Q. 43 A3.5( 4 Votes )

# Product of any four consecutive positive integers is divisible by ……A. 16B. 48C. 24D. 32

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

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

We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.

For, n = 3k

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

= 3k(3k + 1)(3k + 2)(3k + 3)

= 3 3k(3k + 1)(3k + 2)(k + 1)

Therefore it is divisible by 3.

For, n = 3k + 1

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

= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)

= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)

Therefore it is divisible by 3.

For, n = 3k + 2

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

= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)

= 3 (3k + 2)(k + 1)(3k + 4)(3k + 5)

Therefore it is divisible by 3.

n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2

For n = 4p

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

= 4p(4p + 1)(4p + 2)(4p + 3)

= 2 4p(4p + 1)(2p + 1)(4p + 3)

= 8p(4p + 1)(2p + 1)(4p + 3)

Therefore it is divisible by 8.

For n = 4p + 1

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

= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)

= 2 4(4p + 1)(2p + 1)(4p + 3)(p + 1)

= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)

Therefore it is divisible by 8.

For n = 4p + 2

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

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 2 4(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

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

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 2 4(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

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

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 2 4(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 3

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

= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)

= 2 4(4p + 3)(p + 1)(4p + 5)(2p + 3)

= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)

Therefore it is divisible by 8.

Since it is divisible by both 3 and 8 and both are mutually prime numbers .

Therefore, it will be divisible by 3 8 = 24

So, product of four consecutive positive integers is divisible by 24.

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