Question

answer 13number questions fastly.

Answer 1

Final Answer : d) 1


Steps: 1)
Let the four consecutive natural numbers be
(n), (n+1), (n+2),(n+3) where n is natural number.

Their Product =
$n(n + 1)(n + 2)(n + 3)$

2) Their product is increased by (p) I. e natural number.
Then,
S= n(n+1)(n+2)(n+3) + p
=>n(n+3)(n+1)(n+2)
$= > ( {n}^{2} + 3n)( {n}^{2} + 3n + 2) + p \\ \: \: \: let \: t = ( {n}^{2} + 3n) \\ = > t(t + 2) + p \\ = > {t}^{2} + 2t + p$

3)
S is perfect square when p = 1 ,
i .e.
S = (p+1)^2

Therefore,
The Product of four consecutive natural numbers increased by 1 is a perfect square.

Answer 2

Let the numbers be,
n, n+1, n+2 and n+3.

》Now, we need the value of p which will make {n(n+1)(n+2)(n+3)+p} a perfect square.

Now, let product be:
n(n+1)(n+2)(n+3)
So,
=> n(n+3)(n+1)(n+2)
=> (n^2 + 3n) × (n^2 + 3n + 2)

Let, a = n^2 + 3n
So,
=> a × (a +2)
=> a^2 + 2a

•Now we know that since 'n' is a natural number so, n^2+3n will be a natural number too and therefore 'a'will be a natural number too.

》Hence, the product now is in the form
( a^2 + 2a) or 1,
which is a perfect square.

☆ Now,

If p=1,

then the sum of the product and p becomes, a^2 + 2a + 1
=> (a+1)^2.

》Therefore, the value of p = 1.

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