olymmath

RMO 2019 Problem 4

    Consider the following 3×23\times 2 array formed by using the numbers 1,2,3,4,5,61,2,3,4,5,6 , (a11a12a21a22a31a32)=(162534).\begin{pmatrix} a_{11}& a_{12}a_{21}& a_{22} a_{31}& a_{32}\end{pmatrix}=\begin{pmatrix}1& 62& 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×k3\times k array (aij)3×k(a_{ij})_{3\times k} for a suitable kk , adding more columns, using the numbers 7,8,9,…,3k7,8,9,\dots ,3k such that ∑j=1ka1j=∑j=1ka2j=∑j=1ka3j  and  ∑j=1k(a1j)2=∑j=1k(a2j)2=∑j=1k(a3j)2\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

    Dataset text AI-MO AoPS. Quoted with source.

    Official text: Indian Regional Mathematical Olympiad 2019 official paper, page 1.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator in-rmo/2019/p4 · Tag GNSTN7JTTQ · Edit this page on GitHub · Keys: j next, k previous