Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

## Value of Sum – PRMO 2018, Question 16

What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

- $50$
- $53$
- $55$
- $59$
- $65$

**Key Concepts**

Odd-Even

Sum

integer

## Check the Answer

But try the problem first…

Answer:$55$

PRMO-2018, Problem 16

Pre College Mathematics

## Try with Hints

First hint

We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……

Can you now finish the problem ……….

Second Hint

$i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$

$=1-4+10=7$

$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10)$

$=-4$

$i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$

$=3-3+10=10$

$i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$

$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10$

$=13$

$i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$

$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16$

$i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$

$i=9 \Rightarrow$$ 1 \times 9+(10)=19$

Can you finish the problem……..

Final Step

Therefore $ S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55 $

## Other useful links

- https://www.cheenta.com/ordered-pairs-prmo-2019-problem-18/
- https://www.youtube.com/watch?v=h_x9kS-J1XY