Prove that the product of any three consecutive integers is

[r0bHadz]

Homework Statement

Prove that the product of any three consecutive integers
is divisible by 6.

Homework Equations

The Attempt at a Solution

This doesn't seem true to me for any 3 consecutive ints. For example, let a_0 = 0 a_1 = 1 a_2 = 2

3 is not divisible by six.

Assuming they meant a_x > 0we have a(a+1)(a+2) = a^3 + 3a^2 + 2a = p_1^{x_1}p_2^{x_2}...p_n^{x_n}

3a^2= p_1^{x_1}p_2^{x_2}...p_n^{x_n} - a^3 - 2a

the right side would be divisible by 3, which implies 3 is in the prime factorization

the same logic is used for "2a"

which shows that the product of any 3 consecutive ints is a factor of 6.Is my proof alright for this guys? I have major problems with the fundamental theorem of arithmetic so I'm still uncertain here. It seems like it would suffice but you can't be too sure

 

[Mark44]

Homework Statement

Prove that the product of any three consecutive integers
is divisible by 6.

Homework Equations

The Attempt at a Solution

This doesn't seem true to me for any 3 consecutive ints. For example, let a_0 = 0 a_1 = 1 a_2 = 2

3 is not divisible by six.

But 0 is divisible by 6. 0 * 1 * 2 = 0, not 3

Assuming they meant a_x > 0we have a(a+1)(a+2) = a^3 + 3a^2 + 2a = p_1^{x_1}p_2^{x_2}...p_n^{x_n}

You're making this more complicated than it needs to be. If you have three consecutive integers, a, a + 1, and a + 2, at least one of them has to be even, so you're half way home. All you need to do is to show that one of the three is also divisible by 3.

3a^2= p_1^{x_1}p_2^{x_2}...p_n^{x_n} - a^3 - 2a

the right side would be divisible by 3, which implies 3 is in the prime factorization

the same logic is used for "2a"

which shows that the product of any 3 consecutive ints is a factor of 6.Is my proof alright for this guys? I have major problems with the fundamental theorem of arithmetic so I'm still uncertain here. It seems like it would suffice but you can't be too sure

 

[SammyS]

Homework Statement

Prove that the product of any three consecutive integers
is divisible by 6.

Homework Equations

The Attempt at a Solution

This doesn't seem true to me for any 3 consecutive ints. For example, let a_0 = 0 a_1 = 1 a_2 = 2

3 is not divisible by six.

The product is obtained by multiplication, not by addition.

Zero is divisible by 6 .

 

[r0bHadz]

Whoops don't know what I was thinking there. I think I wrote this late at night and forgot that it was a product or something lol. And I wrote the rest of my post in the morning and just went along with it >.>

Anyways, I initially wanted to prove it by stating that one of the products will be odd as well, but this question was under a chapter relating to primes so I have a feeling like using the F.T.A was more appropriate.

Besides the whole zero thing, is the proof considered valid?

 

[StoneTemplePython]

You're in a discrete math course... can I assume you've seen binomial coefficients, and know e.g. that every number in pascal's triangle is an integer?

If so your problem is to consider $\frac{(n+2)(n+1)(n)}{3!}$

what does that remind you of?

 

[r0bHadz]

You're in a discrete math course... can I assume you've seen binomial coefficients, and know e.g. that every number in pascal's triangle is an integer?

If so your problem is to consider $\frac{(n+2)(n+1)(n)}{3!}$

what does that remind you of?

I have but I would have to review them (I forget about it.) This problem is under the section "primes and greatest common divisors" so I think they want me to focus on the fundamental theorem of arithmetic.

Is my proof sound using primes

 

[StoneTemplePython]

I have but I would have to review them (I forget about it.) This problem is under the section "primes and greatest common divisors" so I think they want me to focus on the fundamental theorem of arithmetic.

Is my proof sound using primes

I'll take a closer look at the primes argument... your candor is appreciated but what you said is worrisome. Mathematics is about synthesis, and to me if you go through a discrete math course without having binomial coefficients locked down cold, it's a failure. I'd go as far as to say binomial coefficients or 'choosing' operations, plus basic set manipulations, plus proof techniques are what you should learn in this course.

 

[StoneTemplePython]

I don't like what you did with primes... it looks like bunch of symbols that confuse the issue.

How about you write the product

$(n)(n+1)(n+2)$

and use modular arithmetic, which you were working on a week or so ago? All you need to do is use two well chosen modulo's to prove that at least one of those terms is even and one must be a multiple of 3, hence the product must be divisible by 2 as well as by 3.

If you really want to relate this to primes, take solace that 2 and 3 are primes.

 

[Mark44]

but this question was under a chapter relating to primes so I have a feeling like using the F.T.A was more appropriate.

No, and that's pretty much what I meant when I said you're making this more complicated that it needs to be. @StoneTemplePython is giving you some good advice about modular arithmetic. Keep in mind that even and odd numbers have to do with mod 2.

All you need to do is use two well chosen modulo's

What he said...

 

[r0bHadz]

I'm just finding it really hard to state that if the first integer is not divisible by three, and the second isn't, then the third is. Or if the first isn't, then the second is, or the first is. It makes sense, but its so damn hard to explain mathematically

no answers please just wanting to update that I am still trying... lol

I want to say its something like this:

6 | (a)(a+1)(a+2) => one of these has to be even, so only need to show divisibility by 3

if a mod 3 = a, 0≤a<3

we have 3 cases for a:
a = 0, which is divisible by 3, proof complete
a = 1 => (a+2) mod 3 = 0,
a = 2 => (a+1) mod 3 = 0

I'm just having a hard time writing this so that it is "formal" -_-

 

[StoneTemplePython]

I'm just having a hard time writing this so that it is "formal" -_-

I think you're basically there. Let me make a suggestion: use a table.

so for mod 3, you could have
$\left[\begin{matrix} \text{ }& \text{case 1} & \text{case 2} & \text{case 3} \\ n\%3& \text{insert value} & \text{insert value} & \text{insert value} \\ (n+1)\%3 & \text{insert value} & \text{insert value} & \text{insert value} \\ (n+2)\%3 & \text{insert value} & \text{insert value} & \text{insert value}\\\end{matrix}\right]$

and circle / highlight the zero in each column
- - - -
as an encore:

i.) You may point out with brute force solutions that it works for edit:$\{$ (cleanup) modulo 2 or modulo 3, but what about modulo 17, etc.?$\}$ Structurally this table looks an awful lot like something that comes up in group theory (which is sometimes touched on with permutation groups in discrete math). Just keep your eyes open for this structure.

ii.) Another deceptively simple idea from discrete math is the pidgeon hole principle. You may have seen it already. My idea isn't fully formed here, but I think you probably could use a pidgeon hole argument for the mod 3 case (which implies the result for the other case by a dominance relation).

 

[r0bHadz]

Hmm great idea about the pigeonhole principle. I can sort of see how expanding it from the most basic case 2 pigeons, 3 holes, or 3 pigeons, 3 holes -> 3 holes 17 pigeons would work. I have free time today so I think I'm going to research binomial coefficients and come back to this problem, I am not really in a hurry yet