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Q1 can't have A as its answer (otherwise it would contradict itself).

If Q1's answer was C (meaning Q4's answer was A), then Q3's answer would be C. However, Q2's answer should now be A, but this isn't allowed by Q1 (as Q4 is the first answer with A). This is a contradiction.

Therefore Q1's answer is B (meaning Q3's answer is A).

For Q2's answer to be C, Q4's answer would have to be A, which would contradict Q4. Therefore Q2's answer is B, making Q4's answer B. QED.

Puzzle 82

Below are thirteen 5 lettered, everyday words, each of which has had two of its letters removed.

In total these 26 letters are A-Z. The remaining letters in each word are in the correct order.

There are no words which are spelled differently based upon location (favour/favor, etc) and there are no plurals.

Can you determine the original words?

APE
BAE
BOD
ANC
ROE
ELL
RAY
BUC
ORA
UMB
SUA
LOL
GES

APE + P + L >> APPLE
BAE + R + K >> BRAKE
BOD + X + E >> BOXED
ANC + M + I >> MANIC
ROE + F + Z >> FROZE
ELL + D + W >> DWELL
RAY + G + V >> GRAVY
BUC + N + H >> BUNCH
ORA + C + B >> COBRA
UMB + J + O >> JUMBO
SUA + Q + T >> SQUAT
LOL + Y + A >> LOYAL
GES + U + S >> GUESS

Puzzle 83

What 5 digit number am I thinking of.

It has two prime digits.

Digit 3 is the highest digit.

Digit 2 is lowest digit.

Digit 1 is higher than the sum of digits 4 and 5.

Digit 5 is half of digit 4.

Digit 1 is one smaller than digit 3.

Digit 5 is between digit 2 and digit 1

There are no duplicates and the digit 0 doesn't appear.

Since digit 5 is half of digit 4 we have four possibilities:

_ _ _ 2 1
_ _ _ 4 2
_ _ _ 6 3
_ _ _ 8 4

But since digit 1 is higher than the sum of digit 4 and digit 5 we can eliminate two possibilities leaving:

_ _ _ 2 1
_ _ _ 4 2

Since digit 5 is between digit 2 and digit 1, we can eliminate _ _ _ 2 1, leaving:

_ _ _ 4 2

Since digit 1 must be higher than the sum of digit 4 and digit 5 we have three possibilities:

7 _ _ 4 2
8 _ _ 4 2
9 _ _ 4 2

But digit 1 is 1 smaller than digit 3 we can eliminate 9 _ _ 4 2, leaving

7 _ 8 4 2
8 _ 9 4 2

Since digit 2 is the lowest number, it must be 1:

7 1 8 4 2
8 1 9 4 2

The answer has exactly 2 prime digits, which only leaves

7 1 8 4 2. QED.

Puzzle 84

You find yourself playing a game 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.

Ace outranks King, which outranks Queen, which outranks Jack - 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:

1. My card could lose to a Blue card.
2. 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.
3. 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.

Hint: What cards can your friend have at the start?

Answer: Your friend.

After #1 you know that you have the Red Queen (RQ), and your friend's card could be beaten by a blue card, so your friend can only have one of the following cards:

RK, RJ, BK, BQ, BJ, OA, OK, OQ, OJ, GA, GK, GQ.

Of these 12 cards there are 6 Aces or Kings, therefore by #2 their card is not orange.

Your friend now has one of the following cards:

RK, RJ, BK, BQ, BJ, GA, GK, GQ.

Of these 8 cards there are 2 Jacks and 1 Ace, therefore by #3 their card is a King, which beats your Queen. QED.

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