Yes, but think of the satisfaction you got from throwing your keyboard around the room for a while.
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Yes, but think of the satisfaction you got from throwing your keyboard around the room for a while.
That is 7 squared = 49Code:2
7 = 49
*Every even digit must be even (even # multiplied by anything is even)Quote:
Originally posted by Elrond
Why? because it follows the rules.
3816547290->
10 * 381654729 = 3816547290
9 * 42406081 = 381654729
8 * 4770684 = 38165472
7 * 545221 = 3816547
6 * 63609 = 381654
5 * 7633 = 38165
4 * 954 = 3816
3 * 127 = 381
2 * 19 = 38
1 * 3 = 3
I just wrote a little program to find it ;)
Even without that, there are a few rules that would make it easier:
The last digit MUST be 0 (divisible by 10).
The 5th digit must therefore be 5 (first five divisible by 5, and 0 is gone).
You only care about finding the correct number up to 8 (if you follow the rule of having only different digits, you have 0 as the tenth, so when you end up with the remaining 9 digits, all numbers can be divided by 9, if you remember the rule that when the sum of the digits can be divided by 9, the number can be divided by 9).
Every 2nd digit must be an even digit (must be divisible by an even number).
I have not found any other rule yet, but it already reduces by a lot the number of possible answers... (about 2401 left if I am correct, but might not be as I am not that good with statistics).
* And consequently every odd digit must be odd
Well I found the hint, but I don't know what it means.Quote:
Originally posted by Doctor Luz
Do you want to try a riddle in the style of SeventhStar?
Well, you have to look for a password hidden in the attachment.
There is a 'poem' which will give you a hint for knowing what to do with the attachment.
I think this is very easy.
Read carefully and sorry for the bad english.
"HELLO GURUS"Quote:
Originally posted by Doctor Luz
Do you want to try a riddle in the style of SeventhStar?
Well, you have to look for a password hidden in the attachment.
There is a 'poem' which will give you a hint for knowing what to do with the attachment.
I think this is very easy.
Read carefully and sorry for the bad english.
:D
am i right? :p :D
Totally right! you are a champ!Quote:
Originally posted by SeventhStar
"HELLO GURUS"
am i right? :p :D
:D
Same here. The hint in the poem is quite obvious. But I don't know what to do with it either. I must definitely not be good to work with images. :(Quote:
Originally posted by solarflare
Well I found the hint, but I don't know what it means.
If you really have the hint, you only have to think about how is the .gif structure and do what the hint says.Quote:
Originally posted by Elrond
Same here. The hint in the poem is quite obvious. But I don't know what to do with it either. I must definitely not be good to work with images. :(
;)
Yep, that's it.Quote:
Originally posted by KevinHall
That is 7 squared = 49Code:2
7 = 49
Since the riddle has been solved, why not tell us how it is done?Quote:
Originally posted by Doctor Luz
If you really have the hint, you only have to think about how is the .gif structure and do what the hint says.
;)
Please? :)
The answer is "HELLO GURUS".
If you load the GIF image into a good photo editor, you can edit the colors in the color table. There are two blue values, two red values, two green values, and two yellow values. If you change the second yellow value to black, you see the message "HELLO GURUS". You can also do this with a hex editor if you know the GIF file format, but using a photo editor is much easier.
- Kevin
Good explanation, I could not explain it better myself.Quote:
Originally posted by KevinHall
The answer is "HELLO GURUS".
If you load the GIF image into a good photo editor, you can edit the colors in the color table. There are two blue values, two red values, two green values, and two yellow values. If you change the second yellow value to black, you see the message "HELLO GURUS". You can also do this with a hex editor if you know the GIF file format, but using a photo editor is much easier.
- Kevin
How about the following:
Take four 9's and some mathematical operators and make them equal to 100.
How about 99 + 9/9 = 100.
Here's a question that came to me while I was watching the clock to much when I was a kid. Take the time on a digital clock and add up the digits. One minute later do the same. What time or times would work so that the first sum is twice the second sum? There are many but one time would work as a valid answer. It's not too hard but it was the only thing I could come up with.
If you have any questions let me know.
-Ben
3:49 / 3:50
4:39 / 4:40
5:29 / 5:30
6:19 / 6:20
7:09 / 7:10
12:49 / 12:50
That's it. So what's next?
-Ben
I don't know. I can't think of a riddle right now.
This is not a riddle but a real question I could not explain very well to my kid.
Sir Isaac Newton said that "for every action, there is always an opposite and equal reaction" (I think that's the exact word, just correct me if i'm wrong).
We know that if we move a small object, we can easily do it. But how does that statement apply? We cannot, supposed to be, move the object because it will always react according to the force we apply?
When you answer, treat me like a kid please.:)
Well, I think the small object reacts according to the 3rd newton law, however this reaction is not in the small object, is in the object which comunicates the force to the small object (your hand, your finger...).
You comunicate a force to the object and the object comunicates the same force with opposite sense to you.
Then the object moves. If the force you comunicate is instantaneous, the object moves accordingly with the 2nd Newton's law with an acceleration F=m*a.
When thinking about forces it's important to remember that you may not be aware of all the forces being applied. Here are a couple of examples that might help.
First imagine you’re standing on solid flat ground and you try to push a 3 kg box. It shouldn't be that hard to move. Now try and push a 300 kg box. That should be almost impossible to move. Now imagine doing this on a frozen lake. If you tried to push the 300 kg box what would happen? The box wouldn't move you would. You would be pushed back away from the box. When you push something, something has to be pushing on you if you don't move. Many times the thing pushing on you is the earth. It's hard to see that when you’re on solid ground but on the ice it becomes much more apparent.
The next question might be if the earth is pushing back why doesn’t it move? Well it does. Now this will really mess with your head but like the good Doctor Luz said Force=mass*acceleration. If you apply a 10 Newton (N) force to an object the earth applies a 10N force to you (assuming you don't move and the object does.) Since the earth is so large compared to the object its acceleration will be very small. And usually completely insignificant. I think I'm starting to ramble now so I'll shut up.
Does that help? If not let me know and I'll try to explain a little better.
-Ben
To complete the example of bmacri, if you push a 3 kg box hard enough while standing on ice, it will move forward, but you will also move a little backward. This shows the the box, even small, applies some reactive force upon you.
Have your child put on some roller blades and throw something heavy. Have someone behind him (or her) though ready to catch him (or her) b/c they could easily fall. But the point is that both the object thrown and your child will move -- just in an opposite direction. Your child pushes on the object, thus it moves. The object pushes back on your child thus he (or she) moves. They move in the opposite direction.
I tutored for university physics for 3 years. Let me know if you need some more examples or if you have some other questions.
- Kevin
I actually worked this out in my head once. I believe the earth's center of mass moves a distance of only a fraction of a width of a proton (very small fraction if I remember correctly) when we throw a heavy object as far as we can. Just in case anyone was wondering.Quote:
The next question might be if the earth is pushing back why doesn’t it move? Well it does. Now this will really mess with your head but like the good Doctor Luz said Force=mass*acceleration. If you apply a 10 Newton (N) force to an object the earth applies a 10N force to you (assuming you don't move and the object does.) Since the earth is so large compared to the object its acceleration will be very small. And usually completely insignificant. I think I'm starting to ramble now so I'll shut up.
And as it is likely enough that other people will throw the same kind of object in the opposite direction somewhere on Earth, it gives an even weaker result... :pQuote:
Originally posted by KevinHall
I actually worked this out in my head once. I believe the earth's center of mass moves a distance of only a fraction of a width of a proton (very small fraction if I remember correctly) when we throw a heavy object as far as we can. Just in case anyone was wondering.
And given that the earth is not truely a rigid object, the effect that you have on the earth will be mostly local decaying with distance rather quickly (within a couple of yards of where you stand for any significant measurements). ;)Quote:
Originally posted by Elrond
And as it is likely enough that other people will throw the same kind of object in the opposite direction somewhere on Earth, it gives an even weaker result... :p
So if everyone on one side of the earth was to jump up and down at the same time would the earth go flying off into space? Or better yet how many people jumping up and down on the same side of the earth would it take for the earth to be thrown out of orbit?
I'm not really expecting an anwser it's just a question that seems to come up everytime when a discusion like this occures.
To be calculated, but it might be more that the number of people that have ever lived on Earth so far. It means that the risk is small enough ;) :p :DQuote:
Originally posted by bmacri
So if everyone on one side of the earth was to jump up and down at the same time would the earth go flying off into space? Or better yet how many people jumping up and down on the same side of the earth would it take for the earth to be thrown out of orbit?
I'm not really expecting an anwser it's just a question that seems to come up everytime when a discusion like this occures.
No. People (even collectively) do have the strength to overcome the gravity of the earth. The center of mass of the earth would follow the same orbit around the sun and in one second, everything would be back to normal.Quote:
Originally posted by bmacri
So if everyone on one side of the earth was to jump up and down at the same time would the earth go flying off into space?
It would take an aweful lot. You would need the collective mass of people to much greater (thrown out of orbit) than the mass of the earth. Say the average person weighs 100kg (perhaps a little on the heavy side) and "much greater" would be a factor of at least 100. Then we have 100*(6e24 kg / 100kg) = 6e24 people or 6 septillion people (6 trillion trillion people).Quote:
Originally posted by bmacri
Or better yet how many people jumping up and down on the same side of the earth would it take for the earth to be thrown out of orbit?
- Kevin
1
11
21
1211
111221
312211
13112221
What row of numbers comes next?
1113213211Quote:
Originally posted by KevinHall
1
11
21
1211
111221
312211
13112221
What row of numbers comes next?
Better yet, how 'bout prove that there can never be a 4 in this sequence.
You're right. There can never be a 4 in the sequence because the first instance of any number other than 1 will be preceded by a 1 in the next sequence. Thus the longest chain of a single number will be "111" which will become "31" in the next line. Since "111" is the longest sequence, there will never be a "41" or "4x" (fill in the x). Ok, so this is not a proof that would pass in a set theory class, but I think it should do for this thread.
- Kevin
Now how about the following (this one I actually created myself -- hope it isn't too hard):
C15
2046
5652
133332
What line comes next? Let me know if you need a hint.
Ooops... That last one is not answerable. Here is something identical in design, but that is answerable:
D15
1A24
5007
40232
What line comes next? (Extra credit if you can tell me why the puzzle in the last post is not answerable.) Again, let me know if you need a hint.
- Kevin
Ok, here is a hint. Each line is a representation of the number two thousand five hundred sixty seven.
My guess is
101000000111
Correct! Can you tell me why the first puzzle puzzle I gave could not be answered?
- Kevin
Well it starts with base 13, then base 10, base 7, and finally base 4. If we were to continue it farther then the next entry would be base 1. From my understanding there is no way to represent the number 2046 in base 1.
By the way good question Kevin.
The only questions I can think of are sports related so unless you want some crazy sports question like who played 3rd base for the Pirates in 1975 someone else can give it a try.
-Ben
P.S. If you get the joke about the question above then you watch way too much ESPN. I think the year is 1975.
But there is:Quote:
Originally posted by bmacri
Well it starts with base 13, then base 10, base 7, and finally base 4. If we were to continue it farther then the next entry would be base 1. From my understanding there is no way to represent the number 2046 in base 1.
000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Good answer! :pQuote:
Originally posted by solarflare
But there is:
000...000
I'm not so sure about that. For example in oct
2046 = 6*8^0 + 7*8^1 + 7*8^2 + 3*8^3.
or in base 13
2046 = 5*13^0 + 1*13^1 + 12*13^2.
so in base 1 you are saying
2046 = 0*1^0 + 0*1^1 + 0*1^2 + ... + 0*1^2045.
I'm not sure that works. Of course you may have been kidding, I don't know.;)
Well in base one, the digit doesn't matter. I could just as easily have written ***...*** for a clearer meaning. It's the same thing as tally marks.Quote:
Originally posted by bmacri
I'm not so sure about that. For example in oct
2046 = 6*8^0 + 7*8^1 + 7*8^2 + 3*8^3.
or in base 13
2046 = 5*13^0 + 1*13^1 + 12*13^2.
so in base 1 you are saying
2046 = 0*1^0 + 0*1^1 + 0*1^2 + ... + 0*1^2045.
I'm not sure that works. Of course you may have been kidding, I don't know.;)
Quote:
Original question from aio:
Sir Isaac Newton said that "for every action, there is always an opposite and equal reaction" (I think that's the exact word, just correct me if i'm wrong).
We know that if we move a small object, we can easily do it. But how does that statement apply? We cannot, supposed to be, move the object because it will always react according to the force we apply?
When you answer, treat me like a kid please.
Quote:
Doctor Luz:
Well, I think the small object reacts according to the 3rd newton law, however this reaction is not in the small object, is in the object which comunicates the force to the small object (your hand, your finger...).
You comunicate a force to the object and the object comunicates the same force with opposite sense to you.
Then the object moves. If the force you comunicate is instantaneous, the object moves accordingly with the 2nd Newton's law with an acceleration F=m*a.
Quote:
bmacri:
When thinking about forces it's important to remember that you may not be aware of all the forces being applied. Here are a couple of examples that might help.
First imagine you’re standing on solid flat ground and you try to push a 3 kg box. It shouldn't be that hard to move. Now try and push a 300 kg box. That should be almost impossible to move. Now imagine doing this on a frozen lake. If you tried to push the 300 kg box what would happen? The box wouldn't move you would. You would be pushed back away from the box. When you push something, something has to be pushing on you if you don't move. Many times the thing pushing on you is the earth. It's hard to see that when you’re on solid ground but on the ice it becomes much more apparent.
The next question might be if the earth is pushing back why doesn’t it move? Well it does. Now this will really mess with your head but like the good Doctor Luz said Force=mass*acceleration. If you apply a 10 Newton (N) force to an object the earth applies a 10N force to you (assuming you don't move and the object does.) Since the earth is so large compared to the object its acceleration will be very small. And usually completely insignificant. I think I'm starting to ramble now so I'll shut up.
Does that help? If not let me know and I'll try to explain a little better.
Quote:
Elrond
To complete the example of bmacri, if you push a 3 kg box hard enough while standing on ice, it will move forward, but you will also move a little backward. This shows the the box, even small, applies some reactive force upon you.
Quote:
KevinHall
Have your child put on some roller blades and throw something heavy. Have someone behind him (or her) though ready to catch him (or her) b/c they could easily fall. But the point is that both the object thrown and your child will move -- just in an opposite direction. Your child pushes on the object, thus it moves. The object pushes back on your child thus he (or she) moves. They move in the opposite direction.
I tutored for university physics for 3 years. Let me know if you need some more examples or if you have some other questions.
I actually worked this out in my head once. I believe the earth's center of mass moves a distance of only a fraction of a width of a proton (very small fraction if I remember correctly) when we throw a heavy object as far as we can. Just in case anyone was wondering.
Quote:
Elrond
And as it is likely enough that other people will throw the same kind of object in the opposite direction somewhere on Earth, it gives an even weaker result...
Quote:
KevinHall
And given that the earth is not truely a rigid object, the effect that you have on the earth will be mostly local decaying with distance rather quickly (within a couple of yards of where you stand for any significant measurements).
Quote:
bmacri
So if everyone on one side of the earth was to jump up and down at the same time would the earth go flying off into space? Or better yet how many people jumping up and down on the same side of the earth would it take for the earth to be thrown out of orbit?
I'm not really expecting an anwser it's just a question that seems to come up everytime when a discusion like this occures.
Quote:
Elrond:
To be calculated, but it might be more that the number of people that have ever lived on Earth so far. It means that the risk is small enough
Sorry I wasn't able to return immediately.Quote:
KevinHall
No. People (even collectively) do have the strength to overcome the gravity of the earth. The center of mass of the earth would follow the same orbit around the sun and in one second, everything would be back to normal.
It would take an aweful lot. You would need the collective mass of people to much greater (thrown out of orbit) than the mass of the earth. Say the average person weighs 100kg (perhaps a little on the heavy side) and "much greater" would be a factor of at least 100. Then we have 100*(6e24 kg / 100kg) = 6e24 people or 6 septillion people (6 trillion trillion people).
Anyway, thank you friends for the enthusiastic replies. But I think I have to restate (or clarify) the question. To quote myself partly in the original question; "But how does that statement apply?". In other words, it's not a question what will happen after striking the object. It's the application of the statement of the law based on what we know will happen.
Put it the other way around, it seems that the statement is wrong based on what we know (or observed) will happen because based on that statement, the reaction will always be equal. In other words no matter how strong the force we will apply, the object will not move because their is always an equal opposite reaction.
Put it the other way around again, the question probably is that, do I really understand English? :D :D
k
let me post a simple one :D
than i'll continue with some hard ones :p :p :p
continue the following sequence (each line is one item)
(warning cyrillic characters ;) )
qaz!-10VIÀ10
wsx@/11VÁ9
edc#|13IVÂ10
rfv$\17IIIÃ11
tgb%-25IIÄ10
don't argue :) learn the cyrillic alphabet
AIO. Yes the forces are equal and opposite but they act on different bodies, so they do not cancel each other out.
I think I get it now. I was having trouble with the idea of what 0 would represent. I was thinking it would have to represent zero and one in base one. But if I'm following what your saying the representation of zero in base 1 is nothing. 0 is ONE thing and 00 are TWO things and so on.Quote:
Originally posted by solarflare
Well in base one, the digit doesn't matter. I could just as easily have written ***...*** for a clearer meaning. It's the same thing as tally marks.
That's pretty cool. Thanks for the explaination.
-Ben
YeahQuote:
Sir Isaac Newton said that "for every action, there is always an opposite and equal reaction" (I think that's the exact word, just correct me if i'm wrong).
Well I think to make the statement right we have to remove the words "and equal":D.. Newtons didn't think that the statement is obviously one sided - the doer is always weak.
:confused: :confused:
There is no problem with the statement.
souldoug, what I mean is that does the statement applies in real life? What about everytime you press the key in the keyboard, does it react equaly with the same force as you did? I don't think so:D:D
Yes it does, that is why you feel yourself push on it
Let me just state what I think the basic misunderstanding is here.
If you take a single body and apply equal, but opposite forces to it, then it will not move. TRUE. If two people push with equal force on a box from oppostie sides, then it is going nowhere.
The action-reaction pairs of Newton's Third Law are always two separate bodies. The forces do not cancel each other because they are not acting on the same body. PERIOD. If the two bodies have different masses then they react to the equal in magnitude forces differently.