## Challenge #13 – J. Bozo

We’re back with another challenge! This is in collaboration with Agent Venture who provide amazing online escape room experiences – make sure you check them out!

## The Challenge

Agent Venture needs your top-notch bank robbery skills to help bring J. Bozo to justice for their numerous crimes.

The incriminating files against J. Bozo are locked behind a room with a keyboard and a 3×3 digital grid with letters on it… The passcode is in the form of a pattern which only a few employees know, can you work out what the pattern-passcode of the grid is?

## The Solution

Let’s start by looking at Employee 1, we can see that there are multiple ways we can reach the typed passcode. See the images below to see some of the possible methods. How many ways of getting B, A, B, A, B can you find from Employee 1’s grid?

Similarly for Employee 2, we can see that there are also multiple ways in which we can reach the typed passcode. However, we can deduce that the pattern must start on the middle right square since there is no other starting point for C,C,B in a row (C,C,B,A,B).

Lastly, for Employee 3, there are also multiple ways of reaching the typed passcode C, A, C, B, C.

Using the information we gathered from Employee 2, we can deduce that if we start on the middle right square using Employee 1’s grid, then the next move is to go down to the bottom right square. If we instead try to go up to the top right square, then we reach a dead-end and end up with B, A, C even though the passcode for Employee 1 starts off as B, A, B (B,A,B,A,B).

It is still ambiguous what the next part of the pattern is by looking at the grids for Employee 1 and Employee 2. However, looking at the grid for Employee 3, we can see that the pattern must go through the middle square to get the last part of the pattern-passcode.

Putting all this information together and overlaying the pattern over the grid, we can deduce that the solution is F, I, H, E, D!

How did you do? Comment below how you reached the solution!

## The Challenge

Finally, you have your hands on the blueprints to the bank. You can see that on the blueprints there are only certain routes which you can take in order to avoid triggering the security alarms at 3:23AM (prime bank robbing time).

You need to plan your route carefully to get to the vault as quickly as possible and avoid detection. Luckily, someone has added the number of footsteps it takes along the routes on the blueprints!

First you need to go into the reception cabinet to obtain the keys to the HR office which contain employee records. Inside the records you will find the password to the computer of your accomplice who works in foreign exchange. Your accomplice has left you a key to access the cupboard by the Business Analyst’s desk. You need to find a file containing information about the top-secret floor – including the passcode to the vault.

Can you find a route that takes no more than 217 steps?

## The Solution

Did you manage to find a solution under 217 steps? You’ve just solved a graph theory problem! Graph theory is the mathematical study of graphs which consist of vertices (or nodes, such as the Coffee Machine in the maze) which are connected by edges.

This problem in particular is known as the shortest path problem – we want to find a path between two nodes so that the total is minimised!

See the image below for the shortest path which takes exactly 217 steps.

The shortest route (with a total of 217) is: Entrance, Reception Desk (12), Reception Cabinet (9), Fish Tank (18), Microwave1 (7), Recruitment Materials (16), Employee Records (6), Training Materials (18), Head of HR Seat (10), Foreign Exchange (25), Microwave2 (36), Fridge (8), Frontend Engineers (11), Business Analysts (3), Data Analysts (19), Drinks Machine (14), VAULT (5).

## The Challenge

All that planning has earnt you a break, so you decide to take a nap at the library. As you make your way there, you overhear some cops discussing when the last secret bank-heist occurred in the city.

Because you’re a nosey little thief, you manage to pickpocket 2 pieces of paper from them. You need to use the 2 notes to give you the information of the date of the last bank-heist…

When was the last bank-heist?

## The Solution

### The puzzle bit

The challenge is called The Library and upon closer inspection of the golden note, you may notice that the first letter of each word cycles through the letters ISBN: ISBN = International Standard Book Number!

### The algebra bit

The 2nd note has a string of letters and numbers, but we do not know what the letters stand for. However we have some clues to solve them…

What is A?

The first equation (K + D)2 = A shows us that A must be a square number. The only possibilities between 1-9 are 1,4, or 9.

K and D must be also be between 1-9 and they must be different. The smallest possible choice here is that K and D are equal to 1 and 2 in some order, but A=1 or A=4 are too small to accept even the smallest valid values for K+D. Therefore A=9 which implies that K + D = 3 (incidentally also the smallest possible sum).

What are K and D?

Since we know that K+D=3, we have that either:

• K=1 and D=2 or;
• K=2 and D=1

If K=1, by the second equation C/F = 1, but this means that C=F which is a contradiction, since each letter must represent a different digit. Therefore, K=2 and D=1.

What is H?

Using the 3rd equation, with A=9 and K=2, we see that A-H=K implies that H=7.

What is C?

By the 4th equation: H+D=C, with H=7, D=1, we have that C=8

What is F?

By the 2nd equation: C/F = K, with C=8, K=2 we have that F=4.

Conclusion:

Now that we have found what each letter represents, we can see that the string of numbers is 9781444727296! This is the ISBN number for the book 11/22/63 (Stephen King) which is the date of the last bank heist! (Apologies for the Americanness of the date).

## The Challenge

Difficulty: 5/5.

Everything was going to plan, when you realise your friend is missing.

You notice three robot guards, and you think she may be disguising herself as one of them.

All robot guards are programmed exactly the same:

• Each day is a truth day (so they must tell the truth the whole day) or a lie day (they must lie the whole day).
• They all follow the pattern of one truth day followed by two lie days (T L L T L L T L L…)
• They are not independent: They all have truth/lie days on the same days as each other.

The three guards say the following phrases:

• Guard A: “In 2 days time, I will tell the truth”
• Guard B: “Tomorrow, I will tell the truth”
• Guard C: “One year* ago, I lied”

Which guard is secretly your friend?

*Assume one year = 365 days

## The Solution

We know all guards will work in the same way, so we need to look for an “odd one out”. Let’s start by figuring out what day it is today. Is it a truth day, the first lie day (L1), or the second lie day (L2)?

#### Guard A

Guard A will tell the truth in 2 days time. Let’s look at the 3 cases of today being a Truth day, the first Lie day (L1), or the second Lie day (L2):

• Truth day today: With the cycle T L L T L L…, in 2 days time it will be a lie day. But the guard said they would tell the truth, so this is a contradiction.
• L1 today: With the cycle T L L T L L…, in 2 days time it will be a truth day. But the guard should be lying about it being a truth day, so this is another contradiction.
• L2 today: With the cycle T L L T L L…, in 2 days time it will be a lie day. The guard said they will tell the truth which is a lie as it should be.

So Guard A must be speaking on the 2nd lie day: L2.

#### Guard B

Guard B will tell the truth tomorrow. Let’s look at the cases again, similar to above:

• Truth day today: With the cycle T L L T…, we can see it is a lie day tomorrow. But the guard said they would tell the truth, so we have a contradiction.
• L1 today: With the cycle T L L T…, we can see it is a lie day tomorrow. Since the guard is lying about telling the truth, this is consistent!
• L2 today: With the cycle T L L T… ,we can see it is a truth day tomorrow: But the guard said they would tell the truth which is true on a lie day! A contradiction!

So Guard B must be speaking on the 1st lie day: L1.

#### Guard C

Either Guard A or Guard B must be the friend since we have shown that the day they are programmed to is different. This means there is no shortcut: We must work out which day it is today according to Guard C.

Guard C lied a year ago. This is clearly the hardest part of the puzzle. Let’s break this down into smaller chunks. We have a cycle of 3 repeating: T L L, T L L, T L L, etc.

If we go back 3 days ago, we land in the same place we are today. In fact any “multiple of 3” days ago, we get back to the same day.

366 is a multiple of 3, so if we go back 366 days, we land on the same day. We want to go back only 365 days, so if we add one day “forwards” from 366, we are actually saying that 365 days ago is equivalent to tomorrow!

Formally, we can write this as -365(mod 3) = 1 .

So one year ago is equivalent to one day in the future from our starting point (tomorrow). You could stop here since you can actually tell that B and C cannot co-exist given their mutually exclusive statements, but let’s conduct a final check that A and C are the guards. Recall that Guard A is speaking on the 2nd Lie day.

• L2 today: With the cycle T L L T L L…, one year ago (tomorrow) was a truth day. The guard said they lied which is a lie as it should be.

### Conclusion

To conclude, A and C are the guards, today is the second lie day (L2), and your friend is B! Were you able to deduce who your friend was?

### Common errors

There were a couple of different elements to this puzzle making it challenging. As a result, very few people (out of ~60 submissions in total) sent in a correct solution on the first attempt. Here are some of the most common errors that were made:

• Treating both lie days as the same. In reality, they operate differently due to their relative position in the cycle. Referring to “tomorrow” from L1 will give a different outcome than L2.
• Not using a repeating cycle of 3 (T L L).
• Calculating 365(mod 3) instead of -365(mod 3).
• Calculating 365(mod 7), or using the days of the week in some way. This would lead to the wrong conclusion as we should think in terms of the 3-cycle.

## The Challenge

Difficulty: 1/5

You arrive at the door to steal the bank blueprints, when you see Gary the guard.

“Do you have a drink?” he asks. “Guarding blueprints makes me so thirsty,”

He seems like a nice guy, but if you make him the perfect drink, perhaps you could put him to sleep for long enough to steal the blueprints.

You make your way to the bar and you see this:

The question is, which 3 bottles should you choose to make the perfect drink which will:

• Put Gary to sleep
• Won’t poison him
• Be perfectly balanced

## The Solution

There are many ways to do this, but here we can proceed through deduction.

We have 3 “elements” per bottle, which means that 9 elements will go into the draught in total. But we can only put 4 + 2 + 1 = 7 items in total. To account for the remaining 2 elements, we can deduce that they must be sunflowers.

Because we need 2 sunflowers, we must choose bottle F. The only alternative way to obtain 2 sunflowers is choosing both bottles B and E – but neither of these are purple, and since we can only choose a total of 3 bottles, even if the remaining bottle were purple, we would still be left with an extra sunflower; an unbalanced result.

To balance the 2 sunflowers, the remaining 2 bottles must be purple so we must choose from A, C, D, and H. Since the recipe only calls for 1 snail shell, we can deduce that we need bottle A as this is the only bottle to have only one snail in it.

Finally, to balance the recipe, we can conclude that the remaining 2 monkeys and elephant can only come from bottle H, so the final answer is that we should choose A, F and H.

The benefit of deduction is that it becomes clear that there exists no other solution.

Why is this algebra?

Because we have a system of 8 linear equations (the 8 bottles) with 4 variables (monkey, elephant, snail, sunflower; note that a purple bottle is equivalent to ” – sunflower”).

## Challenge #3 – The Devil’s Tail – Solution

This is the solution to Challenge #3 – The Devil’s Tail, make sure you check that post out before looking at the solution!

This Challenge is a modification of a famous problem which confused many people – The Boy-Girl paradox.

So, it turns out that “Heads” is the most probable at 67%. But why?

The devil could have any of the 3 combinations: {HT, TH, TT}, and 2 of those include a Heads! A common answer is “Equally Likely” as it is easy to assume that the devil showed us a specific coin, meaning the remaining coin is equally likely to be Heads or Tails, but this is not the case as we do not know which coin the devil showed us. Some people’s intuition may also have lead them to think it might be Tails as we have already had a Tails, but this is also not true. Hopefully this problem illustrates how probability is very deceptive and counter-intuitive!

## Challenge #3 – The Devil’s Tail

The devil has two fair coins with one side Heads; the other, Tails. He tosses both of them, has a look, and tells you:

“At least one of the coins is Tails”

He then asks you if you’d like to bet if both of the coins are Tails.

What is the most likely?

• Both tails?
• Can’t tell?

SOLUTION
The solution can be found on this post: Challenge #3 – The Devil’s Tail – Solution.

## Challenge #2 – The Elements – Solution

This is the solution to Challenge #2 – The Elements, make sure you check out the challenge before looking at the solution!

Trust the devil. Why?

The first clue given in the challenge is the note left behind by Doctor Aro.

The key piece of information on the note says that we have to reflect on the odd things in life.

Reflecting the position of the digits in the odd numbers we get:
71 6 53 9 68
(Reflecting the position of the digits for 9 leaves it the same)

But how do we decipher the numbers?

Dr Aro was ‘dabbling in the chemical arts’ and this challenge is called ‘The Elements’, written in a similar way to the elements in the periodic table.

Using the periodic table we get:
Lu-C-I-F-Er

Lucifer i.e. the devil 😈

Were you able to save Doctor Aro?

## Challenge #2 – The Elements

Can you save Doctor Aro?

THE CHALLENGE
Now that you and your accomplice, Ez have escaped, you have one more friend to free, known as “Doctor Aro” 🧑‍🔬 before you can proceed.

Doctor Aro had a near-death experience from her dabbling in the chemical arts. 💀 As a result, she is stuck at the gates of heaven waiting for a decision to be made.

Guarding the gates are the angel and the devil. “I can help you,” they both offer. Since you’re a badass thief, you’re not sure who to trust…

Suddenly in a chemical burst, a piece of paper appears in Doctor Aro’s handwriting pleading for your help:

“Sometimes, you have to reflect on the odd things in life. 17 6 35 9 68”

Who should you trust?
The angel 😇
or
The devil 😈?

TIP
You may need google/pen and paper.

SOLUTION
The solution to this challenge can be found at Challenge #2 – The Elements – Solution.

## Challenge #1 – Prison Break – Solution

This is the solution to Challenge #1 – Prison Break – make sure you check that post out if you haven’t already!

It turns out that Guard C is your friend!

We are given the information that each robot guard always lies for the whole day, or always tells the truth for the whole day. This information does not indicate that the guards have truth/lie days in any particular order.The robot guards could have truth/lie days as follows: Truth, Lie, Truth, Lie, …
But equally could have days as below:
Truth, Truth, Truth, Lie, …

There is no reason to presume that the guards all follow the same orders, so we can assume that each guard is independent of the others. This means a guard’s truth/lie days can be different to the other guards.

So why is Guard C our friend? Guard C said “Today, I lie.”

• If Guard C is telling the truth, then it means Guard C is lying as indicated from the statement.
• – If Guard C is lying, then Guard C has just told us the truth about lying.

Guard C’s statement is a paradox!

Paradoxes like the one presented in this challenge, also known as “Liar’s paradox”, are attributed to the ancient Greek seer Epimenides (fl. c. 6th century BCE). Epimenides, an inhabitant of Crete, famously declared that “All Cretans are liars”.

This paradox will arise whenever the statement refers to whoever is claiming the lie themself.

Consider: “This sentence is a lie”.

This circular logic is important in part because it creates severe difficulties for logically rigorous theories of truth; it was not adequately addressed (which is not to say solved) until the 20th century.