Tag Archives: brain melter

Have You Been Playing Uno Wrong All Along?

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unocards

The holiday season is one of the few times in a given calendar year I play games with most of my family members.

The necessity of gathering for multiple events — it takes two or three days to see everyone around Christmastime, based on geography, family obligations, and such — creates opportunities for group gameplay that simply don’t exist other times of the year.

This got me thinking about house rules.

Every family has house rules for games and activities. Maybe it’s where you stand and throw in a round of cornhole, or what’s fair in a game of Horse, or how many do-overs younger kids get during a trivia game. It could be whether you call all shots during a pool game or only the 8-ball shot. That sort of thing.

I virtually guarantee that every household has some house rules for Monopoly, whether it’s doubling your $200 if you land directly on Go or collecting previously-paid fees when you land on Free Parking.

As it turns out, a lot of us have been playing Uno with house rules as well.

Get this:

unocards3

That’s from the official Uno Twitter account, which I didn’t know was a thing.

This was also a total surprise to me. Growing up, I learned that you can stack Draw 2 cards or Draw 4 cards. Apparently, in some households, you can add to Draw 2 with a Draw 4 or a Draw 4 with a Draw 2, making a Draw 6 for an opponent.

In any case, that sort of stacking has never been allowed in the official rules.

Gasp! That means many heartbreaking Uno moments from my childhood could have been avoided!

So, I decided to dig a little further. Were there other rules I didn’t know about?

As it turns out… there were.

unocards2

In this Facebook post from January of 2018, an astonished Uno player discovered this little gem in the Uno rule book:

Did y’all know that you can only play the Draw 4 Wild card IF you have NO other cards of the same color that can be played??! AND if you suspect that someone has illegally played this card, they have to show you their hand. AND if they in fact played the card illegally they must draw 4, but if not, the person who challenged the play must DRAW 6?

How am I only learning about these rules now?! I, for one, never knew that you could force someone to show you their hand if they broke the honor system Go Fish-style.

Have these revelations changed the way you play Uno, fellow puzzlers? Or am I in the minority as part of a group that thought we knew the rules, but were very much mistaken?

Let us know in the comments section below! We’d love to hear from you.

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Four Dimensional Hats: A Visual Wonder!

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[Image courtesy of Brilliant.org.]

The mobius strip is one of the simplest objects in the world, and yet, the most mind-bending. If you take a strip of paper, add a single twist, and tape the ends together, you transform a two-sided object into a single-sided object. It becomes one continuous surface.

(We’ve discussed the concept briefly in the blog before, but in bagel form.)

But did you know that you can take that idea a step further and end up with this?

[Image courtesy of math.union.edu.]

This is a Klein bottle, an object with one continuous surface. If you trace a path along the surface, you will traverse from the “inside” to the “outside” and back again without breaking stride.

Yes, the word “bottle” is a bit of a misnomer, since this won’t actually hold any liquids; they would just flow along the surface, going “inside” and back “out” without pooling anywhere. This is a result of a mistranslation, as the German word “flache” (surface) was translated as “flasche” (bottle).

This limerick sums up the Klein bottle nicely:

A German topologist named Klein
Thought the Mobius Loop was divine.
Said he, “If you glue
The edges of two,
You get a weird bottle like mine.”

[Image courtesy of Pinterest.]

Although the Klein bottle can’t quite exist as a three-dimensional object — since the object has to pass through itself, which can only happen in four dimensions — we can come close enough to create some impressive approximations, like the glass “Klein bottles” pictured above.

YouTuber and physics student Toby Hendy has even managed to create a technique to knit yourself a Klein bottle hat! Check it out:

Although it’s not an optical illusion, it’s still a visual puzzle for the eyes and the mind, one that has captured the imaginations of mathematicians, artists, and many others throughout the years.

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A Barrage of Brain Teasers!

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[Image courtesy of SharpBrains.com.]

One of our favorite pastimes here on PuzzleNation Blog is cracking brain teasers. From riddles to logic problems, we accept challenges from all comers, be they TV detectives or fellow PuzzleNationers!

I got an email a few days ago from a reader who needed help unraveling a few brain teasers from a list she found online. She was proud to have solved most of them, but a few had eluded her.

We’re always happy to assist a fellow puzzler, so let’s take a look at those brain teasers!

In case you want to try them for yourself before we reveal the answers and how to solve each puzzle, I’ll list the original puzzles here and put a nice spoiler-safe break between the questions and the answers.

QUESTION 1: If you have a 7-minute hourglass and an 11-minute hourlass, how can you boil an egg for exactly 15 minutes?

QUESTION 2: Name the next number in the following sequence: 1, 11, 21, 1211, 111221, 312211, _____.

QUESTION 3: Four people want to cross a river, but the only option is a narrow bridge. The bridge can only support two people at a time. It’s nighttime, and the group has one torch, which they’ll need to use every time they cross the bridge. Person A can cross the bridge in 1 minute, Person B in 2 minutes, Person C in 5 minutes, and Person D in 8 minutes. When two people cross the bridge together, they must move at the slower person’s pace. Can all four get across the bridge in 15 minutes or less?

QUESTION 4: During a recent census, a man told a census taker that he had three children. When asked their ages, he replied, “The product of their ages is 72. The sum of their ages is the same as my house number.” The census taker ran to the man’s front door and looked at the house number. “I still can’t tell,” she complained. The man replied, “Oh, that’s right, I forgot to tell you that the oldest one likes chocolate pudding.” The census taker then promptly wrote down the ages of all three children. How old are they?

QUESTION 5: There are five bags of gold that all look identical, and each contains ten gold pieces. One of the five bags has fake gold, though. All five bags are identical, and the real gold and fake gold are identical in every way, except the pieces of fake gold each weigh 1.1 grams and the pieces of real gold each weigh 1 gram. You have a perfectly accurate digital scale available to you, but you can only use it once. How do you determine which bag has the fake gold?

[Image courtesy of AwesomeJelly.com.]

Okay, here’s your spoiler alert warning before we start unraveling these brain teasers.

So if you don’t want to see them, turn away now!

…

…

…

…

Last chance!

…

…

…

…

Ready? Okay, here we go!

[Image courtesy of Just Hourglasses.com.]

QUESTION 1: If you have a 7-minute hourglass and an 11-minute hourlass, how can you boil an egg for exactly 15 minutes?

A variation on the two jugs of water puzzle we’ve covered before, this puzzle is basically some simple math, though you need to be a little abstract with it.

  • Step 1: Start boiling the egg and flip over both hourglasses.
  • Step 2: When the 7-minute hourglass runs out, flip it over to start it again. (That’s 7 minutes boiling.)
  • Step 3: When the 11-minute hourglass runs out, the 7-minute hourglass has been running for 4 minutes. Flip it over again. (That’s 11 minutes boiling.)
  • Step 4: When the 7-minute hourglass runs out, another 4 minutes has passed, and you’ve got your 15 minutes of egg-boiling time.

QUESTION 2: Name the next number in the following sequence: 1, 11, 21, 1211, 111221, 312211, _____.

The answer is 13112221. This looks like a math or a pattern-matching puzzle, but it’s far more literal than that.

Each subsequent number describes the number before it. 11, for instance, isn’t eleven, it’s one one, meaning a single one, representing the number before it, 1.

The third number, 21, isn’t twenty-one, it’s “two one,” meaning the previous number consists of two ones, aka 11.

The fourth number, 1211, translates to “one two, one one,” or 21. The fifth number, 111221, becomes “one one, one two, two one.” And the sixth, 312211, becomes “three one, two two, one one.”

So, the number we supplied, 13112221, is “one three, one one, two two, two one.”

[Image courtesy of Do Puzzles.]

QUESTION 3: Four people want to cross a river, but the only option is a narrow bridge. The bridge can only support two people at a time. It’s nighttime, and the group has one torch, which they’ll need to use every time they cross the bridge. Person A can cross the bridge in 1 minute, Person B in 2 minutes, Person C in 5 minutes, and Person D in 8 minutes. When two people cross the bridge together, they must move at the slower person’s pace. Can all four get across the bridge in 15 minutes or less?

Yes, you can get all four across the bridge in 15 minutes.

This one’s a little tougher, because people have to cross the bridge in both directions so that the torch remains in play. Also, there’s that pesky Person D, who takes so long to get across.

So what’s the most time-efficient way to get Person D across? You’d think it would be so send D across with Person A, so that way, you lose 8 minutes with D, but only 1 minute going back with the torch with A. But that means only 6 minutes remain to get A, B, and C across. If you send A and C together, that’s 5 minutes across with C, and 1 minute back with A, and there’s your 15 minutes gone, and A and B aren’t across.

So the only logical conclusion is to send C and D across together. That’s 8 minutes down. But if you send C back down, that’s another 5 minutes gone, and there’s no time to bring A, B, and C back across in time.

So, C and D have to cross together, but someone faster has to bring the torch back. And suddenly, a plan comes together.

  • Step 1: A and B cross the bridge, which takes 2 minutes. A brings the torch back across in 1 minute. Total time used so far: 3 minutes.
  • Step 2: C and D cross the bridge, which takes 8 minutes. B brings the torch back across in 2 minutes. Total time used so far: 13 minutes.
  • Step 3: A and B cross the bridge again, which takes 2 minutes. Total time used: 15 minutes.

(It technically doesn’t matter if A returns first and B returns second or if B returns first and A returns second, so long as they are the two returning the torch.)

QUESTION 4: During a recent census, a man told a census taker that he had three children. When asked their ages, he replied, “The product of their ages is 72. The sum of their ages is the same as my house number.” The census taker ran to the man’s front door and looked at the house number. “I still can’t tell,” she complained. The man replied, “Oh, that’s right, I forgot to tell you that the oldest one likes chocolate pudding.” The census taker then promptly wrote down the ages of all three children. How old are they?

Their ages are 3, 3, and 8.

Let’s pull the relevant information from this puzzle to get started. There are three children, and the product of their ages is 72.

So let’s make a list of all the three-digit combinations that, when multiplied, equal 72: 1-1-72, 1-2-36, 1-3-24, 1-4-18, 1-6-12, 1-8-9, 2-2-18, 2-3-12, 2-4-9, 2-6-6, 3-3-8, 3-4-6. We can’t eliminate any of them, because we don’t know how old the man is, so his children could be any age.

But remember, after being told that the sum of the children’s ages is the same as the house number, the census taker looks at the man’s house number, and says, “I still can’t tell.” That tells us that the sum is important.

Let’s make a list of all the sums of those three-digit combinations: 74, 39, 28, 23, 19, 18, 22, 17, 15, 14, 14, 13.

The census taker doesn’t know their ages at this point. Which means that the sum has multiple possible combinations. After all, if there was only one combination that formed the same number as the house number, the census taker would know.

And there is only one sum that appears on our list more than once: 14.

So the two possible combinations are 2-6-6 and 3-3-8.

The chocolate pudding clue is the deciding fact. The oldest child likes chocolate pudding. Only 3-3-8 has an oldest child, so 3-3-8 is our answer.

[Image courtesy of Indy Props.com.]

QUESTION 5: There are five bags of gold that all look identical, and each contains ten gold pieces. One of the five bags has fake gold, though. All five bags are identical, and the real gold and fake gold are identical in every way, except the pieces of fake gold each weigh 1.1 grams and the pieces of real gold each weigh 1 gram. You have a perfectly accurate digital scale available to you, but you can only use it once. How do you determine which bag has the fake gold?

With only one chance to use the scale, you need to maximize how much information you can glean from the scale. That means you need a gold sample from at lesst four bags (because if they all turn out to have real gold, then the fifth must be fake). But, for the sake of argument, let’s pull samples from all five bags.

How do we do this? If we pull one coin from each bag, there’s no way to distinguish which bag has the fake gold. But we can use the variance in weight to our advantage. That .1 difference helps us.

Since all the real gold will only show up before the decimal point, picking a different number of coins from each bag will help us differentiate which bag has the fake gold, because the number after the decimal point will vary.

For instance, if you take 1 coin from the first bag, 2 coins from the second, 3 coins from the third, 4 coins from the fourth, and 5 coins from the fifth, you’re covered. If the fake gold is in the first bag, your scale’s reading will end in .1, because only one coin is off. If the fake gold is in the second bag, your scale’s reading will end in .2, because two coins are off. And so on.

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The Diabolical Long Division Brain Teaser!

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From time to time, I’ll receive an email with a brain teaser I’ve never seen before. Sometimes they come from friends, or fellow puzzlers. Other times, PuzzleNationers will ask for my assistance in solving a puzzle that has flummoxed them.

That was the case with today’s puzzle, and I’ll admit, this one was a bit of a doozy to unravel.

longdiv1

Yup, an entire long division problem with only a single digit set. No letters or encryption to let us know which digits were repeated, as there are in Word Math puzzles published by our friends at Penny Dell Puzzles.

Just a 7 and a bunch of asterisks. “Is this doable?” the sender asked.

Yes, this is entirely doable, friend. Let’s break it down step by step.

First, we need to know our terminology. The 8-digit number being divided is our dividend. The 3-digit number we’re dividing into it is the divisor. The 5-digit number on top is our quotient.

For the other lines, let’s label them A through G for ease of reference later.

longdiv2

There we go. Now, where do we go from here? We start with what we know.

We know that 7 is the second digit in our quotient.

So our divisor, times 7, equals the number on line C. That’s a 3-digit number, which means the first number in our divisor is 1. Why? Because if it was 2, 2 times 7 would give us 14, which would be a 4-digit number on that line.

longdiv3

That means the quotient is somewhere between 100 and 142. (Why 142? Easy. I divided 1000 by 7, and 142 is the last 3-digit number you can multiply 7 against and still end up with a 3-digit answer for line C. 143 times 7 is 1001, which is too high.)

What else do we know from the puzzle as it stands?

Well, look at lines E and F. We bring both of the last two digits in the dividend down for the final part of the equation. What does that mean?

Remember how long division works. You multiply the divisor by whatever number gets you closest to the given digits of the dividend, subtract the remainder, bring down the next digit from the dividend, and do it all over again until you get your answer.

You multiply the first digit of the quotient times the divisor to get the number on line A. You multiply 7 times the divisor to get the number on line C. You multiply the third digit of the quotient times the divisor to get the number on line E.

Following this route, you would multiply the fourth digit of the quotient against the divisor to get the number on line G. But bringing just one digit down didn’t give us a number high enough to be divided into. Instead of needing more lines (H and I, in this case), we bring the last digit of the dividend down and press onward.

That means the fourth digit of the quotient is 0, because the divisor went into the dividend zero times at that point.

longdiv4

And there’s more we can glean just from the asterisks and what we already know. We know that every one of those 4-digit numbers in the equation begin with the number 1.

How do we know that? Easy. That first number in the divisor. With a 1 there, even if the divisor is 199 and we multiply it times 9, the highest possible answer for any of those 4-digit numbers is 1791.

So let’s fill those numbers in as well:

longdiv5

Now look at lines D, E, and F. There’s nothing below the 1 on line D. The only way that can happen is if the second digit in line D is smaller than the first digit on line E. And on line F, you can see that those first two columns in lines D and E equal zero, since there’s nothing on line F until we hit that third column of digits.

That means the second digit on line D is either a 0 or a 1, and the first digit on line E is a 9. It’s the only way to end up with a blank space there on line F.

longdiv6

I realize there are a lot of asterisks left, but we’re actually very close to knowing our entire quotient by now.

Look at what we know. 7 times the divisor gives us a 3-digit answer on line C. We don’t yet know if that’s the same 3-digit answer on line E, but since it’s being divided into a 4-digit number on line E and only a 3-digit number on line C, that means the third digit in our quotient is either equal to or greater than 7. So, it’s 7 or 8.

Why not 9? Because of the 4-digit answers on lines A and G. Those would have to be higher than the multiplier for lines C and E because they result in 4-digit answers, not 3. So the digit in the first and fifth places in the quotient are higher than the digit in the third. So, if the third digit in the quotient is 7 or 8, the first and fifth are either 8 or 9.

So how do we know whether 7 or 8 is the third digit in the quotient?

Well, if it’s 7, then lines C and E would have the same 3-digit answer, both beginning with 9. But line C cannot have an answer beginning with 9, because line B is also 3 digits. The highest value the first digit in line B could have is 9, and 9 minus 9 is zero. But the number on line D begins with 1, ruling out the idea that the numbers on lines C and E are the same.

That makes the third digit in the quotient 8, and the first and fifth digits in the quotient 9.

longdiv7

We know our quotient now, 97809. What about our divisor?

Well, remember before when we narrowed it down to somewhere between 100 and 142? That’s going to come in handy now.

On line F, we know those first two digits are going to be 141 or below, because whatever our divisor is, it was larger than those three digits. That’s how we ended up with a 0 in our quotient.

So, the number on line D minus the number on line E equals 14 or below. So we need a 900-something number that, when added to a number that’s 14 or below, equals 1000 or more. That gives us a field from 986 to 999.

And that number between 986 and 999 has to be divisible by 8 for our quotient to work. And the only number in that field that fits the bill is 992. 992 divided by 8 gives us 124, which is our divisor.

longdiv8

From that point on, we can fill out the rest of the equation, including our lengthy dividend, 12128316.

longdiv9

And there you have it. With some math skills, some deduction, and some crafty puzzling, we’ve slain yet another brain teaser. Nice work everyone!

[After solving the puzzle, I did a little research, and apparently this one has been making the rounds after being featured in FiveThirtyEight’s recurring Riddler feature, so here’s a link.]

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It’s Follow-Up Friday: Math Puzzle Madness edition!

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Welcome to Follow-Up Friday!

By this time, you know the drill. Follow-Up Friday is a chance for us to revisit the subjects of previous posts and bring the PuzzleNation audience up to speed on all things puzzly.

And today, I’d like to revisit the world of viral puzzles and discuss two that have been making the rounds on Facebook recently.

If you’ve been on social media recently, you’ve no doubt seen one or both of these puzzles:

math

14962617_1223950097651735_5609824945112529117_n

Each was probably accompanied by some hyperbolic phrasing like “95% of people and most dogs can’t solve this puzzle! Heck, they can’t even agree on an answer! CAN YOU?!?!?!?!”

Well, duh. Of course they can’t agree on an answer. There’s plenty of room to make different assumptions.

Let’s look at the first puzzle again.

math

Now, if you take the puzzle at face value, the chain would appear to be this:

1 + 4 = 5

2 + 5 (+5) = 12 (We’ve added the previous answer, which is where the +5 comes from.

3 + 6 (+12) = 21

8 + 11 (+21) = 40

So the answer is 40.

But wait. if you assume that the pattern continues for the digits between 3 and 8, you end up with this:

1 + 4 = 5

2 + 5 (+5) = 12

3 + 6 (+12) = 21

4 + 7 (+21) = 32

5 + 8 (+32) = 45

6 + 9 (+45) = 60

7 + 10 (+60) = 77

8 + 11 (+77) = 96

And, in truth, it could be either. You’re not given enough information to know for sure how to proceed. It’s a coin toss whether the last line immediately follows the third line, or whether there’s a whole bunch of lines in between and you need to “get the pattern” to extrapolate the 8th line.

Now let’s look at that second puzzle again.

14962617_1223950097651735_5609824945112529117_n

This one also has the potential for alternate answers, but instead of inferences, it depends on whether you follow the traditional order of operations (parentheses, exponents, multiplication/division, addition/subtraction) or you simply read left to right.

If you use traditional order of operations, you end up with:

Horse + Horse + Horse = 30, so Horse = 10.

Horse + Horseshoes + Horseshoes = 18, so Horseshoes = 4 and Horseshoe = 2.

Horseshoes – Boots = 2, so Boots = 2 and Boot = 1.

Boot + Horse x Horseshoe = Boot + (Horse x Horseshoe) = 1 + (10 x 2) = 21.

But if you simply read the last equation from left to right, you end up with:

Boot + Horse x Horseshoe = 1 + 10 x 2 = 11 x 2 = 22.

So, in fairness, there is no right answer to either puzzle, given the information we have.

Which, to me, doesn’t seem like a great puzzle, but it probably makes for great clickbait.

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Cracking the GCHQ Christmas Card!

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As you may recall, my fellow puzzlers and PuzzleNationers, a few months ago, a government organization in England called the GCHQ — Government Communications Headquarters — released a puzzly Christmas card designed to tax even the savviest puzzle solvers.

They’ve finally released the answers to this mind-blowing series of puzzles, and I’d like to go over some of them with you. Partly to marvel at the puzzle wizardry necessary to solve this challenging holiday gift, and partly to gloat about the parts I managed to solve.

So let’s get to it!

Part 1 was a logic art puzzle where you have to deduce where to place black squares on an open grid in order to form a picture.

Each column and row has a series of numbers in it. These numbers represent runs of black squares in a row, so a 1 means there’s one black square followed by a blank square on either side and a 7 means 7 black squares together with a blank square on either side.

This is mostly a deduction puzzle — figuring out how to place all the strings of black squares with white spaces between them within the space allotted — but no image immediately emerged, which was frustrating. Once the three corner squares started to form though, I realized the answer was a QR code, and the puzzle started to come together nicely.

Part 2 was a series of six multiple-choice brain teasers. I’ll give you the first three questions, along with answers.

Q1. Which of these is not the odd one out?

A. STARLET
B. SONNET
C. SAFFRON
D. SHALLOT
E. TORRENT
F. SUGGEST

Now, if you stare at a list of words long enough, you can form your own patterns easily. Here’s the rationale the GCHQ used to eliminate the odd ones out:

STARLET is an odd one out because it does not contain a double letter.
SONNET is an odd one out because it has 6 letters rather than 7.
SAFFRON is an odd one out because it ends in N rather than T.
TORRENT is an odd one out because it starts with T rather than S.
SUGGEST is an odd one out because it is a verb rather than a noun.

SHALLOT is our answer.

Q2. What comes after GREEN, RED, BROWN, RED, BLUE, -, YELLOW, PINK?
A. RED
B. YELLOW
C. GREEN
D. BROWN
E. BLUE
F. PINK

After playing around with some associative patterns for a while, I realized that somehow these colors must equate to numbers. First I tried word lengths, but 5-3-5-3-4-___-6-4 didn’t make any sense to me. But then, it hit me: another time where colors and numbers mix.

Pool balls. Of course, the colors and numbers didn’t match, because this is a British puzzle, and they don’t play pool, they play snooker.

So the colored balls in snooker become the numbers 3, 1, 4, 1, 5, -, 2, 6. The numbers of Pi. And now the blank makes sense, because Pi reads 3.1415926, and there’s no 9 ball in snooker.

So the next number in the chain is 5, and 5 is the color BLUE.

Q3. Which is the odd one out?
A. MATURE
B. LOVE
C. WILDE
D. BUCKET
E. BECKHAM
F. SHAKA

This one came pretty quickly to me, as the names Oscar Wilde and Charlie Bucket leapt out. And if you follow the phonetic alphabet, you also get Victor Mature, Romeo Beckham, and Shaka Zulu. (I didn’t get Mike Love, however.)

Since Shaka Zulu was the only one where the phonetic alphabet word was the surname, not the first name, SHAKA is the odd one out.

(The other three questions included an encryption puzzle, a number pattern (or progressions puzzle), and a single-letter puzzle.)

Granted, since you could retake this part as many times as you wanted, you could luck your way through or brute force the game by trying every permutation. But managing to solve most of them made this part go much faster.

Part 3 consisted of word puzzles, and was easily my favorite section, because it played to some strengths of mine.

A. Complete the sequence:

Buck, Cod, Dahlia, Rook, Cuckoo, Rail, Haddock, ?

This sequence is a palindrome, so the missing word is CUB.

B. Sum:

pest + √(unfixed – riots) = ?

This one is a little more involved. To complete the formula, you need to figure out what numbers the words represent. And each word is an anagram of a French number. Which gives you:

sept + √(dix-neuf – trois) = ?

Dix-neuf is nineteen and trois is three, so that’s sixteen beneath a square root sign, which equals four. And sept (seven) plus four is eleven.

The French word for eleven is onze, and ZONE is the only anagram word that fits.

C. Samuel says: if agony is the opposite of denial, and witty is the opposite of tepid, then what is the opposite of smart?

This is a terrific brain teaser, because at first blush, it reads like nonsense, until suddenly it clicks. Samuel is Samuel Morse, so you need to use Morse Code to solve this one. I translated “agony” and tried reversing the pattern of dots and dashes, but that didn’t work.

As it turns out, you need to swap the dots and dashes, and that’s what makes “denial” read out. This also worked with “witty” and “tepid,” so when I tried it with “smart,” the opposite was OFTEN.

D. The answers to the following cryptic crossword clues are all words of the same length. We have provided the first four clues only. What is the seventh and last answer?

1. Withdraw as sailors hold festive sing-song
2. It receives a worker and returns a queen
3. Try and sing medley of violin parts
4. Fit for capture
5.
6.
7. ?

Now, I’m not a strong cryptic crossword solver, so this part took FOREVER. Let’s work through it one clue at a time.

1. Withdraw as sailors hold festive sing-song

The word WASSAIL both reads out in “withdraw as sailors hold” and means “festive sing-song.”

2. It receives a worker and returns a queen

The word ANTENNA both “receives” and is formed by “a worker” (ANT) and “returns a queen” (ANNE, reading backward).

3. Try and sing medley of violin parts

The word STRINGY is both an anagram of “try” and “sing” and a violin part (STRING).

4. Fit for capture

The word SEIZURE means both “fit” and “capture.”

Those four answers read out like this:

WASSAIL
ANTENNA
STRINGY
SEIZURE

And with three more answers to go, it seemed only natural that three more seven-letter answers were forthcoming. Plus, when you read the words spelling out downward, you notice that the first four letters of WASSAIL, ANTENNA, STRINGY, and SEIZURE were spelling out.

If you follow that thought, you end up with the start of a 7×7 word square:

 WASSAIL
ANTENNA
STRINGY
SEIZURE
ANNU___
INGR___
LAYE___

And the only seven-letter word starting with INGR that I could think of was INGRATE.

WASSAIL
ANTENNA
STRINGY
SEIZURE
ANNU_A_
INGRATE
LAYE_E_

And if the last word is LAYERED…

WASSAIL
ANTENNA
STRINGY
SEIZURE
ANNU_AR
INGRATE
LAYERED

Then the missing word must be ANNULAR. The original question asked for the last word though, so our answer is LAYERED.

This brings us to Part 4, Number Puzzles, where I must confess that I finally tapped out, because I could only figure out the first of the three progressions involved.

Fill in the missing numbers.

A. 2, 4, 8, 1, 3, 6, 18, 26, ?, 12, 24, 49, 89, 134, 378, 656, 117, 224, 548, 1456, 2912, 4934, 8868, 1771, 3543, …

B. -101250000, -1728000, -4900, 360, 675, 200, ?, …

C. 321, 444, 675, 680, 370, 268, 949, 206, 851, ?, …

In the first one, you’re simply multiplying by 2 as you go.

2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, and so on.

But you begin to exclude every other number as you move into double-digits, triple-digits, quadruple-digits, and beyond.

2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, and so on.

So the answer, 512, becomes the real answer, 52.

But, as I said, I couldn’t crack the other two, and I’m already exhausted just running through these four sections!

And, based on the answers they released recently, Part 5 only got more mindbending from there.

As a matter of fact, not a single entrant managed to get every answer in Part 5 correct. Prizes were awarded to the three people who came closest however, and it turns out a staggering 30,000+ people made it to Part 5. Color me impressed!

This was, without a doubt, the most challenging puzzle suite I have ever seen, and I offer heartfelt kudos to anyone in the PuzzleNation Blog readership who even attempted it!

You’re welcome to try it out for yourself, though. I highly recommend using this link from The Telegraph, which allows you to skip to the next part if you get stumped.

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