the product of 4 consecutive even number is always divisible by
well, the numbers are
2k 2k+2 2k+4 2k+6
multiply them and you have
2^4(k(k+1)(k+2)(k+3))
Now, two of those numbers must be even, so the product is divisible by 2^6
In addition, the product of 3 consecutive numbers is divisible by 6, so
2^7*3 = 384 divides
To determine if the product of 4 consecutive even numbers is always divisible by a certain number, we need to understand the properties of even numbers and divisibility.
An even number is any integer that is divisible by 2. This means that when an even number is divided by 2, the remainder is 0.
Now, let's consider four consecutive even numbers. We can represent them as:
n, n+2, n+4, n+6
where n is any even number. Notice that each consecutive even number has a difference of 2 from the previous one.
To find the product of these four consecutive even numbers, we multiply them together:
n * (n+2) * (n+4) * (n+6)
Since every term in this multiplication contains a factor of 2 (due to the even number property), we can factor out 2 from each term:
2 * (n/2) * (n/2+1) * (n/2+2) * (n/2+3)
Now, we can see that this product is divisible by 2 raised to the power of 4, which is 16. Therefore, the product of 4 consecutive even numbers is always divisible by 16.
In general, the product of 'n' consecutive even numbers will be divisible by 2 raised to the power of 'n', or 2^n.