Let me do some research on this ...chwala said: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##.chwala said: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##
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 said: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.
Fair enough -- I didn't read the post that carefully.Delta2 said:No I don't think its incorrect, it is the same as yours with ##x=4k##. He just names it x instead of 4k.
Would that be;Delta2 said: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?
I've no idea why you are using ##x## for a multiple of four, and then never use this fact explicitly.chwala said: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).chwala said: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.##
Well x is a multiple of 2 was usedPeroK said:I've no idea why you are using ##x## for a multiple of four, and then never use this fact explicitly.
How can you "like" a post that says that 2 will always be divisible by a multiple of 4?lurflurf said:Well x is a multiple of 2 was used
x is a multiple of 4 is not needed
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 said:##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.##
Noted @PeroK ...thanks for pointing that out.PeroK said: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.
E hehe typo police here, you mean arbitrary multiple of 4, not power.PeroK said:when dealing with an arbitrary power of 4.
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.PeroK said: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.
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.PeroK said: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 had initially thought of using this approach of finding sum...i will look at this and get back by weekend...WWGD said: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.
A multiple of a number is a number that can be obtained by multiplying that number by any whole number. For example, the multiples of 3 are 3, 6, 9, 12, and so on.
To prove that a number is a multiple of another number, we can use the division algorithm. If the remainder is 0 when we divide the first number by the second number, then the first number is a multiple of the second number.
Factors are numbers that can divide a given number without leaving a remainder, while multiples are numbers that are obtained by multiplying a given number by any whole number. Therefore, the factors of a number are also its multiples.
Yes, a number can have an infinite number of multiples. This is because we can always multiply a number by a larger whole number to get a new multiple.
The LCM (least common multiple) of two numbers is the smallest positive number that is a multiple of both numbers. The GCD (greatest common divisor) of two numbers is the largest positive number that divides both numbers without leaving a remainder. We can use the multiples of the two numbers to find their LCM and GCD by identifying the smallest multiple that is common to both numbers for the LCM, and the largest multiple that is common to both numbers for the GCD.