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After the local BrainBashers Horse Show, some friends were telling everyone who had won.
Unfortunately, the friends were having trouble remembering, and in the course of the afternoon, they changed their minds a number of times.
In summary, they made the following statements:
Andy said: Taylor Tripton won with Evening Sunrise. Alex Ambleton won with Morning Sunset. Frankie Flanton won with Evening Sunrise. Billie said: Sam Spinton won with Midday Night. Frankie Flanton won with Evening Sunrise. Taylor Tripton won with Morning Sunset. Chris said: Alex Ambleton won with Midday Night. Sam Spinton won with Midday Night. Sam Spinton won with Morning Sunset. Dale said: Alex Ambleton won with Evening Sunrise. Frankie Flanton won with Morning Sunset. Sam Spinton won with Midday Night.
However, none of the statements were fully true.
In fact, just six of the statements were exactly half true (either the person won, or the horse did), the rest were false.
Can you determine who won, and which horse they were showing?
Reasoning
Before we begin the reasoning, let's look at the hint: 'Could Alex Ambleton have won?'. There are three horses in the puzzle, and Alex Ambleton is stated to have won with all three of them. We know that none of the statements were fully true, so Alex couldn't have won otherwise one of the statements would have been fully true.
However, we don't need the hint to solve the puzzle.
By grouping the clues together, we have:
1. Frankie Flanton won with Morning Sunset.
2. Frankie Flanton won with Evening Sunrise.
3. Frankie Flanton won with Evening Sunrise.
4. Sam Spinton won with Midday Night.
5. Sam Spinton won with Midday Night.
6. Sam Spinton won with Midday Night.
7. Sam Spinton won with Morning Sunset.
8. Taylor Tripton won with Morning Sunset.
9. Taylor Tripton won with Evening Sunrise.
10. Alex Ambleton won with Midday Night.
11. Alex Ambleton won with Morning Sunset.
12. Alex Ambleton won with Evening Sunrise.
We know that none of these statements were fully true, so, we know that (for example) it wasn't Frankie Flanton with Morning Sunset.
We can create a grid of all the possibilities and place an X where we know that it can't have been that combination:
M-N
M-S
E-S
Frankie
X
X
Sam
X
X
Taylor
X
X
Alex
X
X
X
We now have an X for all but 3 possibilities, and we know that exactly 6 of the clues must be half true.
So …
… if it was Frankie Flanton with Midday Night then exactly 7 clues would be half true (clues 1, 2, 3, 4, 5, 6, 10).
… if it was Sam Spinton with Evening Sunrise then exactly 8 clues would be half true (clues 4, 5, 6, 7, 2, 3, 9, 12).
… if it was Taylor Tripton with Midday Night then exactly 6 clues would be half true (clues 8, 9, 4, 5, 6, 10), and this must be the correct answer.
?
Puzzle 678
Four friends were discussing the colours of the flowers they had for the local flower show, from this year, and last.
There were four different colours: red, green, blue, and yellow.
No-one entered the same colour flower twice. The entrant who currently has the green flowers, used to have yellow. Betty, who is not Mrs Long, currently has the blue flowers. Mrs Kipper, who now has yellow flowers, used to have blue. Neither Mrs Annie Ford, nor Mrs Jester, have ever owned yellow flowers. Cybil is not Mrs Kipper. Diane and Annie have entered four different colours in total.
Can you determine the full names, and colours of the flowers the ladies currently have, and previously had?
Hint
Start with Betty, Mrs Kipper, and Ada, as these are three different people.
Answer
Now Last
Annie Ford Red Green
Betty Jester Blue Red
Cybil Long Green Yellow
Diane Kipper Yellow Blue
Reasoning
By (3), Betty currently has Blue flowers.
Now Last
Betty Blue
By (4), Mrs Kipper currently has Yellow flowers, and used to have Blue.
Now Last
Betty Blue
Kipper Yellow Blue
By (5), we know Annie Ford's name.
Now Last
Betty Blue
Kipper Yellow Blue
Annie Ford
By (6), Mrs Kipper must be Diane, and the remaining person's name is Cybil.
Now Last
Betty Blue
Diane Kipper Yellow Blue
Annie Ford
Cybil
By (2), whoever currently has the Green flowers used to have Yellow, but by (5) this can't be Annie Ford, so this must be Cybil. Which means that Annie must currently have the Red flowers.
Now Last
Betty Blue
Diane Kipper Yellow Blue
Annie Ford Red
Cybil Green Yellow
By (3), Betty isn't Mrs Long, so must be Mrs Jester. Therefore, Cybil must be Mrs Long.
Now Last
Betty Jester Blue
Diane Kipper Yellow Blue
Annie Ford Red
Cybil Long Green Yellow
By (1), Annie Ford must have previously had Green, and Betty Jester had Red.
Now Last
Betty Jester Blue Red
Diane Kipper Yellow Blue
Annie Ford Red Green
Cybil Long Green Yellow
?
Puzzle 679
Three teachers were discussing how long they had each been teaching.
Alex and Billie had been teaching for a combined total of 36 years. Charlie and Billie had been teaching for a combined total of 22 years. Charlie and Alex had been teaching for a combined total of 28 years.
How long had each been teaching?
Hint
If we look at all of the clues, each person appears exactly twice.
Answer
Alex had been teaching for 21 years.
Billie had been teaching for 15 years.
Charlie had been teaching for 7 years.
Reasoning
If we look at all of the clues, each person appears exactly twice.
So the total of the three clues gives us two lots of Alex + Billie + Charlie = 36 + 22 + 28 = 86.
So, Alex + Billie + Charlie = 43 years.
In each clue we are given two of the people, so we can use this to find the missing person:
Charlie and Billie had been teaching for a total of 22 years, which means that Alex must be: 43 − 22 = 21 years.
Charlie and Alex had been teaching for a total of 28 years, which means that Billie must be: 43 − 28 = 15 years.
Alex and Billie had been teaching for a total of 36 years, which means that Charlie must be: 43 − 36 = 7 years.
Reasoning With Algebra
Let Alex = A, Billie = B and Charlie = C, then:
[1] A + B = 36 [2] C + B = 22 [3] C + A = 28
If we use [3] − [2] we have:
[4] A − B = 6
If we use [1] + [4] we have:
2A = 42 A = 21
By [1] B = 15. By [3] C = 7.
???
Puzzle 680
During the recent BrainBashers cipher convention, a binary code contest took place.
The contest consisted of a binary code transmission where the spaces between the letters were missing and there was no punctuation.
Each letter of the alphabet was translated into its binary equivalent based on its position in the alphabet. The resulting code was then blocked in groups of five digits.
a=1, b=10, c=11, d=100, e=101, f=110, g=111, h=1000, i=1001, j=1010, k=1011, l=1100, m=1101, n=1110, o=1111, p=10000, q=10001, r=10010, s=10011, t=10100, u=10101, v=10110, w=10111, x=11000, y=11001, z=11010
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.
