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Posts posted by flamebirde


On 11/28/2019 at 9:41 AM, TimeSpaceLightForce said:All other figures and the boarder design can also be formed from that 5 polyominoes.
@tumb's up rocdocmac
out of curiosity: was there any significance to the translations of the symbols, or was that just a red herring?

I agree with RocDocMac. Justification given below.
SpoilerCarver is male (Jack offers to "adopt" the Carver's son for the banquet), and is not Jasper (since Jasper's mother has passed away, but the Carver mother answered the phone): therefore Carver must be Jack or Jim. But Carver also cannot be Jack, since Jack's father asked the Carvers to go to the Father Son banquet with the Carver son, implying they are two different families. Therefore, it must be Jim Carver. (This assumes that Jasper's father has not remarried.)
"The Clarks and Carters, staunch Republicans who are very good friends..." implies to me that both Carter parents and both Clark parents are still alive (since one would not say "the Clarks" or "The Carters" if only one Clark/Carter parent is still alive), and therefore Jasper is neither Clark nor Carter. Therefore, Jasper must be either a Clayton or a Cramer. I will assume that Jasper is a Cramer.
Clark and Carter have opposite genders, or else their staunchly Republican parents would be unhappy, and the parents know each other.
If Jasper is a Cramer, then Jack must be either Clark or Carter, and dating Janice. If he is a Carter, than Jane cannot be a Clark, and vice versa (he is the only male unassigned). If he were a Carter, then he must be dating Janice Clark, leaving Jane Clayton (if they were dating then the parents know each other, and Jack and Jane's parents have never met). But wait: the Claytons would name a daughter "Janice"; therefore this cannot be the case. If Jack is instead a Clark, then he must be dating Janice Carter, which again leaves a Jane Clayton, again an impossibility. Therefore, Jasper cannot be a Cramer, and must be a Clayton.
If this is the case, then Jack and Janice are dating and hence Jane must be neither a Carter or a Clark, making her a Cramer. Since Jack and Jane's parents have never met, Jack must be a Clark, and Janice a Carter (who has played on the same athletic teams as Jane).
Hence: Jim Carver, Jasper Clayton, Jane Cramer, Jack Clark, and Janice Carter are the solutions to the puzzle.

1 hour ago, mtngoat said:Happened on this site this afternoon, and saw your post. Looks like fun. Here's your answer.
I J C F D E G B A H
SpoilerAhhh... sucks to be beaten to the punch after investing so much time! Congratulations on the answer. How'd you arrive at it? (I guess my very first conclusion was wrong, and it threw everything else off...)

19 hours ago, flamebirde said:From player II to player III, only 3 positions changed: 1, 5, 10. Player III only gained 2 correct, though, so two of those positions changed are correct and one is incorrect.
From III to IV, 3 positions change again: 5, 8, 9. No score changes, however, which could be a result of a few situations:
I) 5A, 8E, 9B (1 right, two wrong)
II)5B, 8A, 9E (1 right, two wrong)
III) 5E, 8B, 9A (1 right, two wrong)
IV) 5E, 8A, 9B (all are wrong in both cases)
E, A, and B must all fit in 5, 8, or 9. Otherwise, the score would be out of necessity lower than 7 (it would have to be 6 or lower).
Therefore, players II, V, and VII all have their spot 5 wrong, since those players chose a letter besides E, A, or B for spot 5. Player I has spot 8 wrong and players I and VI have spot 9 wrong by the same logic for spots 8 and 9.
Also, player I has spot 1 and 2 wrong, player VII has spot 7 wrong, and player V has spot 10 wrong, since E, A, and B must be in either 5, 8, or 9.
so, player I has spots 1, 2, 8, and 9 wrong for sure. player II has spot 5 wrong, player V has spots 5 and 10 wrong, player VI has spot 9 wrong, player VII has both 5 and 7 wrong.
From IV to V, 3 positions change again: 5, 8, 10. No score change again.
But wait: we know for a fact that V has 5 and 10 wrong.
From III to V, 3 positions change again: 5, 9, 10. No score change again.
In fact, between any two players with a score of 7, only three positions change.
IIIIV 5, 8, 9.
IVV 5, 8, 10.
VVI 5, 9, 10.
VIVII 5, 7, 9.
VII is the only one that changes spot 7 between all players. (we also know that VII has both 5 and 7 wrong, as per above).
Comparing V to VII, 5, 7, and 10 change. We know that V is wrong in 5 and 10, and VII is wrong in 5 and 7, and only 3 are wrong in either V or VII. Two options, then:
I)It's possible that both V and VII have 5, 7, and 10 wrong, making the correct sequence 5E, 7H, 10B.
II) otherwise, some number they share in common is incorrect, and V has 7 correct and VII has 10 correct, meaning 7G and 10H.
The next step is to walk through those last two scenarios... but I'm tired, and it's getting late. I'll leave it for someone else, or otherwise when I have time. Am I on the right track, Rocdocmac?
It's been a long, long time since I played mastermind....
Alright, here we go...
SpoilerIn either case, 7 must be either H or G, and 10 must be either B or H. So, to extend: players I, II, and V all have slot 10 wrong.
to recap:
I has 1, 2, 8, 9, 10 wrong.
II has 5 and 10 wrong.
V has 5 and 10 wrong.
VI has 6 wrong.
VII has both 5 and 7 wrong.
But hold on  if case I is true, and the correct sequence is 5E, 7H, 10B, then this contradicts our initial conclusion above: i.e., that E, A, and B must correspond to slots 5, 8, and 9. B cannot fit in slot 10, or else the given scores cannot be correct!
Therefore, case II must be right. V has 7 correct and VII has 10 correct: 7 must be G, and 10 must be H.
This means that a G anywhere besides 7 is wrong, and a H anywhere besides 10 is wrong. so II must also have 1 wrong.
Next, we analyze III vs VI. What we see is really interesting: III and VI differ only in number 9, and both give a result of 7. Therefore, neither E nor I can fit in spot 9. Players II and III have spot 9 wrong.
to recap:
I has 1, 2, 8, 9, 10 wrong.
II has 1, 5, 9 and 10 wrong.
III has spot 9 wrong.
V has 5 and 10 wrong.
VI has 6 wrong.
VII has both 5 and 7 wrong.
So... if E cannot fit in spot 9, and also cannot fit in spot 5, then E must fit in spot 8, forcing 5 to be A and 9 to be B. So, 5A, 7G, 8E, 9B, and 10H must be correct. Putting this together with the initial four letters I J C F, and via process of elimination, I think this is the correct sequence:
I J C F A D G E B H
Except... it can't be, because III only two letters off and only has a score of 7. I've made a mistake somewhere, and I'm not sure where: I think I can't assume that the first four letters are I J C F. If we look back at II and III, then we see that they differ in 1, 5, and 10, for a net change of 2. Since we know for a fact that II has gotten 1, 5, and 10 all wrong, then that means III must have two of those three positions correct. We also know that 10 must be H, and 5 must be A, making the I in position 1 incorrect.
But this leads to a whole slew of problems with Player V's results... I don't trust the idea of I in position 1 being wrong, not yet at least. I think the italicized section above is where my reasoning goes wrong, but I'm not sure. I think I'll leave it here for now.

SpoilerFrom player II to player III, only 3 positions changed: 1, 5, 10. Player III only gained 2 correct, though, so two of those positions changed are correct and one is incorrect.
From III to IV, 3 positions change again: 5, 8, 9. No score changes, however, which could be a result of a few situations:
I) 5A, 8E, 9B (1 right, two wrong)
II)5B, 8A, 9E (1 right, two wrong)
III) 5E, 8B, 9A (1 right, two wrong)
IV) 5E, 8A, 9B (all are wrong in both cases)
E, A, and B must all fit in 5, 8, or 9. Otherwise, the score would be out of necessity lower than 7 (it would have to be 6 or lower).
Therefore, players II, V, and VII all have their spot 5 wrong, since those players chose a letter besides E, A, or B for spot 5. Player I has spot 8 wrong and players I and VI have spot 9 wrong by the same logic for spots 8 and 9.
Also, player I has spot 1 and 2 wrong, player VII has spot 7 wrong, and player V has spot 10 wrong, since E, A, and B must be in either 5, 8, or 9.
so, player I has spots 1, 2, 8, and 9 wrong for sure. player II has spot 5 wrong, player V has spots 5 and 10 wrong, player VI has spot 9 wrong, player VII has both 5 and 7 wrong.
From IV to V, 3 positions change again: 5, 8, 10. No score change again.
But wait: we know for a fact that V has 5 and 10 wrong.
From III to V, 3 positions change again: 5, 9, 10. No score change again.
In fact, between any two players with a score of 7, only three positions change.
IIIIV 5, 8, 9.
IVV 5, 8, 10.
VVI 5, 9, 10.
VIVII 5, 7, 9.
VII is the only one that changes spot 7 between all players. (we also know that VII has both 5 and 7 wrong, as per above).
Comparing V to VII, 5, 7, and 10 change. We know that V is wrong in 5 and 10, and VII is wrong in 5 and 7, and only 3 are wrong in either V or VII. Two options, then:
I)It's possible that both V and VII have 5, 7, and 10 wrong, making the correct sequence 5E, 7H, 10B.
II) otherwise, some number they share in common is incorrect, and V has 7 correct and VII has 10 correct, meaning 7G and 10H.
The next step is to walk through those last two scenarios... but I'm tired, and it's getting late. I'll leave it for someone else, or otherwise when I have time. Am I on the right track, Rocdocmac?
It's been a long, long time since I played mastermind....

I'll provide the classic answer for a classic riddle for anyone who hasn't heard it before, but I do like Ed's answer quite a bit!
SpoilerA hole.

Huh. I saw the sequence 10^{2}, 11^{2}, 12^{2}, 11^{2} + 9^{2}, 12^{2}+10^{2}, and so i went with 13^{2}+11^{2}.

I know I'm a little late, but:
Spoiler290?

Is the owner of the gun automatically the murderer?
SpoilerIt was Joe
Initial assumption: If Bill was shot with Joe's gun then that makes Joe the killer. Therefore, if we assume this statement to be true, then Jack didn't kill Bill, making that statement true as well, making the other two false.
Jack's statements:
Jack:
I didn't do it. True  assumption.
Al did it. False  by process of elimination.
Joe and I were at the movies last night when Bill was shot. False  by process of elimination.
Bill was shot with Joe's gun. True  assumption.Now onto Joe's statements:
Joe:
I was asleep when Bill was shot. False  he had to be awake to shoot Bill.
Al lied when he said that Tom killed Bill. True  Joe killed Bill, not Tom.
Jack is the only one of us who owns a gun. False  Joe owns a gun, the one used to kill Bill.
Tom and Bill were pals.True by process of elimination.And for Tom?
Tom:
I've never fired a gun in my life. True  by process of elimination.
I don't know who did it. True  by process of elimination.
Joe doesn't own a gun. False  Joe shot Bill with his own gun.
I never saw Bill until they showed me the body False  Tom and Bill were pals.Lastly, Al:
Al:
I didn't do it. True  Joe did it.
Tom did it. False  Joe did it.
Sure I own a gun. True  by process of elimination.
Joe and I were playing poker last night when Bill was shot. False  Joe could not have played poker at the same time as shooting Bill. 
SpoilerSo the chances that the first two will match is 1/52. That much should be fairly obvious I think.
If they do, then the chances of the next two matching is 1/51, and if they do then the chances of the next two are 1/50, and so on.
If a pair doesn't match, then we're looking at an issue  at least one more pair is guaranteed not to match... but there's some finicky stuff going on there, since you could have a loop of 3 or more unmatched pairs if it chains down a bit. So at minimum if you draw a nonmatching card then you have at least 2 unmatched pairs right there and potentially more. Not to mention that you could spawn different "loops" in one run through.
if a pair doesn't match, then the chances of the next pair matching are... what? That's where the riddle lies next I guess. If we number the first deck 152 and the second 1'52' and the very first pair is 12', then we know that the next pair (the one starting with 2) must also not match. Then there's a 1/51 chance that the loop ends there  a 1/51 chance that the next pair is 21', in which case the probabilities continue as before. But there's also a 50/51 chance that the chain extends, and we draw something like 25', which means that now the fifth pair (the one starting with 5) is guaranteed to be a nonmatch with a 1/45 chance of terminating.
I'm sure there's an easier way to think about it, but I'm gonna let it percolate through the back of my head for now.
Seems similar to that airplane seating problem a while back, I'll see if I can't find and link it here.
Edit: here it is:

Spoilerscrabble?

Spoilermaybe vowels?

Spoiler$0 each  the lottery only has three winners?
It's almost certainly a math question, and maybe to do with the names, but it doesn't hurt to try a first guess.

A more religious guess...
SpoilerJesus Christ?

SpoilerAt the bottom of the ocean, presumably...

That one stumped me for quite a while, and for such a simple pattern too! I'm definitely kicking myself over this cute little puzzle.

19 hours ago, rocdocmac said:Both Captain Ed and Thalia have transmitted vital clues in their previous remarks. Eish man, now you ought to get it!
Spoiler../  .... .. . ./ .. / ... .. . .. .. . ../ ..  /  .. !
What hath God wrought!

D'oh. Not a language...
Spoilerbut more like a code?

1 hour ago, rocdocmac said:Flamebirde, you're going to kick yourself!
Haha, you say that like I'm not already! Am I even on the right track? I think I'm about ready to throw in the towel, Rocdocmac  you've certainly stumped me.

Seems I'm the only one left in the dark now... but I will perservere:
SpoilerLatin/Greek alphabets? I'm assuming this has to do with some other language besides English, potentially an obscure one.

Spoilersomething to do with unicode maybe? it can't have to do with the order of the letters of the alphabet... I have a feeling my mind is too burnt out on organic chemistry to puzzle this one out.

I must admit I am also incredibly lost...
SpoilerDoes it have something to do with the pronunciation of the number with the given letter/phonemes?

Spoilerpancreas?
If not, I would guess the word ends in a double "ss" or something similar.

Spoilera chain?
The JayCees
in New Logic/Math Puzzles
Posted · Report reply
I didn't make that connection, primarily because I assume that Jasper is male and hence that clue couldn't be used here (I wasn't sure about the gender of the name, but google says predominantly male).