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The product of three consecutive integers is …? Odd? Divisible by $4$? by $5$? by $6$? by $12$?
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If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
- My answer is correct
- What is the formal proof of what I said in $ i) $
number-theory
edited Aug 2 at 23:32
Blue
43.6k868141
asked Aug 2 at 23:24
Mattiu
760316
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up vote
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If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
- My answer is correct
- What is the formal proof of what I said in $ i) $
number-theory
edited Aug 2 at 23:32
Blue
43.6k868141
asked Aug 2 at 23:24
Mattiu
760316
1
Looks good.$$
– quasi
Aug 2 at 23:26
2
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
- My answer is correct
- What is the formal proof of what I said in $ i) $
number-theory
edited Aug 2 at 23:32
Blue
43.6k868141
asked Aug 2 at 23:24
Mattiu
760316
If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
- My answer is correct
- What is the formal proof of what I said in $ i) $
number-theory
edited Aug 2 at 23:32
Blue
43.6k868141
asked Aug 2 at 23:24
Mattiu
760316
edited Aug 2 at 23:32
Blue
43.6k868141
edited Aug 2 at 23:32
Blue
43.6k868141
edited Aug 2 at 23:32
Blue
43.6k868141
43.6k868141
asked Aug 2 at 23:24
Mattiu
760316
asked Aug 2 at 23:24
Mattiu
760316
asked Aug 2 at 23:24
Mattiu
760316
760316
1
Looks good.$$
– quasi
Aug 2 at 23:26
2
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33
add a comment |Â
1
Looks good.$$
– quasi
Aug 2 at 23:26
2
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33
1
1
Looks good.$$
– quasi
Aug 2 at 23:26
Looks good.$$
– quasi
Aug 2 at 23:26
2
2
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33
add a comment |Â
7 Answers
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One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2;$mod $3$.
- If $nequiv 0;(textmod;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 1;(textmod;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 2;(textmod;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
answered Aug 2 at 23:32
quasi
32.9k22359
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
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I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
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Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
answered Aug 2 at 23:32
Phil H
1,7862311
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
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Assume $$aequiv k mod 3$$ where $0le kle 2$.
$$k=0implies text$a$ is divisible by $3$$$
$$k=1implies text$a+2$ is divisible by $3$$$
$$k=2implies text$a+1$ is divisible by $3$$$
answered Aug 2 at 23:33
Szeto
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Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
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Let $x=n(n+1)(n+2)$ Then $x$ has some property that holds for all $n$ but in particular that means this property holds for $n=1$ but when $n=1$ this means that $x=6$ and only property $D$ satisfies this.
answered Aug 2 at 23:41
Mason
1,1271223
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
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To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1cdot(1+1)cdot(1+2)=1cdot2cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$beginalign
n(n+1)(n+2)&equiv0mod(3)\
(n+1)(n+2)(n+3)&equiv0mod(3)
endalign$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$beginalign
n(n+1)(n+2)&equiv0mod(6)\
(n+1)(n+2)(n+3)&equiv0mod(6)
endalign$$
This time the rewrite part is a little bit different
$$beginalign
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\
&=(6)(n+1)+n(n+1)(n+5)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\
endalign$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1cdot2cdot3=6$ the divisibilty by $12$ and $4$ fails with $5cdot6cdot7=210$. The divisibility by $5$ fails for example with $2cdot3cdot4=24$.
answered Aug 2 at 23:46
mrtaurho
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Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
answered Aug 3 at 2:43
Keith Backman
37927
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2;$mod $3$.
- If $nequiv 0;(textmod;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 1;(textmod;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 2;(textmod;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
answered Aug 2 at 23:32
quasi
32.9k22359
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
accepted
One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2;$mod $3$.
- If $nequiv 0;(textmod;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 1;(textmod;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 2;(textmod;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
answered Aug 2 at 23:32
quasi
32.9k22359
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2;$mod $3$.
- If $nequiv 0;(textmod;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 1;(textmod;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 2;(textmod;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
answered Aug 2 at 23:32
quasi
32.9k22359
One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2;$mod $3$.
- If $nequiv 0;(textmod;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 1;(textmod;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\[4pt]$
- If $nequiv 2;(textmod;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.
Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.
Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.
Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.
Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
answered Aug 2 at 23:32
quasi
32.9k22359
edited Aug 3 at 1:18
answered Aug 2 at 23:32
quasi
32.9k22359
answered Aug 2 at 23:32
quasi
32.9k22359
answered Aug 2 at 23:32
quasi
32.9k22359
32.9k22359
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
add a comment |Â
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
Thanks quasi for your explanation !
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
2
down vote
I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
add a comment |Â
up vote
2
down vote
I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
answered Aug 2 at 23:33
Cameron Buie
83.4k771152
83.4k771152
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
add a comment |Â
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
Thanks cameron :)
– Mattiu
Aug 2 at 23:39
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
You're welcome! By the way, if you choose your counterexample well, you can also disprove (C)--you've already disproved (A), and you'll be disproving (B) as you disprove (E).
– Cameron Buie
Aug 2 at 23:47
add a comment |Â
up vote
1
down vote
Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
answered Aug 2 at 23:32
Phil H
1,7862311
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
answered Aug 2 at 23:32
Phil H
1,7862311
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
answered Aug 2 at 23:32
Phil H
1,7862311
Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
answered Aug 2 at 23:32
Phil H
1,7862311
answered Aug 2 at 23:32
Phil H
1,7862311
answered Aug 2 at 23:32
Phil H
1,7862311
answered Aug 2 at 23:32
Phil H
1,7862311
1,7862311
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
add a comment |Â
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
Thanks Phil H !!
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
Assume $$aequiv k mod 3$$ where $0le kle 2$.
$$k=0implies text$a$ is divisible by $3$$$
$$k=1implies text$a+2$ is divisible by $3$$$
$$k=2implies text$a+1$ is divisible by $3$$$
answered Aug 2 at 23:33
Szeto
3,8431421
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
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up vote
1
down vote
Assume $$aequiv k mod 3$$ where $0le kle 2$.
$$k=0implies text$a$ is divisible by $3$$$
$$k=1implies text$a+2$ is divisible by $3$$$
$$k=2implies text$a+1$ is divisible by $3$$$
answered Aug 2 at 23:33
Szeto
3,8431421
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume $$aequiv k mod 3$$ where $0le kle 2$.
$$k=0implies text$a$ is divisible by $3$$$
$$k=1implies text$a+2$ is divisible by $3$$$
$$k=2implies text$a+1$ is divisible by $3$$$
answered Aug 2 at 23:33
Szeto
3,8431421
Assume $$aequiv k mod 3$$ where $0le kle 2$.
$$k=0implies text$a$ is divisible by $3$$$
$$k=1implies text$a+2$ is divisible by $3$$$
$$k=2implies text$a+1$ is divisible by $3$$$
answered Aug 2 at 23:33
Szeto
3,8431421
answered Aug 2 at 23:33
Szeto
3,8431421
answered Aug 2 at 23:33
Szeto
3,8431421
answered Aug 2 at 23:33
Szeto
3,8431421
3,8431421
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
add a comment |Â
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
Thanks Szeto :)
– Mattiu
Aug 2 at 23:39
add a comment |Â
up vote
0
down vote
Let $x=n(n+1)(n+2)$ Then $x$ has some property that holds for all $n$ but in particular that means this property holds for $n=1$ but when $n=1$ this means that $x=6$ and only property $D$ satisfies this.
answered Aug 2 at 23:41
Mason
1,1271223
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
add a comment |Â
up vote
0
down vote
Let $x=n(n+1)(n+2)$ Then $x$ has some property that holds for all $n$ but in particular that means this property holds for $n=1$ but when $n=1$ this means that $x=6$ and only property $D$ satisfies this.
answered Aug 2 at 23:41
Mason
1,1271223
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $x=n(n+1)(n+2)$ Then $x$ has some property that holds for all $n$ but in particular that means this property holds for $n=1$ but when $n=1$ this means that $x=6$ and only property $D$ satisfies this.
answered Aug 2 at 23:41
Mason
1,1271223
Let $x=n(n+1)(n+2)$ Then $x$ has some property that holds for all $n$ but in particular that means this property holds for $n=1$ but when $n=1$ this means that $x=6$ and only property $D$ satisfies this.
answered Aug 2 at 23:41
Mason
1,1271223
answered Aug 2 at 23:41
Mason
1,1271223
answered Aug 2 at 23:41
Mason
1,1271223
answered Aug 2 at 23:41
Mason
1,1271223
1,1271223
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
add a comment |Â
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
I don't want my answer to seem to facetious but these type of questions always seem silly to me. This is a formal proof, no?
– Mason
Aug 2 at 23:43
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
en.wikipedia.org/wiki/Formal_proof
– Mattiu
Aug 3 at 20:34
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
I certainly haven't provided a formal proof of $(i)$. A counterexample to options is A,B,C, and E is a "formal" argument that these are not the answer.
– Mason
Aug 3 at 20:39
add a comment |Â
up vote
0
down vote
To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1cdot(1+1)cdot(1+2)=1cdot2cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$beginalign
n(n+1)(n+2)&equiv0mod(3)\
(n+1)(n+2)(n+3)&equiv0mod(3)
endalign$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$beginalign
n(n+1)(n+2)&equiv0mod(6)\
(n+1)(n+2)(n+3)&equiv0mod(6)
endalign$$
This time the rewrite part is a little bit different
$$beginalign
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\
&=(6)(n+1)+n(n+1)(n+5)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\
endalign$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1cdot2cdot3=6$ the divisibilty by $12$ and $4$ fails with $5cdot6cdot7=210$. The divisibility by $5$ fails for example with $2cdot3cdot4=24$.
answered Aug 2 at 23:46
mrtaurho
607117
add a comment |Â
up vote
0
down vote
To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1cdot(1+1)cdot(1+2)=1cdot2cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$beginalign
n(n+1)(n+2)&equiv0mod(3)\
(n+1)(n+2)(n+3)&equiv0mod(3)
endalign$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$beginalign
n(n+1)(n+2)&equiv0mod(6)\
(n+1)(n+2)(n+3)&equiv0mod(6)
endalign$$
This time the rewrite part is a little bit different
$$beginalign
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\
&=(6)(n+1)+n(n+1)(n+5)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\
endalign$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1cdot2cdot3=6$ the divisibilty by $12$ and $4$ fails with $5cdot6cdot7=210$. The divisibility by $5$ fails for example with $2cdot3cdot4=24$.
answered Aug 2 at 23:46
mrtaurho
607117
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1cdot(1+1)cdot(1+2)=1cdot2cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$beginalign
n(n+1)(n+2)&equiv0mod(3)\
(n+1)(n+2)(n+3)&equiv0mod(3)
endalign$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$beginalign
n(n+1)(n+2)&equiv0mod(6)\
(n+1)(n+2)(n+3)&equiv0mod(6)
endalign$$
This time the rewrite part is a little bit different
$$beginalign
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\
&=(6)(n+1)+n(n+1)(n+5)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\
endalign$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1cdot2cdot3=6$ the divisibilty by $12$ and $4$ fails with $5cdot6cdot7=210$. The divisibility by $5$ fails for example with $2cdot3cdot4=24$.
answered Aug 2 at 23:46
mrtaurho
607117
To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1cdot(1+1)cdot(1+2)=1cdot2cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$beginalign
n(n+1)(n+2)&equiv0mod(3)\
(n+1)(n+2)(n+3)&equiv0mod(3)
endalign$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$beginalign
n(n+1)(n+2)&equiv0mod(6)\
(n+1)(n+2)(n+3)&equiv0mod(6)
endalign$$
This time the rewrite part is a little bit different
$$beginalign
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\
&=(6)(n+1)+n(n+1)(n+5)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\
endalign$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1cdot2cdot3=6$ the divisibilty by $12$ and $4$ fails with $5cdot6cdot7=210$. The divisibility by $5$ fails for example with $2cdot3cdot4=24$.
answered Aug 2 at 23:46
mrtaurho
607117
edited Aug 2 at 23:53
answered Aug 2 at 23:46
mrtaurho
607117
answered Aug 2 at 23:46
mrtaurho
607117
answered Aug 2 at 23:46
mrtaurho
607117
607117
add a comment |Â
add a comment |Â
up vote
0
down vote
Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
answered Aug 3 at 2:43
Keith Backman
37927
add a comment |Â
up vote
0
down vote
Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
answered Aug 3 at 2:43
Keith Backman
37927
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
answered Aug 3 at 2:43
Keith Backman
37927
Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
answered Aug 3 at 2:43
Keith Backman
37927
answered Aug 3 at 2:43
Keith Backman
37927
answered Aug 3 at 2:43
Keith Backman
37927
answered Aug 3 at 2:43
Keith Backman
37927
37927
add a comment |Â
add a comment |Â
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1
Looks good.$$
– quasi
Aug 2 at 23:26
2
You must also show that sometimes it is not divisible by $12$. Otherwise, $12$ could also be an answer.
– spiralstotheleft
Aug 2 at 23:33