googon97
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Posts posted by googon97
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sorry apparently pressing space 6 times posts it 6 times.
1 2 3 4 5 6
0 0 0 0 0 0 0
1 2 6 12 20 30 42
2 6 16 30 48 70 96
3 12 30 54 84 120 162
4 20 48 84 128 180 240
5 30 70 120 180 250 330
6 42 96 162 240 330 432
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f(x,y)=(xy mod 4)-2
x\y 1 2 3 4 5 6
0 -2 -2 -2 -2 -2
1 -1 0 1 -2 -1 0
2 0 -2 0 -2 0 -2
3 1 0 -1 -2 1 0
4 -2 -2 -2 -2 -2 -2
5 -1 0 1 -2 -1 0
6 0 -2 0 -2 0 -2
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I believe that there is no statement in the question that describes the road as infinite; I believe it can't be. It also doesn't say the road is tangent against the circle with a radius of 1 mile. If the road is starting at 1 mile away and continuing away, the minimum path is 1+2pi.
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0^0=1
That is undefined
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Using 0's and mathematical operators, create a positive number.
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2,4,6,6,6,8,10,12,11,11,?
33,36,39,47,55,72, ?
6,2,5,5,4,5,7,?
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NO it's impossible. Imagine the chess board pattern. The opposite corners are the same color. One rectangle can cover on tile of each color. That leaves two of the same tile that can't be covered with the last rectangle.
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I do agree it does say spoons.
115 118 113 110 120 117 116 112 121 114 are the spoons. The assistant draws one spoon each time.
Very nice, and I thought you had me on that point.
But I read the OP again, and sure enough, it says ...
You disagree and ask her to replace the spoons and repeat the process,
But you get honorable mention anyway. Your solution works except for the red letter.
However, if you didn't know how many spoons were taken, would you ask please replace the spoon?
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115 118 113 110 120 117 116 112 121 114 are the spoons. The assistant draws one spoon each time.
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Start with ABC. After the operation it is ABC - A - B - C. In other words, add the digits and subtract it from the original number. Therefore 689 = 666
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If Googon has a normal garage opener with buttons 0-9 and enter and a code of 3 digits, it takes 3,000 button pushes to ensure the garage to open. Googon's Garage Gadget Guys builds a garage opener with 10 buttons. Googon realizes that there is no room for an enter button (he can't just take out a number 0-9). His garage opener opens if the code is pushed at any time. For example pushing 0123 could open the garage if the code is 012 or 123. Googon is worried that the garage opener is not safe. He needs your help on the issue (seriously he doesn't know the answer). How many times does someone need to push a button to ensure the garage will open? Can you prove that? Is their a pattern to the button pushing?
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A n-agon can be broken into n-2 triangles. Each triangle need three peces of information. But, after giving the information for the first triangle, every other triangle already has one peice of information, one side. Besides the first triangle, each triangle needs 2 pieces of information. The information needed for this would be 3+(2(n-3)) or 2n-3.
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3333....3 is 3*1111....1
The number consisting of only ones must be a multiple of 3 and 1111...1. For it to be a multiple of 111....1, it must be a multiple of 100 1's long. The first multiple of 100 divisible by 3 is 300. So, the number is 111...1 with 300 1's.-
1
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A sees two 3's. Everyone knows 242 doesn't work.
Right... we know that everyone sees a 3, but we only know that as the readers of the puzzle. C does not know that A sees two 3's, because C thinks he might have a 2 on his forehead. And if C
did have a 2, then B could not know that A sees any 3's at all, since B might have a 4 on his forehead. So, since C doesn't know his own number yet, there is no reason for him to expect (nor is there reason for C to expect that B would expect) that A could deduce that his number was 3. The hard part is keeping track of who knows what in each circumstance.A sees two 3's he knows B and C see at least one. That works for all people.
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C thinks B thinks, if A saw 42, he would either have a 2 or a 3. Everyone already knows that 242 is not an option. So, C thinks B thinks, that if it was 342, A would have said 3.
I'm not sure I follow you there.
If A saw 4 and 2, then A could not know that 2,4,2 is an invalid combination (as you mentioned, he would know that he had a 2 or a 3 but could not tell which).
Everyone knows that there are at least two 3's. Everyone also knows that everyone else can see at least one 3. 242 has no 3's.
This is not true if B has a 4 and C has a 2. To elaborate: C knows he has either a 3 or 2. As you mentioned, C knows that
if he has a 2, then B does not know whether he has a 3 or a 4. Furthermore, if C has a 2, then C knows that B knows that if he has a 4, A must be wondering whether he has a 2 or a 3. But in such event, A would see a 4 and a 2, and therefore A would not have any way of knowing for certain that someone has a 3. In summary: C knows that if he has a 2, then B cannot assume that A is aware that any of them definitely has a 3.A sees two 3's. Everyone knows 242 doesn't work.
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C thinks B thinks, if A saw 42, he would either have a 2 or a 3. Everyone already knows that 242 is not an option. So, C thinks B thinks, that if it was 342, A would have said 3.
I'm not sure I follow you there.
If A saw 4 and 2, then A could not know that 2,4,2 is an invalid combination (as you mentioned, he would know that he had a 2 or a 3 but could not tell which).
Everyone knows that there are at least two 3's. Everyone also knows that everyone else can see at least one 3. 242 has no 3's.
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C knows that if B saw 3 and 2, he would say 3 because 342 was eliminated.
Careful to track who has eliminated what. The combination 342 was eliminated on the basis that C can see that A and B have 3's. But B only knows he doesn't have a 4 if neither A nor C has a 2; and C can't know that he doesn't have a 2. So, from C's perspective, if C has a 2, then B can't eliminate 342 as a possibility. Since C can't infer that B can eliminate that case (even though we know he really can, since C actually has a 3), C cannot proceed to deduce his number from there.
C thinks B thinks, if A saw 42, he would either have a 2 or a 3. Everyone already knows that 242 is not an option. So, C thinks B thinks, that if it was 342, A would have said 3.
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3
Everyone knows that everybody knows that there must be one 3.
That leaves us with:
233
323
332
333
234
243
324
342
423
432
Everyone knows that the other two have 3's and they either have a 2 or 3. Only the person with a 2 in the scenario, doesn't know that two 3's and a 2 is wrong. The other two will know each other know.A will say I don't know. B and C know that A either saw 2 and 3 or 3 and 3. This eliminates 234, 243, 324, 342. This makes the possibilities:
233
323
332
333
423
432
It becomes B's turn. B sees 3 and 3. C knows that B sees either 3 and 3 or 3 and 2. C knows that the possibilities are 332 and 333. C knows that if B saw 3 and 2, he would say 3 because 342 was eliminated. B doesn't know whether he has 3 or 2, so he says I don't know. C can then guess 3.
S & C & S & C
in New Logic/Math Puzzles
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