AMC 8 India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Smallest positive value | Algebra | PRMO-2019 | Problem 13

Try this beautiful problem from Algebra based smallest positive value from PRMO 2019. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra PRMO 2019 based on smallest positive value

Smallest positive value| PRMO | Problem 13

Each of the numbers \(x_1, x_2,……….x_{101}\) is \(±1\). What is the smallest positive value of \(\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) ?

  • $24$
  • $10$
  • $34$

Key Concepts




Check the Answer


PRMO-2019, Problem 13

Pre College Mathematics

Try with Hints

\(S=\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) .

we have \((x_1+x_2+x_3+….+x_{101})^2={x_1}^2+{x_2}^2+…..+{x_{101}}^2+2S\)

\(\Rightarrow 2S\)=\((\displaystyle\sum_{i=1}^{101} {x_i})^2-\displaystyle\sum_{j=1}^{101} {x_j}^2\)

Can you now finish the problem ……….

Since we have \(x_i=\pm 1\) so \({x_i}^2=1\)

so \(2S=(\displaystyle\sum_{i=1}^{101} {x_i})^2-{101}\)

Since \(\displaystyle\sum_{i=1}^{101} {x_i}\) will be an integer

so \((\displaystyle\sum_{i=1}^{101} {x_i})^2\) will be a perfect square .

For smalll positive \(S\), \((\displaystyle\sum_{i=1}^{101} {x_i})^2\)must be smallest perfect square greater than \({101}\)

So \((\displaystyle\sum_{i=1}^{101} {x_i})^2={121}\)

\(\Rightarrow (\displaystyle\sum_{i=1}^{101} {x_i})\)=\({11}\) or \({-11}\)

Can you finish the problem……..

We can verify that the desired sum can be achieved by putting \(45\) \(x_i\)’s to be –1 and \(56\) \(x_i\)’s to be \(1\) So, \(2S = 121 – 101 = 20\)

\(\Rightarrow s=10\)

Subscribe to Cheenta at Youtube

Leave a Reply

Your email address will not be published. Required fields are marked *