Thalia
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Posts posted by Thalia
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Think this one has been posted before.
EDIT:
Divide the coins into 2 groups of 10. One group has X heads and Y tails. The other has Y heads and X tails. Flip all the coins in one group.
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On 3/15/2018 at 5:20 PM, flamebirde said:
Incidentally, the title was a hint too: plain in color, and also plain -->plane, since a piece of paper is flat.
Lol. That's what lead to one of the other guesses. Plain to see--> crystal clear.
Crystal would be fancier drinking cups though. And I couldn't quite call it white.
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Lol. I saw people making it on tv once. The process is pretty impressive. Think it was a specialized kind though. Not good for mass production. Nice riddle.
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Still struggling with the last line.
Paper?
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Got the same letter and number equivalents as Pickett. I'm reading it as
ONE 1(ONE) + 2(ONE) + 4(ONE) + 17 = ONE (1+2+4)(ONE) +17 = ONE 7(ONE) +17 = ONE DONE MD? No clue what that means though... -
Maybe not quite white but
Ice or crystal?
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Assuming the 99% refers to weight...
That seems like a ton of water to evaporate off.
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If he's extremely lucky and the first 26 cards are black, then the best is 100%. Odds are greater than 50% any time there's more black than red cards on the table. Don't know if there's an ideal average number of black cards to have over the red cards though.
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Well, there goes your curve estimation. I don't remember the situation but there were certain combinations that I could calculate with some multiplication. I think one of them was a CE combination. But of course the other combinations didn't feel like cooperating. I'll try to work on this over the weekend.
Agree on your pink combinations except for XMEE. Should be the same count as EEM, right?
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Make an X? Length would be 2*sqrt2 mi. Seems too easy though...
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Sorry to say I didn't save my counts. Slightly burnt out from counting, recounting, and rerecounting the previous ones. lol. Maybe I'll try again when I've had more time to recover.
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#1
1/4?
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I did some quick counting for the easier ones. I've noticed that at least up to this point, the number of combinations of C, E, M, and X is (# cubelets removed+1)^2. There are 25 combinations for 4 cubelets removed. The ones including X would be the same as the counts for 3 cubelets without X. I did not count the combinations that have 2 or more E's or CCEM, CCMM, AND CEMM. I got 191 for the rest. I'm guessing the ones I didn't count are going to be pretty high so maybe around 500?
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3 hours ago, rocdocmac said:
@Thalia ...
I actually noticed something in this line yesterday!
T26<>1: If 26 cubelets are removed, one would be left with either X, M (= F), C or E, i.e. T26 = 4 (not "1"!).
We now know that T3 = 139.
Thus T0, T1, T2, T3, ... = 1, 4, 22, 139, ...
The sequence should therefore end with
... T24, T25, T26, T27 = ... 139, 22, 4, 0
Maximum values would be for T13 = T14
When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube.
Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?
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A couple more revisions because I can't tell the difference between counting to 4 and counting to 6 at night.
SpoilerMMM (2) 5.11.23 5.11.15 CCC (3) 1.3.7 1.3.27 1.9.21 CCX (3) 1.3.14 1.9.14 1.14.27 CEX (4) 1.2.14 1.6.14 1.8.14 1.14.18 CMX (2) 1.5.14 5.14.21 EEX (5) 2.4.14 2.8.14 2.14.16 2.14.18 2.14.26 EMX (3) 2.5.14 5.12.14 5.14.20 MMX (2) 5.11.14 5.14.23 CCM (8) 1.3.5 1.3.13 1.3.17 1.5.9 1.9.11 1.9.13 1.9.23 1.5.27 CMM (5) 1.5.11 5.7.11 5.9.11 5.11.25 1.5.23 EMM (9) 2.5.11 4.5.11 5.6.11 5.8.11 5.11.16 5.11.18 5.11.26 2.5.23 5.10.23 EEM (18) 2.4.5 2.5.8 2.5.10 2.5.12 2.5.16 2.5.18 2.5.20 2.5.22 2.5.24 2.5.26 5.10.12 5.10.16 5.10.20 5.10.22 5.10.24 5.10.26 5.20.22 5.20.26 CCE (16) 1.3.2 1.3.4 1.3.6 1.3.8 1.3.16 1.3.18 1.3.22 1.9.2 1.9.4 1.9.10 1.9.12 1.9.20 1.9.22 1.2 .27 1.27..8 1.18.27 1.24.27 EEE (13) 2.4.6 2.4.10 2.4.12 2.4.16 2.4.18 2.4.20 2.4.22 2.4.24 2.4.26 2.8.20 2.8.22 2.16.24 2.18.22 CEE (22) 1.2.4 1.2.6 1.2.8 1.2.12 1.2.16 1.2.18 1.2.20 1.2.22 1.2.24 1.2.26 1.6.8 1.6.12 1.6.16 1.6.18 1.6.22 1.6.24 1.6.26 1.8.12 1.8.18 1.8.24 1.8.26 1.18.24 CEM (24) 1.2.5 1.4.5 1.5.6 1.5.8 1.5.10 1.5.12 1.5.16 1.5.18 1.5.20 1.5.22 1.5.24 1.5.26 2.5.19 4.5.19 5.6.19 5.8.19 5.10.19 5.12.19 5.16.19 5.18.19 5.19.20 5.19.22 5.19.24 5.19.26 New total is 139.
Thanks for checking all these. That's a lot of numbers!
On 1/23/2018 at 12:55 AM, rocdocmac said:I wonder whether an equation exists that one could use to predict the next number (4-cubelet removal or T4) and higher T-numbers in the sequence. Such a sequence should start with:
T0, T1, T2, T3, ..., (i.e. 1, 4, 22, 13?, ..., with T0 = 1, indicating a solid cube resulting from 0-cubelet removal),
and end with:
..., T23, T24, T25, T26, T27 (i.e. 13?, 22, 4, 1, 0 with T27 = 0, which is equivalent to the removal of all 27 cubelets leaving nothing.
The maximum value would be for T13.
T26 would leave 1 cube. But T25 leaving 2 cubelets leaves more than 4 shapes assuming the two cubelets don't have to be connected. . .
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Full guess with numbered cubelets and revision to CEE.
SpoilerMMM (2) 5.11.23 5.11.15 CCC (3) 1.3.7 1.3.27 1.9.21 CCX (3) 1.3.14 1.9.14 1.14.27 CEX (4) 1.2.14 1.6.14 1.8.14 1.14.18 CMX (2) 1.5.14 5.14.21 EEX (5) 2.4.14 2.8.14 2.14.16 2.14.18 2.14.26 EMX (3) 2.5.14 5.12.14 5.14.20 MMX (2) 5.11.14 5.14.23 CCM (8) 1.3.5 1.3.13 1.3.17 1.5.9 1.9.11 1.9.13 1.9.23 1.5.27 CMM (5) 1.5.11 5.7.11 5.9.11 5.11.25 1.5.23 EMM (9) 2.5.11 4.5.11 5.6.11 5.8.11 5.11.16 5.11.18 5.11.26 2.5.23 5.10.23 EEM (18) 2.4.5 2.5.8 2.5.10 2.5.12 2.5.16 2.5.18 2.5.20 2.5.22 2.5.24 2.5.26 5.10.12 5.10.16 5.10.20 5.10.22 5.10.24 5.10.26 5.20.22 5.20.26 CCE (16) 1.3.2 1.3.4 1.3.6 1.3.8 1.3.16 1.3.18 1.3.22 1.9.2 1.9.4 1.9.10 1.9.12 1.9.20 1.9.22 1.2 .27 1.27..8 1.18.27 1.24,27 EEE (11) 2.4.6 2.4.10 2.4.18 2.4.20 2.4.22 2.4.24 2.4.26 2.8.20 2.8.22 2.16.24 2.18.22 CEE (23) 1.2.4 1.2.6 1.2.8 1.2.12 1.2.16 1.2.18 1.2.20 1.2.22 1.2.24 1.2.26 1.6.8 1.6.12 1.6.16 1.6.18 1.6.22 1.6.24 1.6.26 1.8.12 1.8.18 1.8.24 1.8.26 1.18.24 1.18.26 CEM (24) 1.2.5 1.4.5 1.5.6 1.5.8 1.5.10 1.5.12 1.5.16 1.5.18 1.5.20 1.5.22 1.5.24 1.5.26 2.5.19 4.5.19 5.6.19 5.8.19 5.10.19 5.12.19 5.16.19 5.18.19 5.19.20 5.19.22 5.19.24 5.19.26 New total is 138.
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Not sure if these are supposed to go together or be 6 separate things.
SpoilerDid Amos sell his bicycle *TO* an old friend? Sell an old friend to his bicycle? (!?!?!?) Sell his bicycle which he considers an old friend?
When I see something has one letter, I think of an envelope or the words "a" or "I".
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New number for EEE. But I'm still getting the same result for CEE and CEM.
SpoilerMMM (2) 2 CCC (3) 3 CCX (3) 3 CEX (4) 4 CMX (2) 2 EEX (5) 5 EMX (3) 3 MMX (2) 2 CCM (8) 8 CMM (5) 5 EMM (9) 9 EEM (18) 18 CCE (16) 16 EEE (11) 11 CEE (21) 21 CEM (24) 24 Total is 136.
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Only 13 this time.
SpoilerMMM (2) CCC (3) CCX (3) CEX (4) CMX (2) EEX (5) EMX (3) MMX (2) CCM (8) CMM (5) EMM (9) EEM (18) CCE (16) -
CEX
I picked an edge and went through the 8 corners from there. The core doesn't really matter at this point so if we look at it as removing CE...
The corners are 1, 3, 7, 9, 19, 21, 25, and 27. So starting with cubelet number 2 as the edge (numbered starting with the side facing us), you can remove corner 1. But if you number from the top face, that shape can be called 1,3. Each time you remove an corner, there is a second corner that will give you the same shape. 2,7 from the side is 2,21 from the top. 9 pairs with 19. 25 pairs with 27. So I'm counting 4. What am I missing?
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Recount. EEM and EEC intentionally left out.
SpoilerMMM (2) CCC (3) CCX (3) CEX (4) CMX (2) EEX (5) EMX (3) MMX (2) CCM (8) CMM (5) EMM (9) EEE (12) CCE (16) CEM (24) About CEX
SpoilerWhen I had 4, you said to check it. Unless I'm missing something, the shapes involving the core should have the same counts as in the two cubelets question. CE made 4 shapes. CEX should make 4 as well...
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Lol. I'm not sure what my score is anymore. Can you score this?
MMM (2)
CCC (3)
CCX (3)
CEX (4)
CMX (2)
EEX (5)
EMX (3)
MMX (2)
CCM (8)
CMM (5)
EMM (8)
EEE (11)
EEM (18)
CCE (15)
CEE (21)
CEM (24) -
Nevermind the CEX numbers. I really need to find a cube I can hold...
New math
in New Logic/Math Puzzles
Posted
flamebirde - last 2 digits
Subtract b from the sum of ab and ac.
35+14=49
49-5=44