calculate max and min number in range
$begingroup$
I'm writing a script that's supposed to selectively add hidden to div elements based on which page has been selected.
Each section should have 16 images so the sections go as follows
Section 1: 1 - 16
Section 2: 17 - 33
Section 3: 34 - 40
Section 4: 41 - 57
I'm trying to figure out an equation given the page number can derive the range start and end.
My first assumption was from 2 onwards it would be
Section 2: (((section_num - 1) * 16) + 1) - (section_num * 16) + 1
(16 + 1) - (16 * 2) + 1
17 - 33
But it doesn't always follow this line.
algebra-precalculus
edited Dec 22 '18 at 11:46
YiFan
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
$endgroup$
add a comment |
$begingroup$
I'm writing a script that's supposed to selectively add hidden to div elements based on which page has been selected.
Each section should have 16 images so the sections go as follows
Section 1: 1 - 16
Section 2: 17 - 33
Section 3: 34 - 40
Section 4: 41 - 57
I'm trying to figure out an equation given the page number can derive the range start and end.
My first assumption was from 2 onwards it would be
Section 2: (((section_num - 1) * 16) + 1) - (section_num * 16) + 1
(16 + 1) - (16 * 2) + 1
17 - 33
But it doesn't always follow this line.
algebra-precalculus
edited Dec 22 '18 at 11:46
YiFan
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
$endgroup$
$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47
add a comment |
$begingroup$
I'm writing a script that's supposed to selectively add hidden to div elements based on which page has been selected.
Each section should have 16 images so the sections go as follows
Section 1: 1 - 16
Section 2: 17 - 33
Section 3: 34 - 40
Section 4: 41 - 57
I'm trying to figure out an equation given the page number can derive the range start and end.
My first assumption was from 2 onwards it would be
Section 2: (((section_num - 1) * 16) + 1) - (section_num * 16) + 1
(16 + 1) - (16 * 2) + 1
17 - 33
But it doesn't always follow this line.
algebra-precalculus
edited Dec 22 '18 at 11:46
YiFan
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
$endgroup$
I'm writing a script that's supposed to selectively add hidden to div elements based on which page has been selected.
Each section should have 16 images so the sections go as follows
Section 1: 1 - 16
Section 2: 17 - 33
Section 3: 34 - 40
Section 4: 41 - 57
I'm trying to figure out an equation given the page number can derive the range start and end.
My first assumption was from 2 onwards it would be
Section 2: (((section_num - 1) * 16) + 1) - (section_num * 16) + 1
(16 + 1) - (16 * 2) + 1
17 - 33
But it doesn't always follow this line.
algebra-precalculus
algebra-precalculus
edited Dec 22 '18 at 11:46
YiFan
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
edited Dec 22 '18 at 11:46
YiFan
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
edited Dec 22 '18 at 11:46
YiFan
4,8511727
edited Dec 22 '18 at 11:46
YiFan
4,8511727
edited Dec 22 '18 at 11:46
YiFan
4,8511727
4,8511727
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
asked Dec 22 '18 at 11:37
Samuel M.Samuel M.
1204
1204
$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47
add a comment |
$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47
$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since you've taken the first "inteverval" to be $1-16$, the ending page (16 in the first case) for any section will be 16*(section number) and NOT 16*(section number). You can also see it like: Let n be current section number, and $x$ be it highest page #. Then, by what you wrote, $x=16n+1$. Also, the lowest page # in $(n+1)th$ section $=((n+1)-1)*16+1=16n+1=x$, which isn't true right?
Instead, if you take what I said, i.e, $x=16n$, then the lowest page in next section will be $=((n+1)-1)*16+1=16n+1=x+1$, which is obviously correct.
Your formula for lower page number was CORRECT anyway!
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Since you've taken the first "inteverval" to be $1-16$, the ending page (16 in the first case) for any section will be 16*(section number) and NOT 16*(section number). You can also see it like: Let n be current section number, and $x$ be it highest page #. Then, by what you wrote, $x=16n+1$. Also, the lowest page # in $(n+1)th$ section $=((n+1)-1)*16+1=16n+1=x$, which isn't true right?
Instead, if you take what I said, i.e, $x=16n$, then the lowest page in next section will be $=((n+1)-1)*16+1=16n+1=x+1$, which is obviously correct.
Your formula for lower page number was CORRECT anyway!
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Since you've taken the first "inteverval" to be $1-16$, the ending page (16 in the first case) for any section will be 16*(section number) and NOT 16*(section number). You can also see it like: Let n be current section number, and $x$ be it highest page #. Then, by what you wrote, $x=16n+1$. Also, the lowest page # in $(n+1)th$ section $=((n+1)-1)*16+1=16n+1=x$, which isn't true right?
Instead, if you take what I said, i.e, $x=16n$, then the lowest page in next section will be $=((n+1)-1)*16+1=16n+1=x+1$, which is obviously correct.
Your formula for lower page number was CORRECT anyway!
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
$endgroup$
add a comment |
$begingroup$
Since you've taken the first "inteverval" to be $1-16$, the ending page (16 in the first case) for any section will be 16*(section number) and NOT 16*(section number). You can also see it like: Let n be current section number, and $x$ be it highest page #. Then, by what you wrote, $x=16n+1$. Also, the lowest page # in $(n+1)th$ section $=((n+1)-1)*16+1=16n+1=x$, which isn't true right?
Instead, if you take what I said, i.e, $x=16n$, then the lowest page in next section will be $=((n+1)-1)*16+1=16n+1=x+1$, which is obviously correct.
Your formula for lower page number was CORRECT anyway!
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
$endgroup$
Since you've taken the first "inteverval" to be $1-16$, the ending page (16 in the first case) for any section will be 16*(section number) and NOT 16*(section number). You can also see it like: Let n be current section number, and $x$ be it highest page #. Then, by what you wrote, $x=16n+1$. Also, the lowest page # in $(n+1)th$ section $=((n+1)-1)*16+1=16n+1=x$, which isn't true right?
Instead, if you take what I said, i.e, $x=16n$, then the lowest page in next section will be $=((n+1)-1)*16+1=16n+1=x+1$, which is obviously correct.
Your formula for lower page number was CORRECT anyway!
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
answered Dec 22 '18 at 12:01
Ankit KumarAnkit Kumar
1,542221
1,542221
add a comment |
add a comment |
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$begingroup$
This is not linear algebra.
$endgroup$
– YiFan
Dec 22 '18 at 11:39
$begingroup$
think it's basic algebra but that's the only near algebra that was on the tag
$endgroup$
– Samuel M.
Dec 22 '18 at 11:44
$begingroup$
If you looked sown, you would see the [algebra-precalculus] tag, which is more appropriate here.
$endgroup$
– YiFan
Dec 22 '18 at 11:47