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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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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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