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$a$, $b$, and $c$, non-negative consecutive integers with $a + b + c$ odd, prove $abc$ divisible by $24$ [duplicate]

Prove 24 divides $u^3-u$ for all odd natural numbers $u$Prove that $n^3(n^2-1)$ is divisible by 24 for all nProving that $n|mimplies f_n|f_m$Does mathematical induction assume that non-negative integers are infinite?Help with Induction problem?Prove: The product of any three consecutive integers is divisible by $6$.Generalization of “Sum of cube of any 3 consecutive integers is divisible by 3”Induction for a sequence starting with a negative and ending with a positive number.Can mathematical inductions work for other sets?Why do we do mathematical induction only for positive whole numbers?Prove by induction that $n^3-n$ is divisible by $24$ for all odd positive integersProve that all positive integers $n$, $(1-{sqrt 5})^n$ can be written in the form $a-b{sqrt 5}$ where $a$ and $b$ are positive integers

0

$begingroup$

This question already has an answer here:

• 
  Prove 24 divides $u^3-u$ for all odd natural numbers $u$
  
  6 answers
  
  


• 
  Prove that $n^3(n^2-1)$ is divisible by 24 for all n
  
  6 answers
  
  

If $a$, $b$, and $c$ are three non-negative consecutive integers and $a + b + c$ can't be divided by $2$.

Prove by induction that $abc$ can be divided by $24$.

For the question I don't even know where to start proving it, I don't know how to symbolize the $a$, $b$, and $c$ into numbers.

induction proof-explanation divisibility

share|cite|improve this question

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

$endgroup$

marked as duplicate by José Carlos Santos, Bill Dubuque divisibility
Users with the  divisibility badge can single-handedly close divisibility questions as duplicates and reopen them as needed.

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Mar 15 at 15:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  First idea : One of three consecutive integers must be divisible by $3$, hence the product is divisible by $3$. Second idea : If $a$ were odd, $a+b+c$ would be even, which is ruled out. Hene $a$ and $c$ are even and one of them must be divisible by $4$. This shows that the product is also divisible by $8$.
  $endgroup$
  – Peter
  Mar 15 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $,a,b,c = n-1,n,n+1,$ so $,abc = n^3-n,,$ and $,a+b+c = 3n,$ odd $Rightarrow n$ odd, so the dupe applies.
  $endgroup$
  – Bill Dubuque
  Mar 15 at 15:25
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Prove 24 divides $u^3-u$ for all odd natural numbers $u$
  
  6 answers
  
  


• 
  Prove that $n^3(n^2-1)$ is divisible by 24 for all n
  
  6 answers
  
  

If $a$, $b$, and $c$ are three non-negative consecutive integers and $a + b + c$ can't be divided by $2$.

Prove by induction that $abc$ can be divided by $24$.

For the question I don't even know where to start proving it, I don't know how to symbolize the $a$, $b$, and $c$ into numbers.

induction proof-explanation divisibility

share|cite|improve this question

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

$endgroup$

marked as duplicate by José Carlos Santos, Bill Dubuque divisibility
Users with the  divisibility badge can single-handedly close divisibility questions as duplicates and reopen them as needed.

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Mar 15 at 15:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  First idea : One of three consecutive integers must be divisible by $3$, hence the product is divisible by $3$. Second idea : If $a$ were odd, $a+b+c$ would be even, which is ruled out. Hene $a$ and $c$ are even and one of them must be divisible by $4$. This shows that the product is also divisible by $8$.
  $endgroup$
  – Peter
  Mar 15 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $,a,b,c = n-1,n,n+1,$ so $,abc = n^3-n,,$ and $,a+b+c = 3n,$ odd $Rightarrow n$ odd, so the dupe applies.
  $endgroup$
  – Bill Dubuque
  Mar 15 at 15:25
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Prove 24 divides $u^3-u$ for all odd natural numbers $u$
  
  6 answers
  
  


• 
  Prove that $n^3(n^2-1)$ is divisible by 24 for all n
  
  6 answers
  
  

If $a$, $b$, and $c$ are three non-negative consecutive integers and $a + b + c$ can't be divided by $2$.

Prove by induction that $abc$ can be divided by $24$.

For the question I don't even know where to start proving it, I don't know how to symbolize the $a$, $b$, and $c$ into numbers.

induction proof-explanation divisibility

share|cite|improve this question

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

$endgroup$

share|cite|improve this question

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

share|cite|improve this question

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

edited Mar 15 at 9:13

lioness99a

3,9212727

edited Mar 15 at 9:13

lioness99a

3,9212727

asked Mar 15 at 9:10

Fifi12Fifi12

62

asked Mar 15 at 9:10

Fifi12Fifi12

62

2 Answers
2

active

oldest

votes

0

$begingroup$

Because $a,b,c$ are consecutive, then we can say that $b=a+1$ and $c=a+2$.

We know that $a+b+c$ is not divisible by $2$, which is another way of saying that $a+b+c$ is odd.

For the proof, we try with $a=1$ but this gives us $a+b+c=1+2+3=6$ which is even. So we try $a=2$ which gives us $a+b+c=2+3+4=9$ so we can continue

This gives us begin{align}abc&=2times 3times 4\
&=6times4\
&=24end{align}

which is clearly divisible by $24$.

Now consider $a=3$, this gives us $a+b+c=3+4+5=12$ which is even. If we try $a=4$, we get $a+b+c=4+5+6=15$ which is odd. Hopefully you can see that $a$ must always be even. We can write an even number as $2k$ where $k$ can be any number.

Now we do the induction step. If we assume that it works for $a=2k$ then we need to prove that it works for $a=2k+2$. We can write this out as follows:

We know that $2k$ is even (and therefore $2k+2$ is also even). We also know that $2k(2k+1)(2k+2)$ is divisible by $24$ - this is what we are assuming. We want to prove that $(2k+2)(2k+3)(2k+4)$ is also divisible by $24$.

Are you able to continue from here?

share|cite|improve this answer

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So can I write it like this to prove it?$8k^3+36k^2+52k+24=24p$ $4(2k^3+9k^2+13k+6)=4(6p)$
  $endgroup$
  – Fifi12
  Mar 15 at 11:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where has $p$ come from?
  $endgroup$
  – lioness99a
  Mar 15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in $8k^3+36k^2+52k+24$ is can be $24$ multiple by a number I symbolized as $p$
  $endgroup$
  – Fifi12
  Mar 15 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you've not proved anything there. You've shown it is divisble by $4$ by taking out the factor on the LHS, but you haven't shown it is also divisible by $6$ yet
  $endgroup$
  – lioness99a
  Mar 15 at 13:00
  

  
  
  
  

add a comment | 

0

$begingroup$

Because the integers are consecutive and the sum is odd, we know that exactly one of them is odd and it has to be the middle one. So we have $a=2n$, $b=2n+1$, and $c=2n+2,$ for some integer $n$. Then

$$abc = 2n(2n+1)(2n+2) = 4n(2n^2+3n+1).$$

So if $n$ is even, then $8$ divides $abc.$ If $n$ is odd, then $2n^2+3n+1$ is even and again $8$ divides $abd.$ So now we just have to show that $3$ divides $abc.$

So either use the fact that one of three consecutive integers must be divisible by three, or do three cases $n = 3k$, $n=3k+1$ and $3k+2.$

share|cite|improve this answer

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

$endgroup$

add a comment | 

0

$begingroup$

Because $a,b,c$ are consecutive, then we can say that $b=a+1$ and $c=a+2$.

We know that $a+b+c$ is not divisible by $2$, which is another way of saying that $a+b+c$ is odd.

For the proof, we try with $a=1$ but this gives us $a+b+c=1+2+3=6$ which is even. So we try $a=2$ which gives us $a+b+c=2+3+4=9$ so we can continue

This gives us begin{align}abc&=2times 3times 4\
&=6times4\
&=24end{align}

which is clearly divisible by $24$.

Now consider $a=3$, this gives us $a+b+c=3+4+5=12$ which is even. If we try $a=4$, we get $a+b+c=4+5+6=15$ which is odd. Hopefully you can see that $a$ must always be even. We can write an even number as $2k$ where $k$ can be any number.

Now we do the induction step. If we assume that it works for $a=2k$ then we need to prove that it works for $a=2k+2$. We can write this out as follows:

We know that $2k$ is even (and therefore $2k+2$ is also even). We also know that $2k(2k+1)(2k+2)$ is divisible by $24$ - this is what we are assuming. We want to prove that $(2k+2)(2k+3)(2k+4)$ is also divisible by $24$.

Are you able to continue from here?

share|cite|improve this answer

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So can I write it like this to prove it?$8k^3+36k^2+52k+24=24p$ $4(2k^3+9k^2+13k+6)=4(6p)$
  $endgroup$
  – Fifi12
  Mar 15 at 11:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where has $p$ come from?
  $endgroup$
  – lioness99a
  Mar 15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in $8k^3+36k^2+52k+24$ is can be $24$ multiple by a number I symbolized as $p$
  $endgroup$
  – Fifi12
  Mar 15 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you've not proved anything there. You've shown it is divisble by $4$ by taking out the factor on the LHS, but you haven't shown it is also divisible by $6$ yet
  $endgroup$
  – lioness99a
  Mar 15 at 13:00
  

  
  
  
  

add a comment | 

0

$begingroup$

Because $a,b,c$ are consecutive, then we can say that $b=a+1$ and $c=a+2$.

We know that $a+b+c$ is not divisible by $2$, which is another way of saying that $a+b+c$ is odd.

For the proof, we try with $a=1$ but this gives us $a+b+c=1+2+3=6$ which is even. So we try $a=2$ which gives us $a+b+c=2+3+4=9$ so we can continue

This gives us begin{align}abc&=2times 3times 4\
&=6times4\
&=24end{align}

which is clearly divisible by $24$.

Now consider $a=3$, this gives us $a+b+c=3+4+5=12$ which is even. If we try $a=4$, we get $a+b+c=4+5+6=15$ which is odd. Hopefully you can see that $a$ must always be even. We can write an even number as $2k$ where $k$ can be any number.

Now we do the induction step. If we assume that it works for $a=2k$ then we need to prove that it works for $a=2k+2$. We can write this out as follows:

We know that $2k$ is even (and therefore $2k+2$ is also even). We also know that $2k(2k+1)(2k+2)$ is divisible by $24$ - this is what we are assuming. We want to prove that $(2k+2)(2k+3)(2k+4)$ is also divisible by $24$.

Are you able to continue from here?

share|cite|improve this answer

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So can I write it like this to prove it?$8k^3+36k^2+52k+24=24p$ $4(2k^3+9k^2+13k+6)=4(6p)$
  $endgroup$
  – Fifi12
  Mar 15 at 11:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where has $p$ come from?
  $endgroup$
  – lioness99a
  Mar 15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As in $8k^3+36k^2+52k+24$ is can be $24$ multiple by a number I symbolized as $p$
  $endgroup$
  – Fifi12
  Mar 15 at 12:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you've not proved anything there. You've shown it is divisble by $4$ by taking out the factor on the LHS, but you haven't shown it is also divisible by $6$ yet
  $endgroup$
  – lioness99a
  Mar 15 at 13:00
  

  
  
  
  

add a comment | 

$begingroup$

Because $a,b,c$ are consecutive, then we can say that $b=a+1$ and $c=a+2$.

We know that $a+b+c$ is not divisible by $2$, which is another way of saying that $a+b+c$ is odd.

For the proof, we try with $a=1$ but this gives us $a+b+c=1+2+3=6$ which is even. So we try $a=2$ which gives us $a+b+c=2+3+4=9$ so we can continue

This gives us begin{align}abc&=2times 3times 4\
&=6times4\
&=24end{align}

which is clearly divisible by $24$.

Now consider $a=3$, this gives us $a+b+c=3+4+5=12$ which is even. If we try $a=4$, we get $a+b+c=4+5+6=15$ which is odd. Hopefully you can see that $a$ must always be even. We can write an even number as $2k$ where $k$ can be any number.

Now we do the induction step. If we assume that it works for $a=2k$ then we need to prove that it works for $a=2k+2$. We can write this out as follows:

We know that $2k$ is even (and therefore $2k+2$ is also even). We also know that $2k(2k+1)(2k+2)$ is divisible by $24$ - this is what we are assuming. We want to prove that $(2k+2)(2k+3)(2k+4)$ is also divisible by $24$.

Are you able to continue from here?

share|cite|improve this answer

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

$endgroup$

share|cite|improve this answer

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

answered Mar 15 at 9:36

lioness99alioness99a

3,9212727

0

$begingroup$

Because the integers are consecutive and the sum is odd, we know that exactly one of them is odd and it has to be the middle one. So we have $a=2n$, $b=2n+1$, and $c=2n+2,$ for some integer $n$. Then

$$abc = 2n(2n+1)(2n+2) = 4n(2n^2+3n+1).$$

So if $n$ is even, then $8$ divides $abc.$ If $n$ is odd, then $2n^2+3n+1$ is even and again $8$ divides $abd.$ So now we just have to show that $3$ divides $abc.$

So either use the fact that one of three consecutive integers must be divisible by three, or do three cases $n = 3k$, $n=3k+1$ and $3k+2.$

share|cite|improve this answer

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

$endgroup$

add a comment | 

0

$begingroup$

Because the integers are consecutive and the sum is odd, we know that exactly one of them is odd and it has to be the middle one. So we have $a=2n$, $b=2n+1$, and $c=2n+2,$ for some integer $n$. Then

$$abc = 2n(2n+1)(2n+2) = 4n(2n^2+3n+1).$$

So if $n$ is even, then $8$ divides $abc.$ If $n$ is odd, then $2n^2+3n+1$ is even and again $8$ divides $abd.$ So now we just have to show that $3$ divides $abc.$

So either use the fact that one of three consecutive integers must be divisible by three, or do three cases $n = 3k$, $n=3k+1$ and $3k+2.$

share|cite|improve this answer

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

$endgroup$

add a comment | 

$begingroup$

Because the integers are consecutive and the sum is odd, we know that exactly one of them is odd and it has to be the middle one. So we have $a=2n$, $b=2n+1$, and $c=2n+2,$ for some integer $n$. Then

$$abc = 2n(2n+1)(2n+2) = 4n(2n^2+3n+1).$$

So if $n$ is even, then $8$ divides $abc.$ If $n$ is odd, then $2n^2+3n+1$ is even and again $8$ divides $abd.$ So now we just have to show that $3$ divides $abc.$

So either use the fact that one of three consecutive integers must be divisible by three, or do three cases $n = 3k$, $n=3k+1$ and $3k+2.$

share|cite|improve this answer

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

$endgroup$

share|cite|improve this answer

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442

answered Mar 15 at 10:06

B. GoddardB. Goddard

19.8k21442