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Puzzle 217
The other day I was sitting in my local tavern, The Spyglass, which overlooks the sea, when in sailed my old friend the pirate Captain Conan Drum. "Well, shiver me barnacles!" he roared on seeing me. He too is a bit of a puzzle addict and so, after joining me for a glass of milk and telling me about his latest exploits on the high seas, he couldn't resist showing me his latest conundrum.
He reached into one of his jacket pockets and produced seven gleaming £5 coins, which he then proceeded to arrange on the table in front of me exactly as shown below. "Now, me lad." he said, with a mischievous look in his eyes. "I'll wager you'll not be able to solve this one. Take away two coins from this here arrangement to leave five coins across and three coins going down."
It was clear the wily old sea dog still had one or two tricks up his sleeve, as I couldn't for the life of me see how it could be done. Can you see through his skulduggery and solve it?
Reasoning
If we denote Alex by A, Billie by B, Charlie by C, and Drew by D and number the clues.
If Alex had three more sheep, then he'd have one fewer sheep than Billie.Whereas Drew has the same number as the other three shepherds put together.If Charlie had three fewer sheep, he'd have exactly three times the number of Alex.If they were evenly distributed, they'd each have eleven sheep.
By (4): A + B + C + D = 44
And, by (2), these are shared equally between A + B + C and D.
So D = 22, and A + B + C = 22 (*)
By (1): A + 3 = B − 1 A + 4 = B
By (3): C − 3 = 3A C = 3A + 3
Substitute B and C in (*):
A + (A + 4) + (3A + 3) = 22 5A + 7 = 22 5A = 15 A = 3