Barring complex numbers and limiting ourselves to integers...
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#336860 perfect powers
Posted by Prime on 02 December 2013  05:48 AM
#329751 Knight Checker
Posted by Prime on 17 March 2013  04:54 AM
White seems to win in whichever way, provided it has the first move from that position.
If 1...Ne6 x d4, then 2 Nf7 and both d8 and h8 cannot be defended.
If 1...Ne6g5 or d8, then 2 Nd4c6 wins quickly.
If 1 ... Nf6g4, or any other square, then 2 Nd4 x e6
If 1 ... Nd7c5 (or to any other square), then 2 Nd4c6
If 1 ... Nc7anywhere, then again 2 Nd4 x e6 wins.
Obviously, the Knight on e7 cannot move, since it guards c8.
Also, 1 Nb4a6 seems to win quickly.
 1
#329084 Stopping and Turning back hands of time
Posted by Prime on 25 February 2013  08:46 PM
Spoiler forEvery 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..
I see, I mixed up the adavance of the hour hand with the elapsed time for each cycle. And I calculated period incorrectly for the second part in my previous post.
Still, after the corrections, I get a different answer. My hour and minute hands will meet a lot sooner.
Why rejecting the occurences where the clock hands meet at 12 in a midcycle?
I see nothing in the OP prohibiting that. In fact, the clock hands will meet again at 12 on a midcycle before they do so after a whole number of cycles.
1.The full cycle is (12/11 + 12/13 +2) = 574/143 hours elapsed time.
The hour hand advances (12/11+12/13) = 288/143 hours
After 143 such cycles, the hour hand advances 288 hours, which is divisible by 12. (288/12 = 24).
Elapsed time after 143 cycles is 574 hours. And, because the last step of the cycle the hour hand stayed put for exaclty one hour, there was a meeting at the same spot one hour prior. Thus after 573 hours from the start the hour and minute hand meet at the 12hour mark.
2. The hour and minute hands also meet at exact hour mark after 65 full cycles plus 12/11 of an hour. (12/11 + 12/13 + 2)*65 + 12/11 = 132.
It so happens that 132 is also divisible by 12. 132/12 = 11
Thus the first meeting at 12 hour mark will take place exactly after (12/11 + 12/13 + 2)*65 + 12/11 = 262 hours. That is a lot sooner than 573 hours calculated in step 1.
All that is relatively easy to figure out using regular fractions and modular arithmetic. But it does not hurt to verify using the angles in degrees in decimals as calculated by TSLF.
The first step of the cycle the hour hand travels (12/11)*360/12 = 32.72(72).... degrees.
On the third cycle step, the hour hand travels (12/13)*360/12 = 27.69231... degrees.
Making the full cycle: 32.7272(72) + 27.692308 = 60.41958...
After 65 full cycles plus the first step of the next cycle the hour hand travels 60.41958.. * 65 + 32.7272(72) = 3960 (exactly). Which also happens to be divisible by 360. 3960 / 360 = 11.
And so all 12hour mark meetings' elapsed times are as following:
262 + 574*n
263 + 574*n
573 + 574*n
574 + 574*n
The first meeting is after 262 hours.
 1
#329040 Congruent polygons
Posted by Prime on 24 February 2013  09:07 PM
To construct a polygon from given sides and angles, must know where they are in relation to one another.
2). Given n1 angles for ngon, we can figure out the nth angle on our own. It does not represent any additional piece of information.
Or to put it differently, cannot leave more than 2 sides unspecified.
3). It would not hurt to ask whether the shape is not concaved. If it is  must find out all the angles or the pairs of sides that are concaved before embarking on the figure building project.
Given the above, the formula is 2n  3, as has been already noted.
That would be minimum and sufficient information to build an ngon. I.e.:
Shape.......Minimum and sufficient
triangle ...........3
quadrilateral ...5
pentagon.........7
etc.
 1
#327661 All numbers showing
Posted by Prime on 26 January 2013  05:23 AM
Spoiler for had thought...this was an expected value problem with the expected number of rolls needed to get all six numbers equal to the sum of the expected number of rolls needed to get each new number. The average number of rolls to roll a new number being the inverse of the probability to roll a new number. So 14.7?
Yes. Simple concept that cuts thru all the computational complexity.
This concept, does not apply to the question in the OP. The problem called for finding the minimum number of rolls (or dice) to get a winning bet. A winning bet is one, where your probability of winning is better than 0.5. It is not the same concept as average number of rolls (dice) required to get to the objective.
A simple example:
If you roll repeatedly a fair dice, then any given number, say "1", will occur on average once in six rolls. However, if I bet on "1" coming up just after 4 rolls or sooner, I am going to come out ahead. (Same as rolling 4 dice and betting that at least one of them will roll "1".)
P = 1  (5/6)^{4} = 0.5177
The answer to the problem as stated is like Sp and myself have found. I have another equation, which is not recursive, but encompasses all variations. It yields the same result. I'll make a separate post for it.
But of course you still get a winning bet with 14.7 rolls. Your winning chance is better than 60%, but it is not the minimum number of rolls.
 1
#327606 All numbers showing
Posted by Prime on 24 January 2013  12:19 PM
I can get an answer, but no simple explanation. But that should do for everyday gambling needs.
For m=13, the winning chance is approximately 0.51. For m=12, it's about 0.44.
The probability to see just one and the same number after n rolls is 1/6^{n1}.
The probability P(k,n) to get exactly k different numbers (1 < k <= 6) after n rolls is P(k,n1)*k/6 + P(k1,n1)*(7k)/6.
Make a 6column table for k=1 through 6. Populate the first row with 1,0,0,0,0,0. The second row with the formulas above. Then drag the formulas down.
I can't think at the moment of a simple formula instead of all that.
 1
#327530 Ultimate Mate
Posted by Prime on 22 January 2013  09:35 AM
#327396 Insert Coin Here
Posted by Prime on 17 January 2013  08:20 PM
Spoiler for
That's right Prime they were all returned inside the coffee cups.
The trouble with testing the coins in slot machine is ..the proper coins can hit jackpots too.
Also the arcade owner is low on profit so every machine shall be coin operated.
I wouldn’t classify winning with regular coins as “problem”. Just take the winnings home and measure the remaining two coins at leisure using your own precision scale.
But if you insist on using the coin operated balance to the end...
1) If they weigh the same, you’ve got two known regular coins. Pay with one of them and weigh another against one of the remaining 3 returned by the vending machine. If same, the remaining 2 returned coins are your cheating pair. If not  you have found one counterfeit and know which one (heavier or lighter). Set the found counterfeit coin aside pay with your known good coin to weigh the remaining 2 returned coins against one another thus finding the second fake coin. And you still have one spare regular coin, which could be used in another vending machine.
2) If the first trial reveals a counterfeit and its fault (heavy or light), it’s going to cost even less. Use your good coin to pay and weigh 2 of the unknown 3 coins against one another. If they weigh the same, they are all good, so the third coin is the remaining fake. If different, since you know the remaining bogus coin’s fault (lighter or heavier), you can pick it off the balance. Now you have your trick pair, two true coins, and a cup of coffee.
Note, the best case scenario here costs two standard coins for weighing and the worst  three. With my previous method the worst case scenario loses all four standard coins, two for the weighing and two in gambling. The best case costs just one weighing.
All that does not matter, considering you win jackpot, and you have resigned to spend four coins in the coffee machine anyway.
 1
#327335 Chess puzzle
Posted by Prime on 16 January 2013  08:04 AM
#327261 Hitting 137
Posted by Prime on 14 January 2013  09:32 PM
It seems to stand the reason...
I'll try for exact formula later.
 1
#327245 Get the gold coin...
Posted by Prime on 14 January 2013  07:42 AM
A finishing touch...
When you XOR all significant intervals and the result is zero, that is a key position, from which whoever moves must lose the game, provided the opponent plays correctly.
XOR is a binary operation defined as following:
0 xor 0 = 0
0 xor 1 = 1
1 xor 0 = 1
1 xor 1 = 0
Therefore, for ease of calculation convert all significant intervals to binary.
Significant intervals are defined as following:
Pair up neighboring coins starting with the rightmost one. For odd number of coins the leftmost coin must be paired with the left edge of the board. For example, 6 coins must be paired like so: .....c1c2....c3c4......c5c6; whereas 5 coins must be paired: Ec1...c2c3....c4c5, where “E” is the edge of the board.
An interval between the coins of each pair is the number of vacant squares between them. It is called "significant interval" herein. For the leftmost coin (c1) the interval between it and the edge of the board (where applicable) is the number of vacant squares to the left of c1 plus one. However, in the case where the second coin (c2) is the gold coin, and the Ec1 interval is applicable (odd number of coins,) the interval Ec1 shall be just the number of vacant squares to the left of c1.
For example:
_ _ C ..... C _ _G.....C _ C, where the intervals (left to right) are: 3, 2, and 1 (losing on your move);
_ _ C ..... G _ _C.....C _ C, where the intervals (left to right) are: 2, 2, and 1 (must win on your move by moving the very last coin one space).
The intervals outside "pairs" don't matter.
Finally, I must note, there are no significant intervals when the gold coin is the first coin on the left. Just grab it!
The winning strategy is simple: just make sure that after your move all significant intervals XOR to zero. And if you cannot do that – don’t bet money or valuables on the outcome.
Proof:
1). Any terminal position, where your opponent has just one possible move, after which you take the gold coin, conforms to the zero XOR rule.
2). From any position not conforming to zero XOR, you can always obtain a conforming position with a single move.
3). From any position conforming to zero XOR, you can never obtain another conforming position with a single move.
The reason for constructing significant intervals starting with the rightmost coin is, that you cannot leave the last coin hang unpaired. If you did, then one could move it without affecting any significant intervals in violation of the point (3) of the proof above.
Armed with this knowledge, you could make a bet, whereby first your opponent sets up any position and then you choose who moves first.
 1
#326702 Foiled Again
Posted by Prime on 15 December 2012  06:54 PM
I have enjoyed this problem. It looks like something original in the old family of weighting problems. The solution I found seems to be the only solution.
The case where both sets of 8 weight the same is more difficult. There are 3 possibilites: each of the equal sets of 8 have 2 doubly wrapped coins, or they each have one, or both have zero coins. Correspondingly, the set of 11 on the side has zero, 2, or 4 heavy coins.
Lets name one set of 8 W1, the other set of 8  W2, and the set of 11 coins left on the side  S11.
Then the second weighting should be as follows:
Leave W1 in the pan and put 4 more coins from W2 into the same pan. Lets call those 4  W2(4). Put one coin from W2 into the other pan, lets call it W2(1). Add the entire S11 into that pan. That leaves 3 coins from W2 on the side. Lets call them W2(3).
Thus after the second weighting of 12 coins against 12, the solution is as follows:
1. W1 + W2(4) > (weighs more than) W2(1) + S11, then the entire S11 is guarantied to be standard.
2. W1 + W2(4) = W2(1) + S11, then W2(1) + W2(3) have to be standard.
3. W1 + W2(4) < W2(1) + S11, then W2(4) must have standard coins only.
Here is the reasoning behind that solution:
1, If S11 contains standard coins only, its side of the balance should always be lighter. That is because in that case W1 contains 2 doubly wrapped coins and the balance side of S11 + W2(1) can at most have one. Also, this is the only case when that side could be lighter because in the other cases S11 contains either 2 or 4 heavy coins. Thus when S11 + W2(1) is lighter, S11 must be all standard coins.
2. If S11 has 2 doubly wrapped coins then
a). to balance the scale W1 + W2(4) must have the remaining 2 heavy coins, which leaves W2(1) + W2(3) standard.
b). when S11 + W2(1) side is heavier, knowing that W1 contains 1 heavy coin, leaves W2(4) standard.
3. Finally, when S11 contains all 4 doubly wrapped coins making its side of balance heavier, we can safely pick W2(4) as a set of standard coins.
I am curious as to how this problem was constructed.
 3
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