Cryptography

[Fadetoblack]

ok

[vegyjones]

I think your card order is wrong Oldham...

When you take out the 2 matching cards, the other 3 are in number order! :wink :D

[TheOldhamWhisper]

Nope - I'm pretty sure that is the order for No. 7 - The only pair you can remove is the King and Ten to leave JQA in order. It doesn't say anything about taking the matching cards out???

[TheOldhamWhisper]

Of course, I’m now in the position of the other two – because I need to use the picture card in my hand to determine the rest of the cards.

I can’t be holding the Jack – as I can’t deduce anything from doing so. It can’t be a Queen – I can only work out 4 of the cards suits whether it is a club or a spade. So I must be holding a King.

I know that the King (on the board) cannot be red – so I am holding either the spade or the club. If I am holding the Club, I still cannot say for certain what suit the Queen is so I must be holding the Spade.

That means the King on the board is a Club, the Queen must be a Spade (Rule 3) and the A must also be a Club (Rule 4).

Final Order:
K J Q 10 A

So the answer is – Clubs!

[Fadetoblack]

:D

Answer
The two cards on the ends and the card in the middle border a combined number of two suits. So at least one pair of them should be listed as bordering a card of the same suit in clues 1-5. Of the five, only the ace and the queen meet this criterion, so they must be two cards apart. One of them must be in the center, but it can't be the ace (the two cards on the ends would only border hearts - and one of them has to be bordering something else from clues 1-4), so it must be the queen. The ace is not on the left end from clue 8, so it must be on the right end, and the card next to it must be a heart. So far we have this(X represents unknown):
X X Q X A
X X X H X
At least one of the two cards to the left of the queen is lower than the queen (two of the remaining cards are lower, and only one can be right of the queen), so the triplet that is in ascending order is the queen, the ace, and the single card to the left of the queen that is lower than the queen. That means the king is left of the queen, and is also left of both the jack and the ten (otherwise, jack king ace or ten king ace would be a second triplet), so it must be on the far left. The jack must be left of the ten (or else ten jack ace would be a second triplet), so it must be second from the left, and the ten must be second from the right. The only card next to the king is the jack of diamonds (it's a diamond from clue 1). So far we have(T is ten):
K J Q T A
X D X H X
Now we seem to be stuck. We don't have any more clues that can be used. So how did I figure it out?
I held a face card (I said so myself in the intro). I must have known it cannot be that card and deduced the correct answer from there.
But which card did I have? We already know the suit of the jack, so my holding a jack would not help. Could I have held a queen? No, because I know the queen to either be a spade or club (clue 6), so one of the cards next to the queen would have its clue satisfied (see clues 3 and 4), and I could not determine the suit of the card on the other side of it.
Therefore, I must have held a king. But which king? I know the king is not a diamond (clue 6) or a heart (clue 3). If I held the king of clubs, then the king next to the jack would have satisfied the jack's clue, and I could not have determine the suit of either card next to the ten.
Therefore, I held the king of spades. The king of clubs must have been on the far left (only suit available), so the only card next to the jack that could be a spade is the queen, so the only card next to the ten that could be a club is the ace. In summary:
K J Q T A
C D S H C
Therefore there are two CLUBS.
Hide

[vegyjones]

That's how I had it

KQJTA

so that I had 3 cards in a row! Oldham had me worried that I had my crads wrong!

But at least the answers were the same! :)

[Fadetoblack]

Eight and five, last name and given,
We are one six six six even;
The first in cow, the last in oxen
Three in damsel, three in vixen.

Question: What are we called?

[TheOldhamWhisper]

That's how I had it

KQJTA

so that I had 3 cards in a row! Oldham had me worried that I had my crads wrong!

But at least the answers were the same! :)

But Condition No 2 is not fulfilled with the order you had them in Ooo

[vegyjones]

I had the Jack as a heart!

[TheOldhamWhisper]

But that would mean the 10 was a spade and the Ace cannot border a black card!

[vegyjones]

Oops Ooo

[TheOldhamWhisper]

I think I know the riddle but I'm looking for the 1666 connection

Roman Numerals

.

[TheOldhamWhisper]

Can't believe I missed that...

write them in order of value to get 1666

.

[mathare]

Oldham is right and I think I have the connection. Maybe...

Take the C from 'cow' for 100, D, M and L from 'damsel' for a further 1550, X from 'oxen' for a further 10 and V, I and X from 'vixen' for 16 more. Making 1676 So close, that has to be along the right lines

[Fadetoblack]

Very good.

The last name has eight letters (NUMERALS) while the first or given name has five (ROMAN).

Using each of the numerals once gives MDCLXVI, which is 1,666.

C ( which is found in Cow) is first of the seven in the alphabet while X ( which is found in oXen) is last.

Three of seven are found in DaMseL. Three are also found in VIXen.

Explanation of the hint:

I is the only one of five vowels among the seven Roman numerals.

C is found in Cat, M and I in MItten.

I and V are odd numbers. The rest are even numbers.

There are seven Roman numerals. One of them is found in seVen (the V).

[Fadetoblack]

A car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler. The sleuth wasted no time and spared no effort in discovering and carefully examining the available clues. He was able to identify four suspects with certainty that one of them was the culprit.

The four make the statements below. In total, six statements are true and six false.

Suspect A:
1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:
1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:
1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:
1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

Which one is the car thief?

[vegyjones]

D

[TheOldhamWhisper]

.

The Answer is B and A's third question is irrelevant as it could be true or false. There are 2 contradictory statements and we thus need to find 4 more (proven) false statements. Taking each suspect in turn and assuming guilt, we can deduce that only if B is Guilty can we show 6 false statements!

.

[TheOldhamWhisper]

[Fadetoblack]

Consider the lilly. Not really. Consider that six statements are false. A's first statement and C's first statement contradict each other. One of them is false. C's and D's contradict each other. One of them is false. Therefore, there are four additional false statements.

Assume A is guilty. If so, A's second statement, B's second statement, and D's first statement are the additional false statements.

Assume D is guilty. If so, A's second statement, B's first statement, and D's third statement are false. This also only makes five false statements. D did not do it.

Assume C did it. If so, A's second statement, D's first and third statements are false. This again, makes only five false statements.

After ruling out suspects A, C and D, B is the culprit. B's third statement, C's second statement, and D's first and third statements are the additional false statements. This adds up to six.