Puzzle 197
Place all of the digits from 1 to 9 into the circles so that the sum of the numbers in each straight line is the same.
Puzzle Copyright © Cihan Altay
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Hint
Each line adds to 14.
Answer
Puzzle 198
Can you find every occurrence of the word MAY that appears in this grid (horizontally, vertically, or diagonally)?
The hint will reveal the number of times it occurs, but where are they?
Note: this puzzle is not interactive, and the letters cannot be selected.
Puzzle Copyright © Kevin Stone
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Hint
The word MAY appears 39 times.
Answer
The word MAY appears 39 times.
Puzzle 199
Which is larger:
loaves in a dozen baker's dozen
points in the highest snooker break
Puzzle Copyright © Kevin Stone
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Hint
There are 13 loaves in a baker's dozen.
Answer
Loaves in a dozen baker's dozen.
Reasoning
A baker's dozen is 13, so a dozen baker's dozen is:
12 x 13 = 156 loaves
There are 15 red balls in snooker, and each of these can be followed by the black ball, and then the colours in order gives:
15 x (1 + 7) + 2 + 3 + 4 + 5 + 6 + 7 = 147 points
This can rise to 155 if the break follows a foul.
Puzzle 200
In Farmer Stone's hay loft, there were several animals, in particular birds, mice, and cockroaches. One day, feeling bored, I decided to count the animals.
I found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice.
How many of each animal were there?
Puzzle Copyright © Kevin Stone
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Hint
Cockroaches have 6 feet, mice have 4, and birds have 2. They all have one head!
Answer
35 birds, 5 mice, and 10 cockroaches.
Reasoning
Cockroaches have 6 feet, mice have 4, and birds have 2. They each have one head.
We can write down expressions for the heads and the feet, calling birds B, mice M, and cockroaches C.
Counting the heads:
(1) B + M + C = 50
Counting the feet:
(2) 2B + 4M + 6C = 150
We are told that for every mouse there are two cockroaches so, C = 2M. Update (1) and (2) to give:
(3) B + 3M = 50
(4) 2B + 16M = 150
If we double (3) we get:
(5) 2B + 6M = 100
We can now do (4) − (5) to give:
10M = 50
M = 5
So, we have 5 mice (and 10 cockroaches).
We can use M = 5 in (3) to give:
B + 3 x 5 = 50
B + 15 = 50
B = 35
So, C = 10, M = 5 and B = 35.
Double-Checking
10 cockroaches have 10 heads and 60 feet.
5 mice have 5 heads and 20 feet.
35 birds have 35 heads and 70 feet.
The total number of heads = 10 + 5 + 35 = 50.
The total number of feet = 60 + 20 + 70 = 150.
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