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Solving Equation | PRMO 2017 | Question 23

Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation.

Solving Equation - PRMO 2017, Question 23


Suppose an integer r, a natural number n and a prime number p satisfy the equation \(7x^{2}-44x+12=p^{n}\). Find the largest value of p.

  • is 107
  • is 47
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 47.

PRMO, 2017, Question 23

Higher Algebra by Hall and Knight

Try with Hints


First hint

\(7x^{2}-44x+12=p^{n}\)

or, \(7x^{2}-42x-2x+12=p^{n}\)

or, \((7x-2)(x-6)=p^{n}\)

Second Hint

or, \(7x-2=p^{\alpha}\), \(x-6=p^{\beta}\)

or, \((7x-2)-7(x-6)\)=\(p^{\alpha}-7p^{\beta}\)

\(40=p^{\alpha}-7p^{\beta}\)

Final Step

If \({\alpha}, {\beta}\) are natural numbers, p is a divisor of 40

or, p=2 or 5

If p=2, 40=\(2^{\alpha}-7(2^{\beta})\) or, \((2^{3})(5)\)=\(2^{\alpha}-7(2^{\beta})\)

or, \({\beta}\)=3 and \(2^{\alpha}\)=40+56

or, \({\alpha}\) not an integer hence not possible

If p=5 then 40=\(5^{\alpha}-7(5^{\beta})\)

or, \((2)^{3}(5)=5^{\alpha}-7(5^{\beta})\)

or, \({\beta}=1\) and \(5^{\alpha}\)=40+35

or, \({\alpha}\) not an integer hence not possible

so \({\beta}\)=0 or, \(p^{\alpha}\)=47

or, p=47 and \({\alpha}\)=1.

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Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation.

Solving Equation - PRMO 2017, Question 23


Suppose an integer r, a natural number n and a prime number p satisfy the equation \(7x^{2}-44x+12=p^{n}\). Find the largest value of p.

  • is 107
  • is 47
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 47.

PRMO, 2017, Question 23

Higher Algebra by Hall and Knight

Try with Hints


First hint

\(7x^{2}-44x+12=p^{n}\)

or, \(7x^{2}-42x-2x+12=p^{n}\)

or, \((7x-2)(x-6)=p^{n}\)

Second Hint

or, \(7x-2=p^{\alpha}\), \(x-6=p^{\beta}\)

or, \((7x-2)-7(x-6)\)=\(p^{\alpha}-7p^{\beta}\)

\(40=p^{\alpha}-7p^{\beta}\)

Final Step

If \({\alpha}, {\beta}\) are natural numbers, p is a divisor of 40

or, p=2 or 5

If p=2, 40=\(2^{\alpha}-7(2^{\beta})\) or, \((2^{3})(5)\)=\(2^{\alpha}-7(2^{\beta})\)

or, \({\beta}\)=3 and \(2^{\alpha}\)=40+56

or, \({\alpha}\) not an integer hence not possible

If p=5 then 40=\(5^{\alpha}-7(5^{\beta})\)

or, \((2)^{3}(5)=5^{\alpha}-7(5^{\beta})\)

or, \({\beta}=1\) and \(5^{\alpha}\)=40+35

or, \({\alpha}\) not an integer hence not possible

so \({\beta}\)=0 or, \(p^{\alpha}\)=47

or, p=47 and \({\alpha}\)=1.

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