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Algebra Arithmetic Math Olympiad PRMO

Real Numbers and Integers | PRMO 2017 | Question 2

Try this beautiful problem from the Pre-RMO, 2017 based on Real Numbers and Integers. You may use sequential hints to solve the problem.

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

Check the Answer


But try the problem first…

Answer: is 73.

Source
Suggested Reading

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.

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