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 \(\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) ?

  • $24$
  • $10$
  • $34$

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PRMO-2019, Problem 13

Pre College Mathematics

Try with Hints

First hint

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

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 ……….

Second Hint

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……..

Final Step

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, \(2S = 121 – 101 = 20\)

\(\Rightarrow s=10\)

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