Puzzle 105
Below is a very special grid, around each shaded number are 8 white squares. However, each white square should have a number from 1 to 7. Once filled in, these 8 numbers will sum to the shaded number. In addition, once completed correctly, no row nor column will contain a duplicate number within a white square. For example, the top row may be 5 6 4 2 3 1 7, etc. This is a very difficult puzzle, and many people resort to using a computer to help them, which is a challenge in itself.
Puzzle Copyright © Kevin Stone
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Hint
This puzzle almost certainly requires trial-and-error to solve, and there is more than one solution.
Answer
There are a number of correct solutions, this is just one of them.
Puzzle 106
How many days before 17th August is it, if...
...50 days before then, it was four times as many days since March 30th?
Puzzle Copyright © Kevin Stone
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Hint
How many days separate the two dates?
Answer
18 days.
Reasoning
There are 140 days between the two dates, and these 140 days are split in the following way:
4 x Answer | 50 days | Answer
We already know about 50 days, so we have to split the remaining 90 days into the ratio 4:1, which is 72:18.
72 days | 50 days | 18 days
Puzzle 107
Start with a number larger than 0, square it, add 4, double, take away 3, times 4 and finally subtract the original number.
If you were now left with 20, what number did you start with?
Puzzle Copyright © Kevin Stone
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Hint
A little algebra might help.
Answer
1/8.
Reasoning
If we convert the question to algebra, we have:
((n^2 + 4) x 2 – 3) x 4 – n = 20
Expanding the brackets and simplifying gives:
(2n^2 + 8 – 3) x 4 – n = 20
(2n^2 + 5) x 4 – n = 20
8n^2 + 20 – n = 20
8n^2 – n = 0 (*)
8n – 1 = 0
8n = 1
n = 1/8
In the equation marked (*) zero is also a potential solution, but as the question tells us that we "Start with a number larger than 0" we know that n can't be 0, and therefore we can safely divide by n.
Double-Checking
1/8
squared = 1/64
add 4 = 41/64
double = 81/32
take away 3 = 51/32
times 4 = 201/8
subtracting 1/8 gives 20, as required.
Puzzle 108
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 each 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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