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Find the number of $k$ satisfying $frac{n}{3} - 1 9$.

Inequlity $frac{2}{frac{a}{x}+frac{b}{y}}leq ax+by$ for $a+b=1$Find the number of integer solutions to $x_1+x_2+x_3+x_4= 30$ where $0leq x_n <10$ for $1leq n leq 4$Alternate approach to show $|sum_{i leq n} frac{mu(i)}{i}| leq 1$?Prove $| sum_{i leq n} frac{mu(i)}{i} | leq 1$Let $x$ be a real number. Prove the existence of a unique integer $a$ such that $a leq x < a+1$Prove that the number of fractions $a/b$ in lowest terms with $0<a/b leq 1$ and $bleq n$ is $phi(1) + … + phi(n)$.Prove that the nth prime number $p_n$ satisfies $p_nleq 2^{2n-1}$Prove that for every prime number $|{1leq x leq p^2 : p^2 | (x^{p-1}-1) }|=p-1$proof that $frac{m^a}{2}+frac{m}{2}-1$ is not a prime numberHow to find all the solutions for $sumlimits_{i=1}^{n} frac{1}{2^{k_i}}= 1$ for $k_iin Bbb{N}$ and $n$ a fixed positive integer?

1

$begingroup$

Show that the number of $k in mathbb{N}$ satisfying $frac{n}{3} - 1 < k leq frac{n}{2}$ for $n in mathbb{N}, n> 9$ is greater than $frac{n}{6}$.

I looked at the distance $|frac{n}{3} - 1 - (frac{n}{2})| = 1 + frac{n}{6}$ which is $> frac{n}{6}.$ Is this correct or am I missing a proper justification? I'm not sure how to use the $n > 9$ fact either. Thank you.

elementary-number-theory

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asked Mar 20 at 0:39

DarkdubDarkdub

11516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to be careful with using the distance, as you did, in terms of the boundary conditions with the required inequalities and the value of $k$ being an integer. Although the total is $gt frac{n}{6}$, it's possible for there to be fewer than that integers which satisfy the inequalities. Regarding using $n gt 9$, as I show in my answer, doesn't seem to really be needed, so I'm not sure why it's provided. As I state in my answer, perhaps there's another required condition that is not included here.
  $endgroup$
  – John Omielan
  Mar 20 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There were not any additional conditions provided so I was a bit dumbfounded myself. Thank you for your response, I understand now.
  $endgroup$
  – Darkdub
  Mar 20 at 9:33
  

  
  
  
  

add a comment | 

1

$begingroup$

Show that the number of $k in mathbb{N}$ satisfying $frac{n}{3} - 1 < k leq frac{n}{2}$ for $n in mathbb{N}, n> 9$ is greater than $frac{n}{6}$.

I looked at the distance $|frac{n}{3} - 1 - (frac{n}{2})| = 1 + frac{n}{6}$ which is $> frac{n}{6}.$ Is this correct or am I missing a proper justification? I'm not sure how to use the $n > 9$ fact either. Thank you.

elementary-number-theory

share|cite|improve this question

asked Mar 20 at 0:39

DarkdubDarkdub

11516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need to be careful with using the distance, as you did, in terms of the boundary conditions with the required inequalities and the value of $k$ being an integer. Although the total is $gt frac{n}{6}$, it's possible for there to be fewer than that integers which satisfy the inequalities. Regarding using $n gt 9$, as I show in my answer, doesn't seem to really be needed, so I'm not sure why it's provided. As I state in my answer, perhaps there's another required condition that is not included here.
  $endgroup$
  – John Omielan
  Mar 20 at 2:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There were not any additional conditions provided so I was a bit dumbfounded myself. Thank you for your response, I understand now.
  $endgroup$
  – Darkdub
  Mar 20 at 9:33
  

  
  
  
  

add a comment | 

$begingroup$

Show that the number of $k in mathbb{N}$ satisfying $frac{n}{3} - 1 < k leq frac{n}{2}$ for $n in mathbb{N}, n> 9$ is greater than $frac{n}{6}$.

I looked at the distance $|frac{n}{3} - 1 - (frac{n}{2})| = 1 + frac{n}{6}$ which is $> frac{n}{6}.$ Is this correct or am I missing a proper justification? I'm not sure how to use the $n > 9$ fact either. Thank you.

elementary-number-theory

share|cite|improve this question

asked Mar 20 at 0:39

DarkdubDarkdub

11516

$endgroup$

share|cite|improve this question

asked Mar 20 at 0:39

DarkdubDarkdub

11516

share|cite|improve this question

asked Mar 20 at 0:39

DarkdubDarkdub

11516

asked Mar 20 at 0:39

DarkdubDarkdub

11516

asked Mar 20 at 0:39

DarkdubDarkdub

11516

1 Answer
1

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votes

1

$begingroup$

Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 le r le 5$. Thus, the question asks to prove there are more than $frac{n}{6} = m + frac{r}{6} lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.

The inequality becomes

$$2m + frac{r}{3} - 1 lt k le 3m + frac{r}{2} tag{1}label{eq1}$$

In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:

$$r = 0 ; Rightarrow ; 2m - 1 lt k le 3m tag{2}label{eq2}$$
$$r = 1 ; Rightarrow ; 2m - frac{2}{3} lt k le 3m + frac{1}{2}tag{3}label{eq3}$$
$$r = 2 ; Rightarrow ; 2m - frac{1}{3} lt k le 3m + 1 tag{4}label{eq4}$$
$$r = 3 ; Rightarrow ; 2m lt k le 3m + frac{3}{2} tag{5}label{eq5}$$
$$r = 4 ; Rightarrow ; 2m + frac{1}{3} lt k le 3m + 2 tag{6}label{eq6}$$
$$r = 5 ; Rightarrow ; 2m + frac{2}{3} lt k le 3m + frac{5}{2} tag{7}label{eq7}$$

With eqref{eq2} and eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, eqref{eq6} & eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values.

As such, this works for all $n in mathbb{N}$, including for $n gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?

share|cite|improve this answer

edited Mar 20 at 3:10

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

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add a comment | 

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1

$begingroup$

Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 le r le 5$. Thus, the question asks to prove there are more than $frac{n}{6} = m + frac{r}{6} lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.

The inequality becomes

$$2m + frac{r}{3} - 1 lt k le 3m + frac{r}{2} tag{1}label{eq1}$$

In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:

$$r = 0 ; Rightarrow ; 2m - 1 lt k le 3m tag{2}label{eq2}$$
$$r = 1 ; Rightarrow ; 2m - frac{2}{3} lt k le 3m + frac{1}{2}tag{3}label{eq3}$$
$$r = 2 ; Rightarrow ; 2m - frac{1}{3} lt k le 3m + 1 tag{4}label{eq4}$$
$$r = 3 ; Rightarrow ; 2m lt k le 3m + frac{3}{2} tag{5}label{eq5}$$
$$r = 4 ; Rightarrow ; 2m + frac{1}{3} lt k le 3m + 2 tag{6}label{eq6}$$
$$r = 5 ; Rightarrow ; 2m + frac{2}{3} lt k le 3m + frac{5}{2} tag{7}label{eq7}$$

With eqref{eq2} and eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, eqref{eq6} & eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values.

As such, this works for all $n in mathbb{N}$, including for $n gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?

share|cite|improve this answer

edited Mar 20 at 3:10

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

$endgroup$

add a comment | 

1

$begingroup$

Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 le r le 5$. Thus, the question asks to prove there are more than $frac{n}{6} = m + frac{r}{6} lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.

The inequality becomes

$$2m + frac{r}{3} - 1 lt k le 3m + frac{r}{2} tag{1}label{eq1}$$

In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:

$$r = 0 ; Rightarrow ; 2m - 1 lt k le 3m tag{2}label{eq2}$$
$$r = 1 ; Rightarrow ; 2m - frac{2}{3} lt k le 3m + frac{1}{2}tag{3}label{eq3}$$
$$r = 2 ; Rightarrow ; 2m - frac{1}{3} lt k le 3m + 1 tag{4}label{eq4}$$
$$r = 3 ; Rightarrow ; 2m lt k le 3m + frac{3}{2} tag{5}label{eq5}$$
$$r = 4 ; Rightarrow ; 2m + frac{1}{3} lt k le 3m + 2 tag{6}label{eq6}$$
$$r = 5 ; Rightarrow ; 2m + frac{2}{3} lt k le 3m + frac{5}{2} tag{7}label{eq7}$$

With eqref{eq2} and eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, eqref{eq6} & eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values.

As such, this works for all $n in mathbb{N}$, including for $n gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?

share|cite|improve this answer

edited Mar 20 at 3:10

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

$endgroup$

add a comment | 

$begingroup$

Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 le r le 5$. Thus, the question asks to prove there are more than $frac{n}{6} = m + frac{r}{6} lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.

The inequality becomes

$$2m + frac{r}{3} - 1 lt k le 3m + frac{r}{2} tag{1}label{eq1}$$

In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:

$$r = 0 ; Rightarrow ; 2m - 1 lt k le 3m tag{2}label{eq2}$$
$$r = 1 ; Rightarrow ; 2m - frac{2}{3} lt k le 3m + frac{1}{2}tag{3}label{eq3}$$
$$r = 2 ; Rightarrow ; 2m - frac{1}{3} lt k le 3m + 1 tag{4}label{eq4}$$
$$r = 3 ; Rightarrow ; 2m lt k le 3m + frac{3}{2} tag{5}label{eq5}$$
$$r = 4 ; Rightarrow ; 2m + frac{1}{3} lt k le 3m + 2 tag{6}label{eq6}$$
$$r = 5 ; Rightarrow ; 2m + frac{2}{3} lt k le 3m + frac{5}{2} tag{7}label{eq7}$$

With eqref{eq2} and eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, eqref{eq6} & eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values.

As such, this works for all $n in mathbb{N}$, including for $n gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?

share|cite|improve this answer

edited Mar 20 at 3:10

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

$endgroup$

share|cite|improve this answer

edited Mar 20 at 3:10

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215

answered Mar 20 at 2:24

John OmielanJohn Omielan

4,6312215