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If by adding a natural number P in the product of four consecutive even numbers, a perfect square number is obtained, what is the value of P ?
- a)8
- b)2
- c)4
- d)1
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If by adding a natural number P in the product of four consecutive eve...
x(x + 1) (x + 2) (x + 3) + P
= (x2 + 3x) (x2 + 3x + 2) + P (Putting y = x2 + 3x)
= y(y + 2) + P
= y2 + 2y + P
∴ y2 + 2y + P will be a perfect square if P = 1
Most Upvoted Answer
If by adding a natural number P in the product of four consecutive eve...
Explanation:
Given:
Let the four consecutive even numbers be 2n, 2n+2, 2n+4, and 2n+6.
The product of these numbers is (2n)(2n+2)(2n+4)(2n+6) = 16(n)(n+1)(n+2)(n+3).
Let P be the natural number added to the product.
Understanding:
For the product to become a perfect square, the product should be of the form k^2, where k is a natural number.
Therefore, P should be such that 16(n)(n+1)(n+2)(n+3) + P = k^2 for some natural number k.
Solution:
To solve this, let's substitute the values and simplify:
16(n)(n+1)(n+2)(n+3) + P = k^2
=> 16(n^2 + 3n)(n^2 + 3n + 2) + P = k^2
=> 16(n^2 + 3n)^2 + P = k^2
=> 16(n^4 + 6n^3 + 9n^2) + P = k^2
=> 16n^4 + 96n^3 + 144n^2 + P = k^2
Final Calculation:
From here, we can see that for P = 1, the product becomes a perfect square because:
16n^4 + 96n^3 + 144n^2 + 1 = (4n^2 + 6n + 1)^2
Therefore, the value of P is 1.
So, the correct answer is option 'D' (P = 1).
Given:
Let the four consecutive even numbers be 2n, 2n+2, 2n+4, and 2n+6.
The product of these numbers is (2n)(2n+2)(2n+4)(2n+6) = 16(n)(n+1)(n+2)(n+3).
Let P be the natural number added to the product.
Understanding:
For the product to become a perfect square, the product should be of the form k^2, where k is a natural number.
Therefore, P should be such that 16(n)(n+1)(n+2)(n+3) + P = k^2 for some natural number k.
Solution:
To solve this, let's substitute the values and simplify:
16(n)(n+1)(n+2)(n+3) + P = k^2
=> 16(n^2 + 3n)(n^2 + 3n + 2) + P = k^2
=> 16(n^2 + 3n)^2 + P = k^2
=> 16(n^4 + 6n^3 + 9n^2) + P = k^2
=> 16n^4 + 96n^3 + 144n^2 + P = k^2
Final Calculation:
From here, we can see that for P = 1, the product becomes a perfect square because:
16n^4 + 96n^3 + 144n^2 + 1 = (4n^2 + 6n + 1)^2
Therefore, the value of P is 1.
So, the correct answer is option 'D' (P = 1).
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If by adding a natural number P in the product of four consecutive even numbers, a perfect square number is obtained, what is the value of P ?a)8b)2c)4d)1Correct answer is option 'D'. Can you explain this answer?
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