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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

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.