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# Real Numbers and Integers | PRMO 2017 | Question 2

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

PRMO, 2017, Question 2

Elementary Algebra by Hall and Knight

## Try with Hints

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

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

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

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.

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