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From Zeta, first letters of the Greek alphabet, alpha, beta, gamma, delta, epsilon, zeta, eta.
Puzzle 46
Put the characters C  D  M  L  X in the right order...
...but not alphabetical.
[Ref: ZULP]
Direct Link: www.brainbashers.com?ZULP
Hint: Think of Italy.
Answer: X  L  C  D  M.
These are the Roman numerals for 10, 50, 100, 500 and 1000, in ascending order.
Puzzle 47
I was having trouble sleeping last night and I tossed and turned well into the night.
Our local town hall has a clock which strikes on the hour and also strikes just once on the half hour.
During one of my more awake moments I heard the clock strike once, but I could not tell what time it was.
Half an hour later it struck once again, but I still could not tell what time it was.
Finally, half an hour later it struck once again and I knew what the time was.
What time was it?
[Ref: ZEQH]
Direct Link: www.brainbashers.com?ZEQH
Hint: When does the clock strike just once?
Answer: 1.30 in the morning.
The initial single strike was at 12.30.
Puzzle 48
Three countrymen met at a cattle market. "Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."
"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."
"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."
No doubt this was a very primitive way of bartering animals, but it is an interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market.
[Ref: ZUCK] At A Cattle Market. Amusements In Mathematics by Henry Ernest Dudeney (1917).
Direct Link: www.brainbashers.com?ZUCK
Hint: A little algebra might help.
Answer:
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.
This can be solved with a little algebra, where J  Jakes, H = Hodge and D = Durrant:
From the clues:
2 x (H  6 + 1) = J + 6  1 [1]
3 x (D  14 + 1) = H + 14  1 [2]
6 x (J 4 + 1) = D + 4  1 [3]
These can be rearranged to give:
2H  J = 15 [4] 3D  H = 52 [5] 6J  D = 21 [6]
We can now use 2 x [5] + [4] to give:
6D  J = 119 [7]
We can now use [7] + 6 x [6] to give:
35J = 245 J = 7
We can then use J = 7 in [4] and [6] to give H = 11, and D = 21.
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