Hint
Look at the number of sides of each of the shapes.
Answer
Square.
Reasoning
If we add the shape's sides, the total increases by 8, then 10, then 12, etc (2 more each time).
triangle (3) + pentagon (5) + square (4) = 12
Using all of the letters A to Z, each once only, complete these common words. ban--a-i-et-aiz-s--thna-e--eco--ocu--la-e--nkf-xbe--ree-ui-ot-erwis--oast
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 8 go? 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?
3 – no, because the difference between 1 and 3, and the difference between 3 and 5, are both D2.
4 – no, because the difference between 4 and 5 is D1, which we would already have.
6 – no, 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.
