Below, 10 nine-letter words have been broken into chunks of three letters.
The chunks have been moved around, no chunk is used twice, and all of the chunks are used.
Can you determine what the 10 words are?
hol rec ant imp fer con
man dif ice pol acq htn
ort myt new air sag duc
dec lig kch ogy ing tan
uit ent gle tal tor ent
Hints
The first letters of the words are: A, C, D, D, I, L, M, N, P, R.
Answers
acq + uit + tal = acquittal
con + duc + tor = conductor
dec + kch + air = deckchair
dif + fer + ent = different
imp + ort + ant = important
lig + htn + ing = lightning
myt + hol + ogy = mythology
new + sag + ent = newsagent
pol + ice + man = policeman
rec + tan + gle = rectangle
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 20
I recently asked a friend of mine to buy me some ribbon for my latest embroidery project.
He went to the haberdashery shop for a particular length, but he accidentally interchanged the feet and inches.
When I measured the purchased ribbon, I only had 30% of the length I required!
Answer
I asked for 9 foot 2 inches (110 inches) and my friend brought me 2 foot 9 inches (33 inches).
Reasoning
I asked my friend to buy me A feet and B inches, which is a total of 12A + B inches.
He accidentally bought me B feet and A inches, which is 12B + A inches.
We know that 12B + A is actually only 30% of what I wanted, so:
12B + A = 0.3 x (12A + B)
Multiply throughout by 10:
120B + 10A = 3 x (12A + B)
120B + 10A = 36A + 3B
117B = 26A
B = 26A ÷ 117
Don't forget that A and B can only be integers, and between 0 and 11.
Since 26 x A has to be at least 117, the lowest possible value for A is 5. We can check the remaining possibilities.
5 x 26 = 130
6 x 26 = 156
7 x 26 = 182
8 x 26 = 208
9 x 26 = 234 (*)
10 x 26 = 260
11 x 26 = 286
The only number wholly divisible by 117 is 234, which means that A = 9, and B = 2.
