# Understand the problem

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 squares is not the same for each row .Extend the above arrar to a \( 3 \times k \) array \( (a_{ij}){3 \times k} \) for a suitable k , adding more columns , using the numbers \( 7 , 8 , 9 , ….,3k \) such that \( \displaystyle \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} \ and \ \sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 \)

##### Source of the problem

Regional Math Olympiad, 2019 Problem 4

##### Topic

Inequality

##### Difficulty Level

8/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics

# Start with hints

Do you really need a hint? Try it first!

1+2+3+ … 3k = \( \frac{3k(3k+1)}{2}\) So, \( \sum_{j=1}^{k} a_{1j} = \sum_{j=1}^{k} a_{2j}= \sum_{j=1}^{k}a_{3j} = \frac{k(3k+1)}{2}\) \( 1^2 + 2^2 +… (3k)^2 = \frac{k(3k+1)(6k+1)}{2}\) So, \(\sum_{j=1}^{k} (a_{1j})^2= \sum_{j=1}^{k} (a_{2j})^2 = \sum_{j=1}^{k} (a_{3j})^2 = \frac{k(3k+1)(6k+1)}{6}\). This means that 3 | k.

Step 1: Try out with k = 3. Prove that it is not possible to arrange them in the desired order as already some numbers are fixed. We will now try for k = 6. Claim: If 3|k and k > 3 then it is always possible. (This is our conjecture too)

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5) \(n^2 + (n + 5)^2 – (n + 1)^2 – (n + 4)^2 = 8\) \( (m + 1)^2 + (m + 4)^2 – (m + 2)^2 – (m + 3)^2 = 4\) \( (l + 2)^2 + (l + 3)^2 – (l)^2 – (l+ 5)^2 = – 12 \) So, we get \( n^2 + (n + 5)^2 + (m + 1)^2 + (m + 4)^2 + (l + 2)^2 + (l + 3)^2 = (n + 1)^2 + (n + 4)^2 + (m + 2)^2 +(m + 3)^2 + (l)^2 + (l+ 5)^2 \). Also,

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5) Hence putting suitable values of l, m, and n, we get an array like the one below: \( \begin{pmatrix} 1 & 6 & 8 & 11 & 15 & 16 \\ 2 & 5 & 9 & 10 & 13 & 18 \\ 3 & 4 & 7 & 12 & 14 & 17\end{pmatrix} \) \( (n + 1)^2 + (n + 6)^2 + (n + 8)^2 + (n + 11)^2 + (n + 15)^2 + (n + 16)^2

= (n + 2)^2 + (n + 5)^2+ (n + 9)^2 + (n + 10)^2+ (n + 13)^2+ (n + 18)^2

= (n + 3)^2 + (n + 4)^2 + (n + 7)^2+ (n + 12)^2 + (n+14)^2 + (n+17)^2 \). Using this and the above two matrices, you can prove by induction that the claim holds!

n + (n + 5) + (m + 1) + (m + 4) + (l + 2) + (l + 3) = 2n + 2m + 2l + 15 = (n + 1) + (n + 4) + (m + 2) + (m + 3) + (l) + (l+ 5) Hence putting suitable values of l, m, and n, we get an array like the one below: \( \begin{pmatrix} 1 & 6 & 8 & 11 & 15 & 16 \\ 2 & 5 & 9 & 10 & 13 & 18 \\ 3 & 4 & 7 & 12 & 14 & 17\end{pmatrix} \) \( (n + 1)^2 + (n + 6)^2 + (n + 8)^2 + (n + 11)^2 + (n + 15)^2 + (n + 16)^2

= (n + 2)^2 + (n + 5)^2+ (n + 9)^2 + (n + 10)^2+ (n + 13)^2+ (n + 18)^2

= (n + 3)^2 + (n + 4)^2 + (n + 7)^2+ (n + 12)^2 + (n+14)^2 + (n+17)^2 \). Using this and the above two matrices, you can prove by induction that the claim holds!

Using the array found similarly : \( \begin{pmatrix} 1 & 6 & 8 & 11 & 18 & 13 & 21 & 23 & 25 \\ 2 & 5 & 7 & 12 & 15 & 17 & 19 & 22 & 27 \\ 3 & 4 & 9 & 10 & 14 & 16 & 20 & 24 & 26\end{pmatrix} \)

# Watch video

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.