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There are two alternate series, starting with the first two numbers, and each formed by doubling the preceding number in its own series and subtracting 2.
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Puzzle 58
What is the minimum number of queens required on a chessboard such that all squares are attacked?
Hint
The largest possible difference is when 1 is next to 8, a difference of 7. Since we have 7 differences to find, and the largest possible difference is 7, all of the possible differences must exist. Start by looking where can the 8 go.
Answer
Reasoning
The largest possible difference is when 1 is next to 8, a difference of 7. Since we have 7 differences to find, and the largest possible difference is 7, all of the possible differences must exist: 1, 2, 3, 4, 5, 6, 7, and let's call these D1, …, D7.
D7 can only happen when: 1 is next to 8 = D7
D6 can happen when:
1 is next to 7, but these are given numbers that are not next to each other. 2 is next to 8 = D6
Where can we place the 8? If we put 8 above 1, we cannot also satisfy D6 (2 is next to 8).
Therefore, we have two possibilities:
(a) 8 to the left of 1
(b) 8 to the right of 1 (a) 8 to the left of 1
By D6, 2 would be below 8, and this would give us D1, D6, D7. What can we place to the right of 1?
Not 3, because the difference between 1 and 3, and the difference between 3 and 5, are both D2.
Not 4, because the difference between 4 and 5 is D1, which we would already have.
Not 6, because the difference between 5 and 6 is D1, which we would already have.
There are no possible numbers we can place to the right of 1, so 8 can't go to the left of 1. (b) 8 to the right of 1
By D6, 2 would be above 8, and this would give us D3, D6, D7.
4 can't go next to 1, otherwise we'd create another D3. Therefore, 4 goes in the bottom left corner.
We are now left with 3 and 6.
If 6 went above 4, and 3 above 1, these would both be D2.
Therefore, 3 goes above 4, 6 goes above 1.
The final answer is:
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Puzzle 60
Last week my wife and I found ourselves stranded on an island, oops!
Unfortunately, we were also stranded with our two sons, with no clear means of rescue.
Luckily, I came across an old rickety boat which I knew could carry either one adult or two children.
Answer
The two sons rowed to safety first, one son rowed back.
I rowed to safety and the other son rowed back to the island.
The two sons rowed to safety first, one son rowed back again.
My wife rowed to safety and the other son rowed back.
Finally, the two sons rowed to safety.
It is now four years since I first asked for support, and I would like to sincerely thank everyone who has helped me. BrainBashers doesn't show adverts, and I continue to depend on the ongoing support from everyone who enjoys the site.
