 Try this beautiful problem from the PRMO, 2017 based on Real Numbers and Integers.

## Real Numbers and Integers – PRMO 2017

Suppose a, b are positive real numbers such that $a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183$, $a(b)^\frac{1}{2}+b(a)^\frac{1}{2}$=182, find $\frac{9(a+b)}{5}$.

• is 107
• is 73
• is 840
• cannot be determined from the given information

### Key Concepts

Real Numbers

Algebra

Integers

But try the problem first…

Source

PRMO, 2017, Question 2

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here $a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183$ and $a(b)^\frac{1}{2}+b(a)^\frac{1}{2}$=182

This equation gives integer solutions then a and b must be squares.

Let a=$P^{2}$and b=$Q^{2}$

$\Rightarrow a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=P^{3}+Q^{3}=183$ is first equation

Second Hint

and $a(b)^\frac{1}{2}+b(a)^\frac{1}{2}$=$P^{2}Q+Q^{2}P$

=$PQ(P+Q)$=182 is second equation

Final Step

now first equation +3 second equation gives
$P^{3}+Q^{3}+3PQ(P+Q)$=183+$3 \times 182$=729
$\Rightarrow (P+Q)^{3}=9^{3}$
$\Rightarrow(P+Q)=9$
second equation gives PQ(P+Q)=182

$\Rightarrow PQ=\frac{182}{9}$
then $a+b=P^{2}+Q^{2}=(P+Q)^{2}-2PQ$=$\frac{365}{9}$
$\frac{9(a+b)}{5}$=$\frac{9}{5} \times \frac{365}{9}$=73.