence.

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orl

#2 h Sep 30, 2004, 5:30 PM • 2 Y

Y by Adventure10, Mango247

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

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#3 h Feb 8, 2009, 7:15 AM • 2 Y

Y by Adventure10, Mango247

Solution, from IMO Compendium

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riemannHypothesis2.71828

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riemannHypothesis2.71828

#4 h Jan 23, 2011, 9:18 PM • 1 Y

Y by Adventure10

there must be something wrong with your solution, i haven't read it completely but i have found a counter-example.

Consider this for n = 6

3,2,1,1,0,0,0.

This clearly works, and n is bigger than 3. So I think you MIGHT have a slight mistake somewhere.

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JBL

#5 h Jan 23, 2011, 10:44 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Probably you should reread the solution given -- $m$ and $n$ are different letters

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pi37

#6 h May 16, 2016, 4:35 AM • 4 Y

Y by nmd27082001, Wizard_32, DCMaths, Adventure10

Note that
$x_0+x_1+x_2+\cdots = 0\cdot x_0 + 1\cdot x_1 + 2\cdot x_2 + \cdots$ so
$x_0=x_2+2x_3+3x_4+\cdots$ Now we do casework on $x_0$ .

If $x_0=0$ , then there is a $0$ , so $x_0\ge 1$ , a contradiction.
If $x_0=1$ , then it's easy to see that the equation is satisfied only if $x_2=1$ and $x_3=x_4=\cdots = 0$ . There must be at least one $2$ , which means $x_1=2$ . There is one $0$ , so $n=3$ , and $(1,2,1,0)$ is a valid sequence.
If $x_0=2$ , then either $x_2=2$ and $x_3=x_4=\cdots = 0$ , or $x_3=1$ and $x_2=x_4=x_5=\cdots = 0$ . In the first case, since $x_3=x_4=\cdots = 0$ and there are two $2$ s already, $x_1\ge 2$ is impossible. $x_1=0$ and $x_1=1$ yield the valid sequences $(2,0,2,0)$ and $(2,1,2,0,0)$ . The second case is impossible
Finally, suppose $x_0=a$ with $a\ge 3$ . This implies $x_a\ge 1$ , and in fact $x_a=1$ since if $x_a \ge 2$ , we get a contradiction in the equation. This implies the equation can only hold if $x_2=1$ and $x_3=\cdots = x_{a-1}=a_{a+1}=\cdots = 0$ . Now since $x_2=1$ , we need a $2$ somewhere, which can only come from $x_1=2$ . Adjusting the length of the sequence with the correct number of $0$ s yields the sequence $(a,2,1,0,0,\cdots 0,1,0,0,0)$ with $a$ total $0$ s for each $a\ge 3$ .

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#7 h May 16, 2016, 4:39 AM • 3 Y

Y by Wizard_32, Aryan-23, Adventure10

Answer

Solution

First Simple Key Observation See that $\sum_{i = 0}^n x_i = n+1$ This is because every number in the sequence is from { $0,1,..,n$ } as each represent number of occurances of a particular $0<=j<=n$ in the sequence so each $x_i<=n+1$ and if one $x_i=n+1$ this can't hold(as there are only n remaining places to fill and $i$ can't be $n+1$ ). So every number in the sequence is either $0$ , $1$ , ..., or $n$ . And thus $\sum_{i = 0}^n x_i = n+1$ as $n+1$ is the total number of numbers in the sequence.
Second Simple Key Observation Let $x_0=k$ . Observe that $x_0>0$ as if $x_0=0$ implies $0$ is not present in the sequence and simultaneously it is present once (a contradiction).
Main pivotal reasoning Let $x_0=k$ [ $k>0$ ]. Now leaving $x_0$ and the $k$ other $x_i$ which have value $0$ there are $(n+1)-(k+1)=n-k$ other terms of the sequence whose sum is $n+1-k=(n-k)+1$ .
Each of these takes values $>=1$ (else that contradicts definition of $k$ ). After giving each of them a $1$ there is still a $1$ left which must go with one of them making that a $2$ .(argument similar to proving Pigeon-Hole-Principle if you notice

)
So exactly $n-k-1$ values among them are $1$ and only $1$ of them takes value $2$ .( $(n-k+1)=(n-k-1)*1+1*2$ .

"So we see that there are always
i> $n-k-1$ entries of value $1$
ii> $1$ entry of value $2$
iii> $1$ entry of value $k$
iv> $k$ entries of value $0$ ". (Statement I)

Next we consider 3 cases :
Case 1 : $k>=3$
Case 2 : $k=2$
Case 3 : $k=1$

Case 1 : $k>=3$ The whole sequence is filled with $0,1,2$ and $k$ . So for only these " $i$ "s $x_i$ is non-zero. We know there is exactly one $2$ , exactly one $k$ . So $x_2=1$ and $x_k=1$ and $x_0=k>2$ . We know there is a $2$ somewhere and thus $x_1$ must be $2$ . This in turn suggests $n-k-1=2$ or , $k=n-3$ . As we assumed $k>=3$ so here $n>=6$ . Thus for every $n>=6$ there is exactly one sequence $(n-3,2,1,0,...,1,0,0,0)$ . $\blacksquare$

Case 2 : $k=2$ As $k=2$ put that in the Statement I we get that there will be exactly $2$ " $2$ "s in the sequence, exactly $n-2-1=n-3$ " $1$ "s and $2$ zeroes. Again as the sequence is filled with only $0$ , $1$ and $2$ only $x_0,x_1$ and $x_2$ can be non-zero and $x_0=x_2=2$ . What remains is only $x_1$ . $x_1$ can't be $2$ (then $x_2=3$ which is not). So $x_1=0$ or $x_1=1$ .
If $x_1=0$ $n-3=0$ or, $n=3$ and the sequence is $(2,0,2,0)$ . $\blacksquare$
If $x_1=1$ $n=4$ and the sequence is $(2,1,2,0,0)$ . $\blacksquare$

Case 3 : $k=1$ Again put $k=1$ in Statement I and we get that there are exactly $1$ " $2$ ", exactly $1$ " $0$ " and exactly $n-1-1+1=n-1$ " $1$ "s in the sequence. As there is one $2$ somewhere in the sequence and except $x_0,x_1$ and $x_2$ all other $x_i=0$ and $x_0=x_2=1<2$ so $x_1$ must be that $2$ . This in turn gives $n=3$ and our solution is $(1,2,1,0)$ . $\blacksquare$ $\blacksquare$ $\blacksquare$

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yayups

#8 h Jan 18, 2020, 10:54 PM • 1 Y

Y by Adventure10

We claim the sequences that work are $(1,2,1,0)$ , $(2,0,2,0)$ , $(2,2,0,1,0)$ , and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for any $n\ge 6$ . It is easy to see that these work, so we now show that they are the only ones.

Note that $0\le x_k\le n+1$ as $x_k$ is the size of some subset of $\{0,1,\ldots,n\}$ . Furthermore, if $x_k=n+1$ , then all the $x_i$ s are $k$ , so the sequence is $(n+1,\ldots,n+1)$ , which doesn't work. Therefore, $0\le x_k\le n$ for all $k$ . We have the following key claim.

Claim: We have $n+1=x_0+\cdots+x_n=x_1+2x_2+\cdots+(n-1)x_{n-1}+nx_n.$
Proof: Note that $x_0+\cdots+x_n$ is the sum of the number of times $k$ appears in the sequence over all $k$ . Since each term of the sequence is between $0$ and $n$ , this means that $x_0+\cdots+x_n$ is counting the number of terms in the sequence, which is $n+1$ , so $x_0+\cdots+x_n=n+1.$ Furthermore, $x_0+\cdots+x_n$ is the sum of the terms of the sequence, which is $\sum_{k=0}^n k\cdot(\text{number of times }k\text{ appears})=\sum_{k=0}^n kx_k,$ which proves the claim. $\blacksquare$

In particular, we have $\boxed{x_0=x_2+2x_3+\cdots+(n-1)x_n}.$ We now split into cases based on the value of $x_0$ .

Case 1: Suppose $x_0=0$ . However this can't happen, as the sequence then has at least one $0$ , so $x_0\ge 1$ .

Case 2: Suppose $x_0=1$ . Thus, $1=x_2+2x_3+\cdots+(n-1)x_n,$ so $x_2=1$ and $x_3=\cdots=x_n=0$ . But we have exactly one $0$ , so we must have $n\le 3$ . Now, $x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(1+1)=n-1,$ so the sequence is $(1,n-1,1,0)$ or $(1,n-1,1)$ . The first gives to $(1,2,1,0)$ and the second gives $(1,1,1)$ . We see that $(1,1,1)$ doesn't work, so the only solution from this case is $\boxed{(1,2,1,0)}$ .

Case 3: Suppose $x_0=2$ . Thus, we have $2=x_2+2x_3+3x_4+\cdots+(n-1)x_n,$ so $x_4=\cdots=x_n=0$ . We have two subcases.

Case 3.1: Suppose $x_2=2$ and $x_3=0$ . Then, $x_3=\cdots=x_n=0$ , so $n\le 4$ as $x_0=2$ . Now, $x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+2)=n-3,$ so the sequence is $\boxed{(2,0,2,0)}$ or $\boxed{(2,1,2,0,0)}$ .
Case 3.2: Suppose $x_2=0$ and $x_3=1$ . Then, $x_2=x_4=\cdots=x_n=0$ , so $n\le 4$ as $x_0=2$ . Now, $x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+1)=n-2.$ Since $x_3=1$ , we have $x_1\ge 1$ , so $n=3,4$ . These give $(2,1,0,1)$ and $(2,2,0,1,0)$ . Neither of these work, so no solutions here.

Case 4: Suppose $x_0\ge 3$ . Let $a=x_0$ . Then, we have $x_a\ge 1$ , and we have $a=x_2+2x_3+\cdots+(n-1)x_n\ge (a-1)x_a.$ If $x_a\ge 2$ , then we have $a\ge 2(a-1)$ , or $a\le 2$ , which isn't the case. Therefore, $x_a=1$ , so $\sum_{\substack{k=2\\ k\ne a}}^n (k-1)x_k=1.$ This means $x_2=1$ and $x_k=0$ for $k\in\{3,\ldots,n\}\setminus\{a\}$ . Now, $a\ge 3$ , so the sequence has exactly two $1$ s except for possibly $x_1$ . This means that $x_1=2$ . But we have $x_0+\cdots+x_n=n+1,$ so $a+2+1+1=n+1,$ so $a=n-3$ . This gives the sequence $\boxed{(n-3,2,1,0,\ldots,0,1,0,0,0)}$ for any $n\ge 6$ (as $a=n-3\ge 3$ ).

This completes the proof.

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#9 h Apr 28, 2020, 1:52 PM • 2 Y

Y by Mango247, Mango247

We claim that the only sequences are $(1,2,1,0)$ , $(2,0,2,0)$ , $(2,1,2,0,0)$ , and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for all $n\ge 6$ .

First, note that $x_0+\ldots+x_n=n+1$ . As there are $x_0$ $0$ s, there are $n-x_0$ positive numbers with positive index. However, they sum to $n+1-x_0$ , so they must be comprised of all 1s and a single 2. This means that the only positive numbers in our sequence have indices a subset of $\{0,1,2,x_0\}$ . Now, we break into cases based on the value of $x_0$ .

If $x_0=1$ , note that we only have one $2$ , so $x_2=1$ . Now, we must have $x_1=2$ . No other numbers can be positive, so this only yields $(1,2,1,0)$ .

If $x_0=2$ , our sequence now has two $2$ s, so $x_2=2$ . Now, we can have $x_1=0,1$ , leading to the solutions $(2,0,2,0)$ and $(2,1,2,0,0)$ .

Finally, if $x_0\ge 3$ , we must have $x_2=1$ and $x_{x_0}=1$ , so $x_1=2$ . This yields the aforementioned infinite class of solutions.

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#10 h May 9, 2020, 9:04 AM

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Nice problem! Here's my solution: The only possible sequences are $(2,1,2,0,0),(1,2,1,0),(2,0,2,0) \text{ and } (n-3,2,1,\underbrace{0,0, \dots ,0}_{(n-6) \text{ times}},1,0,0,0) \quad \forall n>5$
Notice that the sequence consists of only $0,1, \dots ,n$ . Consider the sets $A$ and $B$ given by $A=\{i \text{ such that }\exists k \text{ with } x_k=i\} \text{ and } B=\{i \text{ such that } x_i \neq 0\}$ Note that the given condition implies that $A=B$ . Now, we have $x_0+x_1+ \dots +x_n=n+1$ as the LHS simply represents the total number of times each number $0,1, \dots ,n$ is present in the sequence. Also, $x_1+2x_2+\dots +nx_n$ is the sum of the elements of the sequence. Thus, we get $x_0+x_1+ \dots+x_n=x_1+2x_2+ \dots +nx_n=n+1$ $\Rightarrow x_0=x_2+2x_3+ \dots +(n-1)x_n$ If there exist $3$ numbers more than $1$ in set $B$ , then let $i>j>k \geq 2$ be the three maximal integers in $B$ . This gives $x_0 \geq (i-1)x_i+(j-1)x_j+(k-1)x_k \geq i+j+k-3 \geq i+3+2-3=i+2$ But then, since $x_0$ lies in set $A$ , so $x_0 \in B$ as well, which contradicts the fact that $i$ is the maximal element in $B$ . So there are at most $2$ elements in $B$ which are more than $1$ . If there are exactly $2$ such elements, then let $i>j \geq 2$ be these numbers. Then $x_0=(i-1)x_i+(j-1)x_j \geq i+j-2 \geq i$ As shown above, $x_0 \in B$ , which means that we must have $x_0 \leq i$ as well. Thus, equality must hold everywhere, which gives $x_0=i,$ $x_i=x_j=1$ and $j=2$ . Then there are $2$ ones in the sequence, and so $x_1=2$ . Since $x_0+x_1+x_i+x_j=n+1$ , so this gives $x_0=n-3$ , and hence $i=n-3$ . As $i>2$ , so we must also have $n-3>2$ , i.e. $n>5$ . This gives the infinite class of sequences mentioned in the beginning.

Now $B$ must have atleast one element more than $1$ , since otherwise $x_0=0$ (which is not possible as this means that $0 \in A$ but $0 \not \in B$ ). So suppose $B$ has exactly $1$ element more than $1$ , and let this element be $i$ . Then $x_0=(i-1)x_i$ . If $i>3$ , then $x_0 \in B$ gives that some multiple of $i-1$ also lies in $B$ , which is contrary to the fact that $i$ is the only element more than $1$ in set $B$ . So we have $i=2$ , and $x_0=x_2$ . Then we get $2x_0+x_1=n+1$ . But $x_3,x_4, \dots ,x_n$ are all zeros, and so $x_0 \geq n-2$ . Also, $x_2 \leq 2$ since it represents the number of times $2$ appears in the sequence, and we cannot have $x_0=x_1=x_2=2$ . Then, as $x_0=x_2$ , so we get $2 \leq x_0 \leq n-2$ , which gives $n \leq 4$ . A simple case bash then yield the remaining answers (in particular, remember that $1 \leq x_0 \leq 2$ ). $\blacksquare$

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awesomeming327.

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awesomeming327.

#11 h Aug 11, 2022, 3:33 AM • 1 Y

Y by Mango247

Let $S$ be the sum $x_0+x_1+\dots+x_n.$ While $S$ is that, $S$ is also zero times the number of times zero appears in the sequence, plus one times the number of times one appears in the sequence, and so on adding to $x_1+2x_2+3x_3+\dots+9x_9.$ Where $x_i=0$ if $x_i$ does not ever happen in the sequence.

$~$
Note that $x_{10}=x_{11}=\dots=0$ because numbers $\ge 10$ aren't digits. We have $9\ge x_0=x_2+2x_3+\dots+8x_9.$ When $x_0=0$ this is impossible. When $x_0=1$ , then $x_2=1$ and that's it for index greater than one. Thus, $(1,2,1,0)$

$~$
Note that $x_0=2$ implies that $(x_2,x_3)=(2,0)$ or $(0,1).$ This gives $(2,0,2,0)$ or $(2,1,2,0,0)$ and none for $(0,1)$ because there's a two.

$~$
For $x_0\ge 3$ , note that $(x_0-1)x_{x_0}\ge x_0-1$ so there is one other two and that's it. This gives $(x_0,2,1,0,\dots,0,1,0,0,0).$

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#12 h Dec 21, 2022, 2:54 PM

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The only solutions are $\boxed{(1,2,1,0), (2,0,2,0), (2,1,2,0,0), (x-3,2,1,0,0,\ldots, 0,1,0,0,0)}$ for any $x\ge 6$ (where the entries in $\ldots$ are zero). We can check that these all work.

Notice that $x_0 + x_1 + \cdots + x_n = 0\cdot x_0 + 1\cdot x_1 + \cdots + n \cdot x_n,$ so
$x_0 = x_2 + 2x_3 + 3x_4 + \cdots + (n-1)x_n$
Clearly $x_0>0$ .

Case 1: $x_0 = 1$ .
Then we must obviously have $x_2 = 1$ and $x_3 = x_4 \cdots = x_n = 0$ . Since there is only one zero in the sequence, we have $n=3$ . Since $x_0 = x_2 = 1$ , we have $x_1 = 2$ . This gives $(1,2,1,0)$ .

Case 2: $x_0 = 2$ .
Then we must have $x_4 = x_5 = \cdots = x_n = 0$ and either $x_3 = 1, x_2 = 0$ or $x_2 = 2, x_3 = 0$ .

If $x_3 = 1, x_2 = 0$ , then we must have $x_i = 3$ for some $i$ , which means $x_1 = 3$ , but this is obviously impossible.

Thus, $x_2 = 2, x_3 = x_4 = \cdots = x_n = 0$ . Notice that $x_i\ne 1\forall i\ne 1$ , so $x_1\in \{0,1\}$ . We must have exactly two zeroes, so $x_1 = 0$ implies $n=3$ and $x_1 = 1$ implies $n=4$ . This gives $(2,0,2,0)$ and $(2,1,2,0,0)$ .

Case 3: $x_0 > 2$ .
Let $x_0 = k$ . Since $k\le n$ , we have $(k-1)x_k \le x_0 = k$ , so $x_k$ is at most $1$ . Since $x_0 = k$ , we have $x_k\ge 1$ , so $x_k = 1$ . Now, $(k-1)x_k = k-1$ , so $\sum_{i=2, i\ne k}^n (i-1)x_i = 1,$ so we must have $x_2 = 1$ and $x_3 = \cdots = x_{k-1} = x_{k+1} = \cdots = x_n = 0$ . Then $x_1\ne 0$ , so there are exactly $n-3$ zeroes in the sequence, which means $x_0 = n-3$ and $n>5$ . Now $x_2 = x_k = 1$ , so $x_1 = 2$ must hold. This gives $(n-3,2,1,0,0,\ldots, 0, 1,0,0,0)$ , as desired.

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#13 h Apr 26, 2025, 11:45 PM

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First note that they should all be nonnegative integers. Observe that $0x_0+1x_1+2x_2+\cdots + nx_n = x_0+x_1+x_2 + \cdots + x_n = n+1.$ The first equation holds because the LHS counts the sum of the numbers in the sequence, while the second equation holds because the LHS counts the number of numbers in the sequence.

Let $x_0=k.$ Clearly $k > 0.$ Then $x_1+x_2+\cdots + x_n = (n-k)+1,$ but the LHS has $n-k$ positive terms. Thus $n-k-1$ of them are $1$ s, there is a $2,$ and the rest are $0$ s. Therefore, at most $x_1, x_2, x_0, x_k$ can be positive, so clearly $n-k+1 \leq 4 \implies n-k \leq 3.$ Meanwhile since we have at least $(n-k-1)$ $1$ s it follows that $1 \leq n-k.$

If $n-k=1,$ we have $x_0=n-1,$ no $1$ s, and $1$ $2$ , with everything else being $0.$ Thus $x_2, x_{n-1}$ are both positive, so they must equal. This yields $(2, 0, 2, 0).$

If $n-k=2,$ we have $x_0=n-2,$ so there is one $1, 2$ in the sequence while the rest is zero. This makes both $x_1, x_2$ positive and casework on which one is $2$ yields $(1, 2, 1, 0), (2, 1, 2, 0, 0).$

If $n-k=3,$ we have $x_0=k=n-3,$ so there are $(n-3)$ $0$ s, $(2)$ $1$ s, $(1)$ $2,$ and $(1)$ $n-3.$ Thus $x_{n-3}, x_1, x_2$ are all positive. Casework on the size of $n$ yields $(k, 2, 1, \{0, \cdots, 0 \}, 1, 0, 0, 0)$ where the set in the middle has $k-3$ $0$ s. Also note that $k \geq 3.$

Thus the solutions are $(2, 0, 2, 0), (1, 2, 1, 0), (2, 1, 2, 0,0), (k, 2, 1, \{0, \cdots, 0 \}, 1, 0, 0, 0)$ where $k \geq 3.$

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