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?