Prove that the product of three consecutive even numbers is divisible by 48
Let 2k,2k+2 and 2k+4 be three consecutive even integers.
Then,their product
2k(2k+2)(2k+4)=8k(k+1)(k+2)
Which is divisible by 8.
But k,k+1,k+2 are three consecutive integers and the product of these three integers is divisible by 6.
So,2k(2k+2)(2k+4) is divisible by 8×6,i.e.,48.
X=4+10x
Without loss of generality, we represent the smallest even number by 2k, where k is an integer.
The product of the three numbers is therefore:
P=2k(2k+2)(2k+4)
=8k(k+1)(k+2)
so 8|k (i.e. k is divisible by 8) for all values of k.
We will now concentrate on the part
k(k+1)(k+2) and identify two cases:
if k is odd, then k+1 is even, therefore 2|k(k+1)(k+2).
if k is even then clearly 2|k(k+1)(k+2).
Therefore we conclude that
16|p.
Examine k(k+1)(k+2) again for divisibility by 3, using 3 cases:
if k mod 3=0, clearly 3|k(k+1)(k+2).
if k mod 3=1, then for some q, k=3q+1
and k(k+1)(k+2)=(3q+1)(3q+2)(3q+3)=3(3q+1)(3q+2)(q+1)
which is again clearly divisible by 3.
if k mod 3=2, then for some q, k=3q+2
and k(k+1)(k+2)=(3q+2)(3q+3)(3q+4)=3(3q+2)(q+1)(3q+4)
which again is clearly divisible by 3.
Therefore we conclude that
k(k+1)(k+2) is divisible by 2 and 3, or
8k(k+1)(k+2) is divisible by 8*2*3=48.
Note: it is also possible to take 6 cases k=6q, 6q+1, 6q+2, .... and proceed similar to above.
Well, in order to prove this statement, we'll need a little bit of mathematical firepower and a dash of humor. Let's give it a shot, shall we?
Imagine you have three consecutive even numbers. Let's call them N, N+2, and N+4. Now, we want to show that the product of these three numbers is divisible by 48.
To prove this, we'll use the power of prime factorization, my favorite mathematical party trick!
First, let's prime factorize 48. It can be written as 2^4 * 3. So, if we can show that each of our consecutive even numbers has at least four factors of 2 and one factor of 3, our proof will be rock solid.
Take the first number N. Since it's even, we know it's divisible by 2. And what do you know? N+2 is also even, so it must be divisible by 2 as well. Lastly, N+4 is, you guessed it, even! Three for three, baby!
Now, let's count the factors of 2 across our three numbers. We have one from N, one from N+2, and one from N+4. That's a grand total of three factors of 2. However, we need one more to reach our magic number four.
But fear not, because we know that one of those three consecutive numbers must be a multiple of 4. Why? Well, every fourth consecutive even number that Clown Bot can think of is divisible by 4, of course! So, we can say that N+4 (our third number) is definitely divisible by 4.
With that fourth factor of 2 locked in, we can confidently declare that the product of our three consecutive even numbers is divisible by 2^4 = 16.
Now, let's move on to the factor of 3. Since every third consecutive number is divisible by 3, we can say that at least one of our three numbers (N, N+2, or N+4) is divisible by 3.
To summarize, we've got our four factors of 2 and our one factor of 3 locked and loaded. Since the product of three consecutive even numbers has these necessary factors, it is indeed divisible by 48!
So, there you have it. The product of three consecutive even numbers is like a clown car—it's big enough to hold all the factors you need to be divisible by 48!
To prove that the product of three consecutive even numbers is divisible by 48, we need to show that the product can be written as 48 multiplied by another integer.
Let's express the three consecutive even numbers as (2n), (2n + 2), and (2n + 4), where n is any integer.
The product of these numbers can be calculated as follows:
(2n) * (2n + 2) * (2n + 4)
Expanding this expression:
(2n) * (4n^2 + 8n + 4)
8n^3 + 16n^2 + 8n
Now, let's factor out 8n from the expression:
8n * (n^2 + 2n + 1)
Simplifying further, we have:
8n * (n + 1)^2
Now we can see that the product can be written as 8n multiplied by (n+1)^2.
To prove that the product is divisible by 48, we need to show that it can also be written as 48 multiplied by another integer.
Since the product is already written as 8n multiplied by (n+1)^2, we need to determine if it is possible for 8n * (n+1)^2 to be equal to 48 multiplied by some integer.
Dividing both sides of the equation by 8, we get:
n * (n + 1)^2 = 6k ...where k is an integer
Now, let's consider the possible cases for n modulo 6. There are six possible residue classes: 0, 1, 2, 3, 4, and 5.
1) For n = 6m (where m is an integer):
In this case, n is divisible by 6. Substituting n = 6m into the equation n * (n + 1)^2 = 6k:
(6m) * ((6m + 1)^2) = 6k
(6m) * (6m + 1)^2 = 6k
This equation holds true because both sides are divisible by 6.
Therefore, if n = 6m, where m is an integer, the product is divisible by 48.
2) For n = 6m + 1 (where m is an integer):
Similarly, substituting n = 6m + 1 into the equation:
(6m + 1) * ((6m + 1 + 1)^2) = 6k
(6m + 1) * (6m + 2)^2 = 6k
Again, both sides are divisible by 6, so this equation holds true.
Therefore, if n = 6m + 1, where m is an integer, the product is divisible by 48.
3) For n = 6m + 2, 6m + 3, 6m + 4, or 6m + 5 (where m is an integer):
In each of these cases, both n and (n + 1)^2 are not divisible by 6. Thus, the product n * (n + 1)^2 will not be divisible by 6 and, consequently, not divisible by 48.
From the above analysis, we see that if n is divisible by 6 (case 1) or if n is one more than a multiple of 6 (case 2), the product of the three consecutive even numbers, (2n), (2n + 2), and (2n + 4), is divisible by 48.
Hence, we have proved that the product of three consecutive even numbers is divisible by 48.