Place letters into the grid such that every row, column, and 2 x 2 block has letters (in any order) that form a common word. Each letter is only used once, and no letter is repeated in the rows / cols / blocks.
Letters allowed: C O L U M B U S
D
K
P
E
T
E
L
A
Note: this puzzle is not interactive, and the squares cannot be clicked.
Other anagrams of these words are OK as long as they don't change the answer grid.
Alternative Answer
This answer uses less common words.
D
U
C
K
S
P
U
L
O
M
E
T
E
L
B
A
Across: DUCK, PLUS, TOME, BALE
Down : DOSE, PLUM, CUBE, TALK
Boxes : SPUD, LUCK, MOLE, BEAT
Other anagrams of these words are OK as long as they don't change the answer grid.
??
Puzzle 206
For being well-behaved at the garden fayre, four children were each given two sweets.
Jesse had an orange sweet. One child who had a red sweet also had a blue one. No child had two sweets of the same colour. A child who had a green sweet also had a red one. Jamie didn't have a red sweet, and Jo had a green one. Jordan didn't have an orange one, and Jesse had no blue sweets.
Knowing that there were two sweets of each colour, can you tell the colours of the sweets each child had?
Answer
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Reasoning
The children were: Jamie, Jesse, Jo, Jordan.
The colours were: Blue, Green, Orange, Red.
There were two sweets of each colour, and by (3) no child had two sweets of the same colour.
By (1), Jesse had an Orange sweet …
… by (6), didn't have Blue …
… by (2) and (4), Red was paired with Blue and Green …
… therefore, Jesse's other sweet can't have been Red, so was Green.
Jamie
Jesse Orange Green
Jo
Jordan
By (4) the other Green sweet was paired with Red, by (5) this must have been Jo.
Jamie
Jesse Orange Green
Jo Green Red
Jordan
By (2) the other Red sweet was paired with Blue, but by (5) this wasn't Jamie, so must have been Jordan.
Jamie
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Leaving Jamie with Orange and Blue.
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
???
Puzzle 207
In this long division, each number has been replaced by another. The answer has also been removed. Can you determine the answer?
....
———————
75 | 796767
791
——
57
30
——
86
82
——
47
47
——
9
Hint
This is a tricky question related to number bases.
Answer
200.
Each number in the sequence is a representation of the number 32 in different bases, starting with base 10.
Base 10: 32 = 3 x 10^1 + 2 x 10^0 = 3 x 10 + 2 = 32
Base 9 : 35 = 3 x 9^1 + 5 x 9^0 = 3 x 9 + 5 = 32
Base 8 : 40 = 4 x 8^1 + 0 x 8^0 = 4 x 8 + 0 = 32
Base 7 : 44 = 4 x 7^1 + 4 x 7^0 = 4 x 7 + 4 = 32
Base 6 : 52 = 5 x 6^1 + 2 x 6^0 = 5 x 6 + 2 = 32
Base 5 : 112 = 1 x 5^2 + 1 x 5^1 + 2 x 5^0 = 1 x 25 + 1 x 5 + 2 = 32
Base 4 : 200 = 2 x 4^2 + 0 x 4^1 + 0 x 4^0 = 2 x 16 + 0 + 0 = 32
