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

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 \(100\)?

- $20$
- $91$
- $13$

Number theory

Perfect square

Integer

But try the problem first...

Answer:\(91\)

Source

Suggested Reading

PRMO-2019, Problem 17

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/sum-of-whole-numbers-amc-10a-2012-problem-8/
- https://www.youtube.com/watch?v=l0k9xtR7bvo

Contents

[hide]

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

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 \(100\)?

- $20$
- $91$
- $13$

Number theory

Perfect square

Integer

But try the problem first...

Answer:\(91\)

Source

Suggested Reading

PRMO-2019, Problem 17

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/sum-of-whole-numbers-amc-10a-2012-problem-8/
- https://www.youtube.com/watch?v=l0k9xtR7bvo

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