on October 5, 2010 by Zeta in News, Comments (0)

# Lock Puzzles Solution Guide

A dome of some blue-greenish, impenetrable metal blocks the way to the malfunctioning SpinDizzy drive:

The lock on the door is large, bulky–easily the size of a dinner table–and complicated, covered in tiny levers and dials and ornate engravings. Its material is some extremely durable and slightly contemptuous metal and its construction irritatingly sound. But closer inspection reveals that it’s some sort of unorthodox combination lock: All those levers and dials are grouped into sections and allow you to enter numbers or symbols. And the creators have been oddly helpful enough to provide you with hints on each one!

Although on closer inspection of the hints, you might begin to wonder if their motivation is charity… or mockery.

[All the puzzles on the Underworlders’s lock have been solving by the adventurers and they’re now free to proceed in their repairs on the navigational drive. Below the break the puzzles posted to the board and their solution guides. If you’re still working on the puzzles, feel free to complete them on your own time. For those who are finished or curious, read on.

Consider this post OOC please. And good luck with the rest of the landing.]

[The first section of the lock has an engraving of a small, curled up, smiling human being petted by a content cat, both unaware of an enlarged, slavering rodent behind the feline, preparing to pounce. The engraving reads:]

**OF CATS AND MICE (and men)**

[Below it are four little dials where you can enter numbers and a block of text beside each.]

1. Cats are much less bothersome than people, especially tourists, so I have a few–but not too many: Fewer than 20 cats. If you pick any two of my cats at random, though, the chance that both are white is exactly one half. How many cats do I have?

SOLUTION: Four cats of whom three are white.

Let’s say the locksmith has *c* cats and *w* of them are white. When picking out two cats, there are *w* cats to pick from for the first feline and *w*-1 for the second, so there are *w*(*w*-1) ways to pick out two white ones. Similarly there are *c*(*c*-1) to pick out any two cats. So we’re looking for *c* and *w* such that:

(*w*(*w*-1))/(*c*(*c*-1)) = 1/2

There are lots of solution to this problem, but even the locksmith doesn’t keep fractional or negative cats, so both *c* and *w* must be whole numbers. The only suitable such numbers which are less than 20 cats are four cats, three of whom are white.

This is a problem which is easily solved by testing out a few numbers intuitively. It’s interesting because while providing you with two seemingly useless pieces of information about his cats, the locksmith has nonetheless revealed their precise number.

*Correct answers:* Ali, Austin, Dragoncat, Findra, fluffy, Jukka, Kefan, Natasha, Patashu, Wolverine

2. Our local carnivorous, rapacious Chthonic Mice are capable of breeding at a prodigious rate, multiplying at alarming speed in the presence of an ample food supply (such as naive Topsiders). They can give birth once a month, to exactly one dozen babies each time. Baby chthonic mice reach mating maturity and can begin giving birth two months after they are born.

You, as a naive Topsider, purchase a baby chthonic mouse from the Inadvisable Pet Emporium and brought it home the day after it was born (they can be rather cute when they’re little). In 10 months from now, how many dangerously predatory chthonic mice will you have menacing the Topside?

SOLUTION: One mouse. Chthonic mice don’t reach *mating* maturity until two months are past. But with no suitable mates, you’ll still only have one Chthonic rodent to manage.

Just a silly trick question, really.

*Correct answers: *Ali, Findra, fluffy, Jukka, Kefan, Natasha, Patashu, Wolverine

3. Karmic Reversal: If five of our Chthonic Mice can catch and devour five of your Topside cats in five minutes, how many minutes will it take 100 Chthonic Mice to catch and devour 100 of your Topside cats?

SOLUTION (Submitted by Natasha): This is a variation on the old ‘If X people can dig X holes in X minutes…’ puzzle, isn’t it? Doesn’t it depend on what size the holes…er, cats are? Oh well, at least the know-it-all bottom-siders aren’t tormenting us with towers of Hanoi. Assuming all Chthonic mice catch and devour any cat at the same rate [*Editor’s note*: Rate of Felines Devoured = 1 cat/mouse/5 minutes] and they all work at once, then any number of mice will devour the same number of cats in five minutes.

(My favorite variation on this cute-but-nonsensical puzzle is: “Q: If X

people dig X holes in X minutes, how many minutes will it take one

person to dig 1/2 hole? A: YOU CAN’T DIG HALF A HOLE, YOU STUPID

TOP-SIDER.”)

*Correct answers:* Ali, Austin, Dragoncat, Findra, fluffy, Jukka, Kefan, Natasha, Patashu, Wolverine

4. Sadly, we don’t deal exclusively with entities as personable as cats and man-eating rodents. To celebrate (ugh) ‘family values’ and propagation of species less wholesome and discreet than chthonic mice, I’ve been charged with inviting a collection of Topsiders to a party. We are required to invite one grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law and one daughter-in-law.

But I hate you all and want to invite as few as possible. Unfortunately, I also don’t trust you–I need your sibling to be there if you want to convince me that you’re a brother, for example. So what’s the smallest number of meddlesome Topsiders I must invite? (And we’ll give you the dubious benefit of the doubt that you’re not, despite appearances, a bunch of inbred genetic mutants. You may assume no incest. I don’t want to know the details of what you get up to up there anyway.)

SOLUTION (submitted by Wolverine): 7 People

A = Grandfather, father of C, father-in-law of D B = Grandmother, mother of C, mother-in-law of D C = Son of A and B, and a Father D = Daughter-in-law of A and B, and a mother E = Daughter of C and D, Granddaughter of A and B, sister of F and G F = Daughter of C and D, Granddaughter of A and B, sister of E and G G = Son of C and D, Grandson of A and B, brother of E and F Needed Grandfathers - A 1 Grandmothers - B 1 Fathers - A C 2 Mothers - B D 2 Children - E F G C 4 Grandchildren - E F G 3 Brothers - G 1 Sisters - E F 2 Sons - C G 2 Daughters - E F 2 Father-in-law - A 1 Mother-in-law - B 1 Daughter-in-law - D 1 A B | | +---------+ | C D | | +---+-----+------+---+ | | | E F G

Some people correctly pointed out that if guests are allowed to vouch for relationships to people outside the party, we need only invite 4.

*Correct answers:* Ali, Austin, Findra, fluffy, Jukka, Kefan, Natasha, Patashu, Wolverine

**COUNTING TO TEN (minus one)**

Perhaps I’m being overly optimistic but I assume that the current generation of Topsiders can still count all the way to ten? If so, here’s to the credit of the Neopolis School District. For this part of the lock, I won’t even ask that you count all the way to ten. You can stop at nine. Indolent.

1. Here’s some (hopefully) familiar digits:

1 2 3 4 5 6 7 8 9

Use the dials to place *exactly* three common mathematical symbols between them so that the result is equal to 100. You can repeat a symbol, but each counts towards the limit of three.

Oh, and rearranging or changing the digits is strictly verboten, so put away the blow torch. They’re in the right order, if you don’t believe me, relearn how to count, Topsider.

SOLUTION: 123 – 45 – 67 + 89

Some adventurers wrote programs to search for an answer. This puzzle is from Henry Ernest Dudeney, perhaps the greatest puzzle writer of the Victorian Age. He was asking people puzzles like these 100 years ago (although he had the common courtesy not to build them into a padlock).

*Correct answers:* Ali, Austin, Dragoncat, fluffy, Jukka, Kefan, Natasha, Patashu

2. Use each of the numbers one through nine exactly once each to fill in the blanks in this equation:

_____ – ____ = 33333

And yes, smarty: The equation has to be correct.

SOLUTION: There are two answers for this one: 41268-7935 and 41286-7953 since we can interchange the last two tens digits. The lock has been built to accept either.

fluffy went into a fair amount of depth on this one.

*Correct answers:* Ali, Austin, Dragoncat, fluffy, Jukka, Kefan, Natasha, Patashu

3. I know that using ALL those numbers was tiring for your rose scent addled brains. I am not without compassion, so to give you a break, you’ll only need four single digits to open this part of the lock. But which?

A + C = D

A x B = C

C – B = B

A x 4 = D

What digits are A, B, C, and D?

SOLUTION (Adapted from Wolverine):

We are presented in this puzzles with four equations and must solve for four unknowns. This is a straightforward problem of simplification and substitution:

A + C = A x 4 -A -A [Subtracting A from both sides] C = A x 3 A x B = C A x B = A x 3 [Substituting in value of C from above] B = 3 C - 3 = 3 C = 6 A = 2 B = 3 C = 6 D = 8

*Correct answers:* Ali, Austin, Dragoncat, fluffy, Jukka, Kefan, Natasha, Patashu, Wolverine

**A COMPLETE WASTE OF TIME (mine, not yours. Yours is worthless, of course)**

1. [This section of the lock has a working digital clock implanted in it. It reads 03:00 at the time Azure recorded the front of the lock.].

Do you see the clock to the left? Good. If you’re unfamiliar with it, it’s a tool developed to help measure time. Those who have evolved beyond dragging their knuckles on the ground have need of such things.

In the course of a day, how many times will this clock display the same digit three or more times in a row? If you need to sit here and watch it and count to figure it out, I won’t tell anyone.

Oh, and undoubtedly many of you are toting fire arms at the moment like barbarians, as Topsiders are wont to do. So to those of you with a higher side arm caliber than mental caliber: This clock uses a standard 12 hour scale, not some uppity military time.

I can tell what you’re thinking, and let me remind you that shooting the clock won’t help you either.

SOLUTION: From midnight to noon, seventeen times fulfilling the criteria appear:

12:22

1:11

2:22

3:33

4:44

5:55

10:00

11:10

11:11

11:12

11:13

11:14

11:15

11:16

11:17

11:18

11:19

These seventeen are repeated again from noon to midnight, for a total of 34 times in the course of the day. There’s nothing really tricky to this problem except clock time makes it easy to miss afew (such as 11:10 or 10:00).

*Correct answers:* Ali, Austin, Dragoncat, Jukka, Kefan, Natasha, Patashu, Wolverine

2. [This section of the lock has a working ANALOG clock implanted in it. But there’s something wrong with it: The minute hand and hour hand are the same size and you can’t tell which is which. One is pointing straight up at 12 and the other straight out to the right at 3. Thinking about, you realize this has to be 3:00–the hour hand would be in a different position at 12:15]

Some of you at the head of your group can probably even read a non-digital clock. Good for you. But even you might have some trouble reading the clock I’ve built to the left. Still, if you think about it, you can probably tell what time it is.

Usually, that is. There are times each day at which it’s impossible to tell with certainty which hand is which and what time it is by looking at the clock.

How many such times exist in the course of the day when you can’t tell what time it is?

SOLUTION: 234 times.

This problem may be solved algebraically, but I usually find it easier to do so with a graphical intuition. Although we can’t tell which hand is the hour hand and minute hand, it helps to be able to differentiate them. For example, by calling one hand the blue hand and one hand the red hand:

We can ‘unwind’ the rim of the clock to make a graph of the various possible combinations of positions for both hands. For the sake of being able to distinguish noon from midnight, we’ll put ‘0’ on this graph rather than having ’12’ appear twice:

Not every point on this graph represents a valid position of the two hands for a clock. For example, you will never have both hands pointing directly to eleven. However, every valid time is represented by a point somewhere on this graph. For example, let us assume for a moment that the blue hand is the minute hand. Then the time 9:30 should have the blue hand pointing to 6 and the red hand halfway between 9 and 10 as follows:

If the red hand is the minute hand and the blue hand is the hour hand, 9:30 is represented by a different point:

For the moment, let us assume the blue hand is the minute hand and the red hand is the hour hand. Then consider what happens from 12:00 to 1:00: The blue hand will travel completely around the clock, while the hour hand only moves from 12 to 1. We may represent this motion with a simple line as follows:

And then from 1:00 to 2:00:

The entire path of the hands from 12:00 to 12:00 if the blue hand is the minute hand looks something like this:

Of course, there’s another possibility: The red hand might be the minute hand and the blue hand the hour hand. In this case, the movement of the clock hands from 12:00 to 12:00 looks something like this:

When we graph both possibilities at once, we find a cross hatch pattern:

Consider the intersections of the two different sets of colored lines in this graph. What do they represent? They represent a valid time regardless of which hand is the minute hand and which is the hour hand — as such we can’t tell which hand is which! There are 144 such intersections, however, 12 of them lie along the identity function (y=x) where the blue hand and red hand overlap. These are the intersections along the purple line below:

Even though we can’t tell which hand is which at these points, we can tell what time it is, and the problem is specifically asking for not being able to tell the correct time as opposed to not being able to tell which hand is which. There are 12 points along this line, leaving 132 ambiguous times from 12:00 to 12:00 or 234 in a full day.

The algebraic approach follows the same intuition, but works symbolically. It boils down to something equivalent to the lines intersecting, however. This may have been the most difficult puzzle on the lock unless it was the last one.

*Correct answers:* Dragoncat, Jukka, Kefan, Patashu

[This section of the lock features a prominent engraving of a group of various species, rendered with hawaiian shirts, flashy cameras, and open mouths as if gabbing loudly. They traverse a rocky landscape in a meandering line, apparently completely unaware of a pit of ravenous… mice? directly in front of them.]

**TOURISM HAZARDS (you probably should just stay at home, really)**

1. A careless coati has gotten herself pricked by a cactus needle whose tip carried a single, but dangerous prokaryote organism. After one day, this germ will have split into two germs. One day later, the two will each split again, forming a total of four germs. Continuing in this fashion, the single bacteria will fill the coati and liquify her internal organs in exactly sixty days.

Knowing this, how long in days would the coati live if she had been pricked by two needles?

And for the record, these puzzles are a work of fiction. Any resemblance between these puzzles and actual events or persons, whether living or soon to be dead, is purely coincidental. Honest.

SOLUTION: 59 days. The growth is exponential, so starting with twice as many needles just moves the progress forward a single day.

Alternative explanation, courtesy of Natasha: Since the doubling time is one day, and since two is double as much as one, that means one less day of doubling, or fifty-nine…um…uh… where was I? Sorry, I’m coming over all sleepy and hazy all of a sudden. I think I’ll skip the last two puzzles and go lie down, since I don’t know how to find expected time anyway…zzzzz.

*Correct answers: *Ali, Austin, Dragoncat, Jukka, Kefan, Natasha, Patashu, Wolverine

2. A wallaby of dubious and often overstated mental capabilities has gotten himself lost in a garden maze. Initially he has to choose one of two directions: If he goes right, then he will wander around listening to our floral music for 3 minutes before returning to his initial position. If he goes left, then with probability 1/3 he will escape the maze after 2 minutes, and with probability 2/3rd, he’ll get lost and return to his initial position after 5 minutes of wandering.

Assume the wallaby isn’t especially endowed in the learning department and is at all times equally likely to pick left or right from the initial position. What is the expected amount of time the wallaby will be trapped in the maze?

SOLUTION: 21 minutes.

This problem can either be very difficult or very easy to solve depending on how one’s approach. The easiest way to solve it is to set up a recursive recurrence. Let W be the expected amount of time it takes the wallaby to escape from his starting position. The key insight is that whenever he returns back to that position, he’s effectively reset, and it will be another W minutes on average before he escapes. So:

W = 1/2 * [Expected time to exit if wallaby goes right] + 1/2 * [Expected time to exit if wallaby goes left] W = 1/2 * (3 + W) + 1/2 * (1/3 * 2 + 2/3 * (5 + W)) 6*W = 9 + 3*W + 2 + 10 + 2*W [Multiplying both sides by 6 for easier terms] W = 9 + 2 + 10 [Subtracting 5*W from both sides] W = 21

One adventurer who solved this problem very quickly was given the additional challenge of calculating the *variance* of the amount of time the wallaby spends trapped in the garden. This is a little trickier.

This problem was adapted from Sheldon Ross’s textbook on Probability Models.

*Correct answers:* Ali, Dragoncat, Jukka, Kefan, Patashu

[The last section of the lock has a beautifully rendered engraving: The entire planetoid ripped apart by abstract representations of a disintegrating gravitational field.]

**EXPECTATION OF THE ESCHATON (grab a dictionary if you need one)**

1. I like to end things on a cheerful note. Our findings indicate that the SpinDizzy’s fault tolerant systems will only suffer a catastrophic failure of apocalyptic proportions if there are ten consecutive drive failures.

Further, we estimate that a drive failure occurs independently each year with probability 1/10, and no more than one failure can occur each year.

With these assumptions, what is the expected number of years before the field collapses and you all reap the consequences of living on such a shoddily created world?

SOLUTION: 11,111,111,110 years. We should be plenty safe.

When approaching this problem, many notice that the probability of getting 10 failures in the next ten years is simply (1/10)^10 and calculate the average as the reciprocal of that number, arriving at 10,000,000,000 years. While intuitive, this is incorrect. Consider flipping coins until one gets two heads in a row. There’s always a chance of 1/4 of getting two heads on the next two flips, but if you try it out, you’ll discover that the average number of flips it takes is not 1/(1/4) = 4, but 6.

Others approached this problem by calculating the first year at which the probability of a failure having occurred is greater than 50% and arrived at a number somewhere above 6 billion years. Again, this is intuitive, but the first year at which the SpinDizzy is more likely than not to have suffered a catastrophic field collapse is not the same as the *expected* amount of time until the SpinDizzy suffers a catastrophic field collapse. Consider a different situation where the only possibilities are the entire SpinDizzy collapses in a ball of flame in year one with probability 51% or year 1,000,000 with probability 49%. Then clearly the first year where the probability of catastrophic failure is greater than half is 1, but the expected or mean time till failure is a little over 49,000. This is a distinction between median and mean.

The insight is that gravity field collapses are actually a lot like wallabies lost in gardens. The situation may be approached recursively:

Let E[k] be the expected time to get k consecutive failures. The answer we’re looking for eventually is E[10]. Let’s assume we know how long it takes to get k-1 consecutive failures, E[k-1], and we want to figure out E[k]. Then our time to get k failures, E[k] is simply the expected time of getting k-1 failures, E[k-1], and then either 1 more year in the case where we get a failure that year (which occurs with probability 1/10) or 1 more year plus E[k] in the case where we don’t get a failure that year and effectively restart (which occurs with probability 9/10). So:

E[k] = E[k-1] + 1*1/10 + 9/10*(1+E[k]) E[k] = E[k-1] + 1/10 + 9/10*+9E[k]/10 E[k]/10 = E[k-1] + 1 E[k] = 10*E[k-1] + 10

Now consider that E[0] is simply 0 (it takes us 0 years to get 0 failures in a row) and looking at that relation, it’s obvious that the answer is just the sum of the first 10 powers of 10 starting with 10^1.

This was a tricky problem, but the answer was cute. I do think the problem could have been worded to be a little bit more clear about what is meant by ‘expected’ (the ‘mean’ or average might have been more precise (although in the latter case, I’m sure someone would point out that ‘average’ can also refer to the median), so apologies for any confusion along those lines.

*Correct Answers:* Ali, Jukka, Kefan, Patashu

**Achievements Section:**

- Completionist: Jukka, Kefan, Patashu
- Secchian Savant: Wolverine (produced clearest solutions)
- Flash of Silver: Ali (fastest puzzle solver)
- Honorary Cyborgs aka Digital Slave Drivers: Natasha, fluffy (completed a non-trivial number of problems by making a machine do the heavy lifting for them. Nothing wrong with that)
- Blue Fox’s Award For AWKward MVP: Natasha (most amusing answers submitted)
- White Squirrel’s Commendation For SED’s Most Valuable Asset: Jukka (first to complete all puzzles correctly)

[I hope these puzzles represented a pleasing change of pace and people enjoyed them. I apologize for any confusion in their specificity. If you’re interested in things like this in the future, or have suggestions for how it could be done better, please contact me on the MUCK or comment below.

Now that a sufficient number of answers have been provided to open the lock with confidence that we won’t simply make the Bottom Siders mock us (much), we’ll be trying to continue the repairs soon, perhaps this week.]

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