The last letters of the words one, two, three, four, five, six.
??
Puzzle 2
At the local sweet shop, three particularly nice sweets are on special offer.
A Nobbler is over three times the price of a Sparkle. Six Sparkles are worth more than a Wibbler. A Nobbler, plus two Sparkles costs less than a Wibbler. A Sparkle, a Wibbler and a Nobbler together cost 40p.
Can you determine the price of each type of sweet?
Reasoning
By (3) a Nobbler, plus two Sparkles costs less than a Wibbler, therefore a Wibbler must be the most expensive sweet.
By (1) a Nobbler is over three times the price of a Sparkle, therefore a Sparkle must be the cheapest sweet.
So the order of sweets, from the least to most expensive, is Sparkle, Nobbler, Wibbler.
If a Sparkle was 1p, by (2) a Wibbler could only be up to 5p, by (4) a Nobbler would cost at least 34p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 2p, by (2) a Wibbler could only be up to 11p, by (4) a Nobbler would cost at least 27p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
If a Sparkle was 3p, by (2) a Wibbler could only be up to 17p, by (4) a Nobbler would cost at least 20p, which is more than a Wibbler and isn't allowed as the Wibbler is the most expensive sweet.
So a Sparkle must be at least 4p.
If a Sparkle was 4p, by (1) a Nobbler must be at least 13p, by (4) a Wibbler would cost 23p. This combination matches all of the clues and is a possible solution.
If a Sparkle was 4p and a Nobbler 14p, by (4) a Wibbler would cost 22p. This would not satisfy (3). And if we increase the price of a Nobbler, (3) is never satisfied.
If a Sparkle was 5p, by (1) a Nobbler must be at least 16p, by (4) making a Wibbler at most 19p. This would not satisfy (3).
If we increase the price of a Sparkle or Nobbler further, (3) is will never be satisfied.
Therefore, the only solution we came across must be the correct one.
?
Puzzle 3
A local school teacher wanted to share 703 sweets equally between the pupils.
If there were more than 20, but fewer than 50 pupils, how many sweets did each receive?
You find yourself playing a game of GreenJack with your friend.
It is played with a deck of only 16 cards, divided into 4 suits:
Red, Blue, Orange, and Green.
There are four cards in each suit:
Ace, King, Queen, and Jack.
All Aces outrank all Kings, which outrank all Queens, which outrank all Jacks, except for the Green Jack, which outranks every other card.
If two cards have the same face value, then Red outranks Blue, which outranks Orange, which outranks Green, again except for the Green Jack, which outranks everything.
Here's how the game is played: you are dealt one card face up, and your friend is dealt one card face down. Your friend then makes some true statements, and you have to work out who has the higher card, you or your friend. It's that simple!
Round 3:
You are dealt the Red Queen and your friend makes three statements:
My card could lose to a Blue card. Knowing this, if I am more likely to have an Ace or a King than a Queen or a Jack, then I have an Orange card. Otherwise, I don't. Given all of the information you now know, if I am more likely to have a Jack than an Ace, then I actually have a King. Otherwise, I don't.
Who has the higher card, you or your friend?
Hint
List all of the cards, and then eliminate some using (1).
Answer
Your friend.
Reasoning
You were dealt the Red Queen.
The possible cards, in order, are:
Green Jack
Red Ace
Blue Ace
Orange Ace
Green Ace
Red King
Blue King
Orange King
Green King
Red Queen (your card)
Blue Queen
Orange Queen
Green Queen
Red Jack
Blue Jack
Orange Jack
By (1), their card could lose to a Blue card (the Blue Ace), leaving:
Orange Ace
Green Ace
Red King
Blue King
Orange King
Green King
Red Queen (your card)
Blue Queen
Orange Queen
Green Queen
Red Jack
Blue Jack
Orange Jack
By (2), their card is not more likely to be an Ace or a King (6) than a Queen or a Jack (6), so their card is not Orange, leaving.
Green Ace
Red King
Blue King
Green King
Red Queen (your card)
Blue Queen
Green Queen
Red Jack
Blue Jack
By (3), their card is more likely to be a Jack (2) than an Ace (1), so their card is a King, leaving:
Red King
Blue King
Green King
Red Queen (your card)
