Can you solve the human cannonball riddle Alex Rosenthal
They call you the human cannonball,
but you’re really more
of a pinball person.
Your act involves flying
through a dozen rings of fire,
bouncing through a trampoline course,
and catching the trapezist
in the grand finale.
Your cannon has metal coils that
accelerate you to the perfect speed.
At least it’s supposed to.
Today’s pre-flight test
fails dramatically,
and upon inspection,
this is a clear act of sabotage:
someone amped the power up to the max.
It’s too late to abort the launch;
the trapezist will plummet if you don’t
catch him in a few minutes.
So you’d better get fixing.
The cannon was invented
by your eccentric mentor,
and as usual, his instructions leave
something to be desired.
The cannon’s electromagnet is powered
by energy cells
located in 16 chambers on two levels.
Each level is a hollowed out square,
with three chambers to a side.
The acceleration is survivable if:
there are twice as many energy cells
in the upper level as in the lower level.
Every chamber has 1 to 3
energy cells.
And each side of the cannon,
made of 6 chambers, has 11 energy cells.
Your mentor designed the cannon to use
a certain number of energy cells,
but when the shipment arrived,
it was 3 short.
So he made all of those conditions
work with this reduced number,
and it fired perfectly.
That’s the amount you’ll need.
Too many, or too few,
and you and the trapezist are doomed.
How many energy cells should you use,
and where?
Pause here to figure it out yourself.
Answer in 3
Answer in 2
Answer in 1
The first step in this puzzle
is to narrow the options.
Let’s focus on rule 3 in isolation.
We could put 11 cells in these two corners
and fulfill it with just 22 total cells,
because adjacent sides share corners.
But if we put 11 in the middle chambers,
we could have 44 cells.
The answer must be in the range
bounded by those extremes.
We can use the other rules
to refine our options further.
Since there are twice as many
cells in the upper level,
the total cells must be
a multiple of 3.
Now, because of rule 4,
we need to find two consecutive multiples
of 3 that can meet all the conditions.
Up to this point we haven’t used the rule
that each chamber must have 1 to 3 cells.
It tells us that the minimum
for the lower level is 8 cells,
in which case, by rule 1,
the upper level would have 16.
On every side, the lower level would
account for 3 of the 11 cells,
so the upper would have to have 8.
But if two opposite sides had 8,
that would be the upper level’s entire 16,
not leaving any for these two chambers.
So 24 is out.
Let’s look at the other extreme.
The upper level can have
at most 3 times 8, or 24 cells,
which would give a total of 36.
That eliminates 39 and 42.
With 36,
if we had 3 cells in each chamber,
each side of the upper level would already
have 9 of its 11 cells,
meaning we’d have to leave empty chambers
on the lower level.
So 36 is also out.
What about 33?
We know we don’t want any sides
with all 3′s,
so we can place 2′s at opposite
corners of the upper level.
Now we have 8 of 11 cells per side.
So we could place exactly 1 cell
in each lower chamber…
but that would fall short
of the lower level’s required 11.
That leaves only two options: 30 and 27,
which must work by process of elimination,
and 27 is the one you should use.
To total 9, the lower level must have
seven 1′s and one 2.
If you put the 2 in a middle chamber,
the upper level would have too many cells.
So the 2 has to go into a corner,
and you can then place the upper level
like this.
You’ve barely snapped
the last energy cell into place
when you hear the ringleader
announcing your act.
You’re pretty sure you’ve also noticed
enough clues to solve the mystery.
Who sabotaged you?