Question

Prove that the product of any 5 consecutive integers is divisible by 120

Answer 1

Answer:

In any sequence of p numbers, exactly one of those values will be divisible by p.  

(In other words, one out of every p values is divisible by p).

Therefore:

At least 1 value is divisible by 5 ( since we hava a sequence of 5 numbers)  

At least one value is divisible by 3 (since you have a sequence of >= 3 values)

At least one value is divisible by 4 (since you have a sequence of 4 values)

Two values are divisible by 2 (since you have two sequences of 2 values). One of those values is divisible by 4, but you still have one other 2 factor.

Thus, the product of any sequence of 5 values is divisible by 120= 5*4*3*2.

Step-by-step explanation:

Answer 2

The product of any 5 consecutive integers is divisible by 120 is proved.

Given:

the product of any 5 consecutive integers is divisible by 120

To find:

To Prove that the product of any 5 consecutive integers is divisible by 120

Solution:

Let us take five consecutive integers such as $1,2, 3, 4, 5,$ .

Multiply these integers, we get

$1$ × $2$ ×$3$ × $4$ × $5 = 120$

Therefore it is divisible by 120.

Hence proved.

But it is not applicable to any other set of integers.

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