Show the proof involving multiples of numbers

[chwala]

Homework Statement

Prove that the sum of four consecutive multiples of $4$ is always a multiple of $8$

Relevant Equations

algebra

Let the first multiple of $4=x$, then it follows that;
$x+(x+4)+(x+8)+(x+12)=4x+24=4(x+6)$ ...where $4$ is a multiple of $8$

 

[PeroK]

Eh?

 

[chwala]

Homework Statement:: Prove that the sum of four consecutive multiples of $4$ is always a multiple of $8$
Relevant Equations:: algebra

Let the first multiple of $4=x$, then it follows that;
$x+(x+4)+(x+8)+(x+12)=4x+24=4(x+6)$ ...where $4$ is a multiple of $8$

Let me do some research on this ...

 

[Delta2]

Hi @chwala! .

4 is not a multiple 8 (instead 8 is a multiple 4)

Your proof needs 1 more final little step. you got to factor an additional factor of 2 out of $4(x+6)$ so u have $4\cdot 2 (...+...)=8(...+...)$ so it is obvious it is a multiple of 8. Got any ideas?

 

[PeroK]

The original problem is equivalent to the sum of any four consecutive integers being even.

 

[chwala]

Thanks I will look at this later...

 

[Mark44]

Homework Statement:: Prove that the sum of four consecutive multiples of $4$ is always a multiple of $8$
Relevant Equations:: algebra

Let the first multiple of $4=x$, then it follows that;
$x+(x+4)+(x+8)+(x+12)=4x+24=4(x+6)$ ...where $4$ is a multiple of $8$

Your interpretation of the problem statement is incorrect. Four consecutive multiples of 4 would be $k \cdot 4, (k+1) \cdot 4, (k+2) \cdot 4, (k+3) \cdot 4$.

Also, if there aren't any relevant equations, just leave that section empty rather than putting in something that is meaningless. "Algebra" is not relevant, let alone being an equation.

 

[Delta2]

Your interpretation of the problem statement is incorrect. Four consecutive multiples of 4 would be k⋅4,(k+1)⋅4,(k+2)⋅4,(k+3)⋅4.

No I don't think its incorrect, it is the same as yours with $x=4k$. He just names it x instead of 4k.

 

[Mark44]

No I don't think its incorrect, it is the same as yours with $x=4k$. He just names it x instead of 4k.

Fair enough -- I didn't read the post that carefully.

 

[chwala]

Hi @chwala! .

4 is not a multiple 8 (instead 8 is a multiple 4)

Your proof needs 1 more final little step. you got to factor an additional factor of 2 out of $4(x+6)$ so u have $4\cdot 2 (...+...)=8(...+...)$ so it is obvious it is a multiple of 8. Got any ideas?

Would that be;

$4\cdot 2(\dfrac{x}{2}+3)=8(\dfrac{x}{2}+3)?$ Where $x$ will always divide $2$...or $2$ will always be divisible by $x.$

 

[PeroK]

Would that be;

$4\cdot 2(\dfrac{x}{2}+3)=8(\dfrac{x}{2}+3)?$ Where $x$ will always divide $2$...or $2$ will always be divisible by $x.$

I've no idea why you are using $x$ for a multiple of four, and then never use this fact explicitly.

 

[Delta2]

Would that be;

$4\cdot 2(\dfrac{x}{2}+3)=8(\dfrac{x}{2}+3)?$ Where $x$ will always divide $2$...or $2$ will always be divisible by $x.$

yes, but you want to say that 2 divides x (because x is a multiple of 4 and 2 divides 4).

 

[lurflurf]

I've no idea why you are using $x$ for a multiple of four, and then never use this fact explicitly.

Well x is a multiple of 2 was used
x is a multiple of 4 is not needed

 

[PeroK]

Well x is a multiple of 2 was used
x is a multiple of 4 is not needed

How can you "like" a post that says that 2 will always be divisible by a multiple of 4?

$4\cdot 2(\dfrac{x}{2}+3)=8(\dfrac{x}{2}+3)?$ Where $x$ will always divide $2$...or $2$ will always be divisible by $x.$

In any case, the OP should be using $4k$ and not $x$ when dealing with an arbitrary multiple of 4. That's just basic, good technique.

 

[chwala]

How can you "like" a post that says that 2 will always be divisible by a multiple of 4?In any case, the OP should be using $4k$ and not $x$ when dealing with an arbitrary power of 4. That's just basic, good technique.

Noted @PeroK ...thanks for pointing that out.

 

[Delta2]

when dealing with an arbitrary power of 4.

E hehe typo police here, you mean arbitrary multiple of 4, not power.

 

[lurflurf]

How can you "like" a post that says that 2 will always be divisible by a multiple of 4?In any case, the OP should be using $4k$ and not $x$ when dealing with an arbitrary multiple of 4. That's just basic, good technique.

I liked it because I liked it. Sometimes I like things that are not perfect. Clearly a simple typo. We all make them. Naming variables conveniently not a the most important technique, but is helpful at times. 4x,4x+4,4x+8,4x+12 is nice I would probably use x-6,x-2,x+2,x+6 instead but it does not matter. Often you would not realize the desired form until after naming the variables. Sure you could then edit your work to appear psychic, but why bother.

 

[Mark44]

In any case, the OP should be using 4k and not x when dealing with an arbitrary multiple of 4. That's just basic, good technique.

For me, doing this is a result of standing in front of a math classroom for many years, trying to make things as self-explanatory as possible. IOW, trying to follow the Principle of Least Astonishment.

 

[WWGD]

This may be a roundabout way. You have a general expression for the sum of 4 terms in an AP (Arithmetic Progression) , with r=4.
When would this sum 4(x+6) be a multiple of 8?
You need a multiple of 2 here, since you already have a 4 available. So you could prove something stronger, of when such a sum is a multiple of 8, and conclude it's true for x a multiple of 4.

 

[chwala]

This may be a roundabout way. You have a general expression for the sum of 4 terms in an AP (Arithmetic Progression) , with r=4.
When would this sum 4(x+6) be a multiple of 8?
You need a multiple of 2 here, since you already have a 4 available. So you could prove something stronger, of when such a sum is a multiple of 8, and conclude it's true for x a multiple of 4.

I had initially thought of using this approach of finding sum...i will look at this and get back by weekend...