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

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

Algebra

Integer

sum

Answer:$$10$$

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, $$2S = 121 – 101 = 20$$

$$\Rightarrow s=10$$

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