Consider the following $3\times 2$ array formed by using the numbers $1,2,3,4,5,6$, $$\begin{pmatrix} a_{11}& a_{12}\\a_{21}& a_{22}\\ a_{31}& a_{32}\end{pmatrix}=\begin{pmatrix}1& 6\\2& 5\\ 3& 4\end{pmatrix}.$$Observe that all row sums are equal, but the sum of the square of the squares is not the same for each row. Extend the above array to a $3\times k$ array $(a_{ij})_{3\times k}$ for a suitable $k$, adding more columns, using the numbers $7,8,9,\dots ,3k$ such that $$\sum_{j=1}^k a_{1j}=\sum_{j=1}^k a_{2j}=\sum_{j=1}^k a_{3j}~~\text{and}~~\sum_{j=1}^k (a_{1j})^2=\sum_{j=1}^k (a_{2j})^2=\sum_{j=1}^k (a_{3j})^2$$
Problem
Source: RMO 2019 P4
Tags: combinatorics, number theory, RMO
20.10.2019 14:27
Thank you so much for posting P4 @Purple_Planet I know it must have taken a lot of time of yours to do so
20.10.2019 14:28
Thanks Purple planet. Use \left( and \right) to make brackets
20.10.2019 14:31
Anyone knows the k for it?
20.10.2019 14:33
I somehow managed to prove that it is compatible for $k=6m$ where $m\in\mathbb{N}$.
20.10.2019 15:04
We just need to find one $k$, right? $$\left(\begin{array}{cccccc} 1& 6 & 9 & 10 & 14 & 17\\2& 5 & 7 & 12 & 15 & 16\\ 3& 4 & 8 & 11 & 13 & 18\end{array}\right).$$
20.10.2019 15:06
biomathematics wrote: We just need to find one $k$, right? $$\left(\begin{array}{cccccc} 1& 6 & 9 & 10 & 14 & 17\\2& 5 & 7 & 12 & 15 & 16\\ 3& 4 & 8 & 11 & 13 & 18\end{array}\right).$$ What? That's it? We needed to find just $k$? Oof! I found this config and left thinking it will be useless to extend
20.10.2019 17:29
Math-wiz wrote: biomathematics wrote: We just need to find one $k$, right? $$\left(\begin{array}{cccccc} 1& 6 & 9 & 10 & 14 & 17\\2& 5 & 7 & 12 & 15 & 16\\ 3& 4 & 8 & 11 & 13 & 18\end{array}\right).$$ What? That's it? We needed to find just $k$? Oof! I found this config and left thinking it will be useless to extend Ahh!! I guess we had to extend it for every suitable $k$...
20.10.2019 17:30
What score do i expect for proving $k=6t$?
20.10.2019 17:30
u might get 13-14 for it coz thats most of the people wrote. Sorry i wrote something reallly wrong earlier
20.10.2019 17:31
Can someone find a $3\times 12$ array. (We might find some pattern).
20.10.2019 17:32
REDACTED......
20.10.2019 17:39
Math-wiz wrote: What score do i expect for proving $k=6t$? Can k = 6t + 3? Sum of squares and sum must be divisible by 3. Sum = 3(6t +3)[3(6t+3)+1]/2 is divisible by 3, so is the sum of squares.
20.10.2019 17:44
Nope. Each one must be $\frac{(k)(3k+1)(6k+1)}{6}$. So $6|k$
20.10.2019 17:47
Purple_Planet wrote: Math-wiz wrote: biomathematics wrote: We just need to find one $k$, right? $$\left(\begin{array}{cccccc} 1& 6 & 9 & 10 & 14 & 17\\2& 5 & 7 & 12 & 15 & 16\\ 3& 4 & 8 & 11 & 13 & 18\end{array}\right).$$ What? That's it? We needed to find just $k$? Oof! I found this config and left thinking it will be useless to extend Ahh!! I guess we had to extend it for every suitable $k$... I mean, just do the same thing. Like, take this array and extend it by adding the same number to each entry.
20.10.2019 17:47
Math-wiz wrote: Nope. Each one must be $\frac{(k)(3k+1)(6k+1)}{6}$. So $6|k$ k = 3 (mod 6) => 3k + 1 = 10 = 4 (mod 6) So k(3k +1) = 3*4= 0 (mod 6). Sorry dont know Latex, = represents congruent to.
20.10.2019 17:49
biomathematics wrote: We just need to find one $k$, right? $$\left(\begin{array}{cccccc} 1& 6 & 9 & 10 & 14 & 17\\2& 5 & 7 & 12 & 15 & 16\\ 3& 4 & 8 & 11 & 13 & 18\end{array}\right).$$ Once you have this, add $18$ to everything and attach it to the right of it.
20.10.2019 17:49
amaanmathbuddy_2006 wrote: u might get 13-14 for it coz thats most of the people wrote. Sorry i wrote something reallly wrong earlier I don't think you'll get anything near that score for just proving $6 \mid k$.
20.10.2019 17:52
REDACTED....
20.10.2019 18:12
Claim: A construction for $k$ exists if and only if $k$ is of the form $3n$ with $n \geq 2$. Proof: We begin with the following constructions for $k=6, k=9$: $k=6$: We use the one which has been already found: Row $1$: $1,6,8,11,15,16$ Row $2$: $2,5,9,10,13,18$ Row $3$: $3,4,7,12,14,17$. Construction for $k=9$: Row $1$: $1,6,8,11,18,13,21,23,25$ Row $2$: $2,5,7,12,15,17,19,22,27$ Row $3$: $3,4,9,10,14,16,20,24,26$ From construction $1$, note that $(x+1)^2+(x+6)^2+(x+8)^2+(x+11)^2+(x+15)^2+(x+16)^2$ $ =(x+2)^2+(x+5)^2+(x+9)^2+(x+10)^2+(x+13)^2+(x+18)^2=$ $ (x+3)^2+(x+4)^2+(x+7)^2+(x+12)^2+(x+14)^2+(x+17)^2$ so if we can make a construction for $k$, we can also do so for $k+6$. This finishes the if part. Note that $1^2+2^2+...(3k)^2=\frac{(3k)(3k+1)(6k+1)}{6}$ should be a multiple of $3$ trivially if a construction exists, which gives $3|k$. It is trivial to notice that $k=3$ has no construction, hence proved.
20.10.2019 18:44
I am kind of confused by the phrasing of the question. Do they expect us to find all k for which you can construct the required array? Or do they just want us to provide a construction for some values/ singular value of k (for example, k = 6m )?
20.10.2019 18:56
I have just written for one k. How much should I expect? Coz that’s what the question said.
20.10.2019 18:59
GogetaBlue wrote: I am kind of confused by the phrasing of the question. Do they expect us to find all k for which you can construct the required array? Or do they just want us to provide a construction for some values/ singular value of k (for example, k = 6m )? Even I felt they meant to ask only one $k$ lmao
20.10.2019 19:06
Right. Most of us would have given construction for k = 6m, but if Mathotsav's construction for k = 9 is correct , then the answer would indeed be k = 6m + 3. I don't know if this is what they are expecting. But if they are expecting a construction for just one value of k, then the sum would literally be a one liner. Hopefully the benefit of doubt goes to us.
20.10.2019 19:56
Wow I trusted you all when you told me $6 \mid k$ smh
20.10.2019 20:27
Math-wiz wrote: Nope. Each one must be $\frac{(k)(3k+1)(6k+1)}{6}$. So $6|k$ This is wrong, wth? $k=6r+3$ works too
20.10.2019 20:50
I didn't realise Mathotsav uploaded it already, redacted.
21.10.2019 10:32
I'm really really happy this has not turned out to be a copied problem from some other MO.
21.10.2019 17:59
This was question 82 in USSR mathematics olympiad problems by dover publications
21.10.2019 18:08
but isnt problem 82 somewhat different or im seeing the wrong book
Attachments:
problem 82.docx (411kb)
21.10.2019 18:08
MathPassionForever wrote: Kiddos, in case you aren't convinced, see what Mathotsav has for you in store: $$\left(\begin{array}{ccccccccc} 1& 6 & 8 & 11 & 13 & 18 & 21 & 23 & 25\\2& 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27\\ 3& 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{array}\right).$$ How did he get this? Hexagrammum16 wrote: I'm really really happy this has not turned out to be a copied problem from some other MO. Gurunadham wrote: This was question 82 in USSR mathematics olympiad problems by dover publications Savage, but if you could share the source pdf / link
21.10.2019 18:28
Delta0001 wrote: MathPassionForever wrote: Kiddos, in case you aren't convinced, see what Mathotsav has for you in store: $$\left(\begin{array}{ccccccccc} 1& 6 & 8 & 11 & 13 & 18 & 21 & 23 & 25\\2& 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27\\ 3& 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{array}\right).$$ How did he get this? Hexagrammum16 wrote: I'm really really happy this has not turned out to be a copied problem from some other MO. Gurunadham wrote: This was question 82 in USSR mathematics olympiad problems by dover publications Savage, but if you could share the source pdf / link It's available on Library Genesis ... check it from there ...
22.10.2019 18:05
@above I have already downloaded the book and seen problem 82. However, as told in #32, the mentioned problem is not that. I would be delighted if you share the correct source. Which you probably don't have
25.10.2019 16:15
I was able to find my $3$x$6$ array using $\pmod{3}$ and $\pmod{9}$. It was a little lengthy though.
11.05.2020 23:20
Observe that: $1^2+6^2=25+12, 2^2+5^2=25+4, 3^2+4^2=25$. $12^2+7^2=181+12,11^2+8^2=181+4 ,10^2+9^2=181$. $13^2+18^2=481+12, 14^2+17^2=481+4, 15^2+16^2=481$. So, now let $k_1=25$, $k_2=181$ and $k_3=481$. All the row sums are equal, but the sum of the squares row-wise is not the same for each row. So, in order to meet the desired conditions we arrange(distribute) this $k_i + x$ terms such that the sum of the squares (row-wise) also becomes same. ($i \in {1,2,3}; x \in {0,4,12}$) \begin{tabular} {|c|c|c|c|} \hline Row & $(a_{11})^2+(a_{12})^2$ & $(a_{13})^2+(a_{14})^2$ & $(a_{15})^2+(a_{16})^2$ \\ \hline 1&$k_1+12$&$k_2$&$k_3+4$\\ \hline Row & $(a_{21})^2+(a_{22})^2$ & $(a_{23})^2+(a_{24})^2$ & $(a_{25})^2+(a_{26})^2$ \\ \hline 2&$k_1+4$&$k_2+12$&$k_3$\\ \hline Row & $(a_{31})^2+(a_{32})^2$ & $(a_{33})^2+(a_{34})^2$ & $(a_{35})^2+(a_{36})^2$ \\ \hline 3&$k_1$&$k_2+4$&$k_3+12$\\ \hline \end{tabular}Now arrange the terms corresponding to this distribution, so, that the sum of squares (row-wise) becomes same along with the row-wise sum (which is already same for each row). After that you will obtain the array as given below $\begin{pmatrix} 1 & 6 & 9 & 10 & 14 & 17\\ 2 & 5 & 7 & 12 & 15 & 16\\ 3 & 4 & 8 & 11 & 13 & 18 \end{pmatrix}$ The least value of $k$ is $3$.
21.10.2023 11:33
biomathematics wrote: GogetaBlue wrote: I am kind of confused by the phrasing of the question. Do they expect us to find all k for which you can construct the required array? Or do they just want us to provide a construction for some values/ singular value of k (for example, k = 6m )? Even I felt they meant to ask only one $k$ lmao As far as I remember, it was to find only one $k$. I found only one such $k$ and I assume I got full. I just checked the official solution file, it also just finds one such $k$.