In normal reading, every non-top comment has two links: parent, and reply. This makes sense. When you click reply, the comment is shown again above the input textbox, which still makes sense. Below the comment, there's one link. And this link says: reply. Now why should I want to use the reply link, when my reply form is already right before my nose? Noreover, why remove the link to parent (which, unlike the reply link, I actually would use from time to time)?
Not that it's a big problem, not clicking a pointless link is more than easy, and if I want to get to the parent, it's just one extra mouse click (once on the comment number to get a normal display of the message, where I then have the parent link). It's just that it looks stupid to me every time again to see a pointless reply link on the reply page, while obviously the coders have done some work to remove the (actually useful) parent link.
Are we going to suddenly see completely transparent vehicles driving around?
I doubt it. After all, you probably don't want the enemy seeing all the actions you do in the vehicle ("Oh, they are loading the weapon now, maybe this would be a good time to attack...")
Easy: I remember from math that a line is a degenerated triangle. Then I note that each stick already represents a line. Therefore I don't have to do anything to get six (degenerated) triangles.
Rule for n: Multiply the digit representing 10^k by (10-n)^k, and add the results together. The resulting number is a multiple of n exactly if the original number was. Instead of (10-n)^k of course any number which differs from that by a multiple of n can be used. (The proof and the generalization for bases other than 10 is left as exercise to the reader:-))
Examples: n = 2: Pure rule: * 124 -> 1*64 + 2*8 + 4 = 12. 12 is a multiple of 2, therefore 124 is, too. * 123 -> 1*64 + 2*8 + 2 = 11. 11 is not a multiple of 2, therefore 123 isn't either. Improved rule: Since 8^n is a multiple of 2 for n>=1, it can be subtracted from itself, giving 0. Since 0 times anything is 0, this way only the last digit remains, and we get the well-known rule that a number is even exactly if the last digit is even.
n=3: Pure rule: * 123 -> 1*49 + 2*7 + 3 = 66. 66 is a multiple of 3, therefore 123 is, too. * 124 -> 1*49 + 2*7 + 4 = 67. 67 is not a multiple of 3, therefore 124 isn't either. Improved rule: 7 is 2*3+1, therefore 7^k = (2*3+1)^k = sum_i binomial(k,i) 2^i*3^i. All summands except for i=0 are multiples of 3 and can therefore be eliminated. The summand for i=0 is just 1. Therefore we get the well known rule that a number is a multiple of 3 exactly if the sum of it's digits is. Note that there is not really the need to carry out the general calculation, you can do it step by step: Since 7 can be replaced with 1, 7*7 can be replaced with 7*1, which again is 7 and can be replaced by 1. That is, you always can multiply the "reduced" number instead of the original one for determining the next "reduced power". I'll do this in my following examples.
n=4: Pure rule: 124 -> 1*36 + 2*6 + 4 = 52. 52 is a multiple of 4, therefore 124 is, too. Improved rule: 6-4 = 2, therefore the next-to-last digit can be multiplied by 2 instead of 6. For n>=2, 6^n is a multiple of 4 and therefore can be reduced to 0 (alternatively: for the third-last daigit, 6*2=12, 6*2-3*4=0, and 0*6 = 0 for all other digits). Therefore you need only look at the last two digits, and the number is multiple of 4 exactly if 2*(next-to-last digit)+(last digit) is.
In the following examples, I'll directly skip to the improved rule.
n=5: Since 10-5 is 5, again only the last digit matters.
n=6: 1*4 = 4 4*4 = 16, 16-2*6 = 4 Therefore all digits but the first have to be multiplied by 4, and then all of them added together. Example: 1296 -> 4*1 + 4*2 + 4*9 + 6 = 54 = 6*9. In practice, it's probably simpler to remember that a number is a multiple of 6 exactly if it is both a multiple of 2 and a multiple of 3, and test those two separably.
n=7: Now we come to the first interesting case (i.e. the first one which gives a non-trivial rule). Consecutive multiplication with 3 and eliminating multiples of 7 gives the series: 1, 3, 9-7=2, 6, 18-14=4, 12-7=5, 15-14=1,... You see, it's a cycle of 6 numbers. Let's see the rule in action, with a sufficiently large number: Is 14663586 a multiple of 7? Well, let's see: 6*1 = 6 8*3 = 24 5*2 = 10 3*6 = 18 6*4 = 24 6*5 = 30 4*1 = 4 1*3 = 3 -------- sum = 119 Now, it's not hard to check that 119 = 70+49 is a multiple of 7. Otherwise, we could just do a second iteration: 9*1 = 9 1*3 = 3 1*2 = 2 -------- sum = 14 Ok, now it's obvious, isn't it?
Now you mentioned 11. So 10-11 = -1, therefore you have to multiply the digits with (-1)^k, i.e. subtract the sum of the even-position digits from the sum of the odd-position ones. Alternatively, you can of course alter by 11 to get a positive result (which in this case is 10 again), and then for the next step you get 10*(-1) = -10, which then gets altered to 1 again. Which means an alternate rule is to split the number in groups of two digits and sum them up. Let's try both rules on 1051138: Alternation rule: 1-0+5-1+1-3+8 = 11, which clearly is a multiple of 11- Digit-pair rule: 1+05+11+38 = 55, also obviously a multiple of 11.
What if the light bulb doesn't work? But since there was no requirement that additional tools may not be used, there's of course a simple solution: Step 1: Go into the room and connect a voltmeter to the light bulb. Connect that voltmeter to a sender which sends you the result out of the room. There's no need to enter the room again now. Step 2: Try all the switches and watch the voltage.
One of the problems he exploited was that IE obviously eliminates newline characters before searching for the tags. That is, if you write
"Javas cript"
inside an attribute, then IE will interpret that as "Javascript" anyway, but it evaded the filter at MySpace (which did scan for "Javascript"). Now I'm not sure that it's really the website's fault not to know about an IE bug which is that stupid.
Well, just remove the "++i", and you can also remove the "almost" part of your description. If you find a corresponding browser vulnerability, you just might switch off JavaScript with your JavaScript code, of course:-)
I'm sure it would be there, and it basically would be Linux. It just wouldn't compete with Windows, but maybe with an OS based on GEM (remember, originally Windows also just was a GUI on top of DOS), or with MacOS, or maybe with some OS which doesn't even exist in our world but would have been written if Windows had not existed.
Microsoft made computing cheap? So Microsoft created all those IBM PC clones which were in competition to IBM's own products and therefore forced the prices down?
The reason why computers got cheap wasn't Microsoft. It was IBM building a PC from standard parts without making exclusivity deals on any vital component, thus opening up the market for PC clones (something they most certainly didn't intend:-)). Another important factor IMHO was IBM demanding a second source for the 8086/8088 (AMD), which also caused competition on that front and therefore kept the processor price down.
And are you really sure if MS would have had relatively low DOS prices if there had been no competition (DR DOS)?
Did you know that there's recent evidence to suggest that the assumption that print media and online media differ in the effect line length has on readability?
Slashdot Mirror
One possibility would be to allow selection of different skins from the user preferences, like Wikipedia does.
In normal reading, every non-top comment has two links: parent, and reply. This makes sense.
When you click reply, the comment is shown again above the input textbox, which still makes sense.
Below the comment, there's one link. And this link says: reply.
Now why should I want to use the reply link, when my reply form is already right before my nose?
Noreover, why remove the link to parent (which, unlike the reply link, I actually would use from time to time)?
Not that it's a big problem, not clicking a pointless link is more than easy, and if I want to get to the parent, it's just one extra mouse click (once on the comment number to get a normal display of the message, where I then have the parent link). It's just that it looks stupid to me every time again to see a pointless reply link on the reply page, while obviously the coders have done some work to remove the (actually useful) parent link.
I prefer Platinium, the noble metal
I doubt it. After all, you probably don't want the enemy seeing all the actions you do in the vehicle ("Oh, they are loading the weapon now, maybe this would be a good time to attack
I didn't know there's a DOS port of Firefox.
You'll make kphone kalls instead.
Easy: I remember from math that a line is a degenerated triangle. Then I note that each stick already represents a line. Therefore I don't have to do anything to get six (degenerated) triangles.
Rule for n: :-))
...
Multiply the digit representing 10^k by (10-n)^k, and add the results together. The resulting number is a multiple of n exactly if the original number was. Instead of (10-n)^k of course any number which differs from that by a multiple of n can be used. (The proof and the generalization for bases other than 10 is left as exercise to the reader
Examples:
n = 2:
Pure rule:
* 124 -> 1*64 + 2*8 + 4 = 12. 12 is a multiple of 2, therefore 124 is, too.
* 123 -> 1*64 + 2*8 + 2 = 11. 11 is not a multiple of 2, therefore 123 isn't either.
Improved rule:
Since 8^n is a multiple of 2 for n>=1, it can be subtracted from itself, giving 0. Since 0 times anything is 0, this way only the last digit remains, and we get the well-known rule that a number is even exactly if the last digit is even.
n=3:
Pure rule:
* 123 -> 1*49 + 2*7 + 3 = 66. 66 is a multiple of 3, therefore 123 is, too.
* 124 -> 1*49 + 2*7 + 4 = 67. 67 is not a multiple of 3, therefore 124 isn't either.
Improved rule:
7 is 2*3+1, therefore 7^k = (2*3+1)^k = sum_i binomial(k,i) 2^i*3^i. All summands except for i=0 are multiples of 3 and can therefore be eliminated. The summand for i=0 is just 1. Therefore we get the well known rule that a number is a multiple of 3 exactly if the sum of it's digits is.
Note that there is not really the need to carry out the general calculation, you can do it step by step: Since 7 can be replaced with 1, 7*7 can be replaced with 7*1, which again is 7 and can be replaced by 1. That is, you always can multiply the "reduced" number instead of the original one for determining the next "reduced power". I'll do this in my following examples.
n=4:
Pure rule:
124 -> 1*36 + 2*6 + 4 = 52. 52 is a multiple of 4, therefore 124 is, too.
Improved rule:
6-4 = 2, therefore the next-to-last digit can be multiplied by 2 instead of 6. For n>=2, 6^n is a multiple of 4 and therefore can be reduced to 0 (alternatively: for the third-last daigit, 6*2=12, 6*2-3*4=0, and 0*6 = 0 for all other digits). Therefore you need only look at the last two digits, and the number is multiple of 4 exactly if 2*(next-to-last digit)+(last digit) is.
In the following examples, I'll directly skip to the improved rule.
n=5:
Since 10-5 is 5, again only the last digit matters.
n=6:
1*4 = 4
4*4 = 16, 16-2*6 = 4
Therefore all digits but the first have to be multiplied by 4, and then all of them added together.
Example:
1296 -> 4*1 + 4*2 + 4*9 + 6 = 54 = 6*9.
In practice, it's probably simpler to remember that a number is a multiple of 6 exactly if it is both a multiple of 2 and a multiple of 3, and test those two separably.
n=7:
Now we come to the first interesting case (i.e. the first one which gives a non-trivial rule). Consecutive multiplication with 3 and eliminating multiples of 7 gives the series:
1, 3, 9-7=2, 6, 18-14=4, 12-7=5, 15-14=1,
You see, it's a cycle of 6 numbers. Let's see the rule in action, with a sufficiently large number:
Is 14663586 a multiple of 7? Well, let's see:
6*1 = 6
8*3 = 24
5*2 = 10
3*6 = 18
6*4 = 24
6*5 = 30
4*1 = 4
1*3 = 3
--------
sum = 119
Now, it's not hard to check that 119 = 70+49 is a multiple of 7. Otherwise, we could just do a second iteration:
9*1 = 9
1*3 = 3
1*2 = 2
--------
sum = 14
Ok, now it's obvious, isn't it?
Now you mentioned 11. So 10-11 = -1, therefore you have to multiply the digits with (-1)^k, i.e. subtract the sum of the even-position digits from the sum of the odd-position ones. Alternatively, you can of course alter by 11 to get a positive result (which in this case is 10 again), and then for the next step you get 10*(-1) = -10, which then gets altered to 1 again. Which means an alternate rule is to split the number in groups of two digits and sum them up.
Let's try both rules on 1051138:
Alternation rule: 1-0+5-1+1-3+8 = 11, which clearly is a multiple of 11-
Digit-pair rule: 1+05+11+38 = 55, also obviously a multiple of 11.
You want to maximize the money.
Indeed I'd expect both to come from the Village of Life, because if they were from the Village of Death, they certainly would be dead!
Which the liar will probably answer with the direction you came from :-)
And of course the truth-teller could say: "I don't know. Otherwise I'd already be there, instead of standing here and thinking about what to do."
What if the light bulb doesn't work?
But since there was no requirement that additional tools may not be used, there's of course a simple solution:
Step 1: Go into the room and connect a voltmeter to the light bulb. Connect that voltmeter to a sender which sends you the result out of the room. There's no need to enter the room again now.
Step 2: Try all the switches and watch the voltage.
inside an attribute, then IE will interpret that as "Javascript" anyway, but it evaded the filter at MySpace (which did scan for "Javascript"). Now I'm not sure that it's really the website's fault not to know about an IE bug which is that stupid.
Well, just remove the "++i", and you can also remove the "almost" part of your description. :-)
If you find a corresponding browser vulnerability, you just might switch off JavaScript with your JavaScript code, of course
Well, having over 1 million foes is also an achievement ...
I'm sure it would be there, and it basically would be Linux. It just wouldn't compete with Windows, but maybe with an OS based on GEM (remember, originally Windows also just was a GUI on top of DOS), or with MacOS, or maybe with some OS which doesn't even exist in our world but would have been written if Windows had not existed.
Microsoft made computing cheap? So Microsoft created all those IBM PC clones which were in competition to IBM's own products and therefore forced the prices down?
:-)). Another important factor IMHO was IBM demanding a second source for the 8086/8088 (AMD), which also caused competition on that front and therefore kept the processor price down.
The reason why computers got cheap wasn't Microsoft. It was IBM building a PC from standard parts without making exclusivity deals on any vital component, thus opening up the market for PC clones (something they most certainly didn't intend
And are you really sure if MS would have had relatively low DOS prices if there had been no competition (DR DOS)?
20 years ago, you could safely ignore it.
It's a good complement to Google's Pidgeon Rank system.
That's the NSA trying to prevent widespread knowledge about the creation of Death Rays. Just imagine what happens if terrorists get to read that!
You think one week is not enough? God created the whole universe in one week! Ok, it's not exactly bug-free ... :-)
They don't just cripple the phones, they also cripple their customers? I didn't know that they are that bad
Parse error: Expected verb, got '?'.
First Subtask: Understand what this task actually means. Edit the EXIF data of a JPEG file? Add a symlink to the file? Or what?