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May 14, 2020

Largest Possible Value | PRMO-2019 | Problem 17

Try this beautiful problem from PRMO, 2019 based on Largest Possible Value.

Largest Possible Value | PRMO | Problem-17


Let a, b, c be distinct positive integers such that \(b + c – a\),\( c + a – b\) and \(a + b – c\) are all perfect squares.
What is the largest possible value of \(a + b + c\) smaller than \(100\)?

  • $20$
  • $91$
  • $13$

Key Concepts


Number theory

Perfect square

Integer

Check the Answer


Answer:\(91\)

PRMO-2019, Problem 17

Pre College Mathematics

Try with Hints


Let \(b + c – a = x^2\) … (i)
\(c + a – b = y^2\) … (ii)
\(a + b – c = z^2\) … (iii)

Now since \(a\),\( b\), \(c\) are distinct positive integers,
Therefore, \(x\), \(y\), \(z\) will also be positive integers,
add (i), (ii) and (iii)
\(a + b + c = x^2 + y^2 + z^2\)
Now, we need to find largest value of \(a + b + c or x^2 + y^2 + z^2\) less than \(100\)
Now, to get a, b, c all integers \(x\),\( y\), \(z\) all must be of same parity, i.e. either all three are even or all three
are odd.

Can you now finish the problem ..........

Let us maximize\(x^2 + y^2 + z^2\), for both cases.
If \(x\), \(y\), \(z \)are all even.
Therefore,

\(b + c – a = 8^2 = 64\)
\(c + a – b = 42 = 16\)
\(a + b – c = 22 = 4\)
Which on solving, give\( a = 10\),\( b = 34\), \(c = 40\) and \(a + b + c = 84\)
If x, y, z are all odd
\(\Rightarrow b + c – a = 92 = 81\)
\(c + a – b = 32 = 9\)
\(a + b – c = 12 = 1\)
Which on solving, give \(a = 5\) ,\(b = 41\), \(c = 45\) and\( a + b + c = 91\)

Can you finish the problem........

Therefore Maximum value of \(a + b + c < 100 = 91\)

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