Cheenta
How Cheenta works to ensure student success?
Explore the Back-Story

# 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

$$7x^{2}-44x+12=p^{n}$$

or, $$7x^{2}-42x-2x+12=p^{n}$$

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

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

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

$$40=p^{\alpha}-7p^{\beta}$$

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.

## Subscribe to Cheenta at Youtube

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

$$7x^{2}-44x+12=p^{n}$$

or, $$7x^{2}-42x-2x+12=p^{n}$$

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

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

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

$$40=p^{\alpha}-7p^{\beta}$$

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.

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.