Colin James Physics - Logic puzzles.
Last updated: 30th June 2013.
Logic puzzles.
These are general logic puzzles - not Physics puzzles, but they should be challenging and fun (just like Physics).
If you know what you are looking for in a puzzle or have done a similar one before you may find an easier puzzle harder or a harder one straightforward.
So it its worth scanning through the different categories.
When you click on a puzzle it will appear below (and the screen will move down to show the puzzle) with a solution further down the screen.
Easier puzzles.
Running puzzle.
Tennis puzzle.
Water puzzle.
Walking puzzle.
Harder puzzles.
Spending puzzle.
Cakes puzzle.
Ages puzzle.
Chess puzzles.
Chess puzzle 1.
Chess puzzle 2.
Chess puzzle 3.
Chess puzzle 4.
Chess puzzle 5.
Cakes puzzle.
At the market Client L goes to Baker E to buy some cakes.
Although it is early Baker E is packing up because there are no cakes left.
Client L says, "did you have many cakes?"
"Oh yes," said Baker E. "I had plenty but 4 clients came and each one asked for 6/7ths of what I had plus 1/7th of a cake
and now I have none at all."
"How did you cut exactly 1/7th of a cake?" asked Client L.
"I didn't need to cut any," replied the baker.
How many cakes did Baker E have to start with?
Solution
Solution
Solution
Solution
Solution: Cakes puzzle.
The solution is 400 cakes and this is one way to derive it:
Variables.
x0 = original number of cakes before any were sold.
x1 = number of cakes after (6/7)x0 + 1/7 were sold to client 1.
x2 = number of cakes after (6/7)x1 + 1/7 were sold to client 2.
x3 = number of cakes after (6/7)x2 + 1/7 were sold to client 3.
x4 = number of cakes after (6/7)x3 + 1/7 were sold to client 4.
x1 = x0 - (6x0/7 + 1/7) = 7x0/7 - 6x0/7 - 1/7 = = (x0 - 1)/7
x2 = x1 - (6x1/7 + 1/7) = 7x1/7 - 6x1/7 - 1/7 = = (x1 - 1)/7
x3 = x2 - (6x2/7 + 1/7) = 7x2/7 - 6x2/7 - 1/7 = = (x2 - 1)/7
x4 = x3 - (6x3/7 + 1/7) = 7x3/7 - 6x3/7 - 1/7 = = (x3 - 1)/7
Finding x4 in terms of x0 and equating x4 to zero (as there are no cakes left after the 4th client has bought what they want):
x4 = (x3 - 1)/7
x4 = ((x2 - 1)/7 - 1)/7
x4 = (((x1 - 1)/7 - 1)/7 - 1)/7
x4 = ((((x0 - 1)/7 - 1)/7 - 1)/7 - 1)/7
x4 = (((x0 - 1 - 7)/49 - 1)/7 - 1)/7
x4 = ((x0 - 1 - 7 - 49)/343 - 1)/7
x4 = (x0 - 1 - 7 - 49 - 343)/2401
x4 = (x0 - 400)/2401
x4 = 0 =>
(x0 - 400)/2401 = 0 =>
x0 = 400
An alternative way is to work backwards from x4 = 0 and find how many cakes there were at each previous stage until you reach the point before the first client bought any:
x4 = 0
(x3 - 1)/7 = 0
x3 - 1 = 0
x3 = 1
x3 = (x2 - 1)/7
(x2 - 1)/7 = 1
x2 - 1 = 7
x2 = 8
x2 = (x1 - 1)/7
(x1 - 1)/7 = 8
x1 - 1 = 56
x1 = 57
x1 = (x0 - 1)/7
(x0 - 1)/7 = 57
x0 - 1 = 399
x0 = 400
End of page.