Tag Archives: Puzzle to solve

One Letter Makes All the Difference

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I spend a lot of time on this blog talking about the power of words, and rightly so. Words are the foundation of civilization. They’re how we communicate, how we express ourselves, how we interpret and process and quantify the world around us.

And playing with them is a cornerstone of entertainment. Jokes and puns depend on wordplay, as do riddles, brain teasers, and so many puzzles. Whether they’re being crossed, anagrammed, or shared with friends on an online Scrabble board, words are puzzle power.

That’s true even of letters. A single letter can not only speak volumes, it can be the key to unlocking an entire puzzle.

For instance, let’s talk crosswords. Knowing one across entry is a plural often allows you to place an S, giving you an anchor for the down entry that crosses it.

Cryptograms often offer a single letter as a hook to get you started. In addition, anytime you see a lone letter in a quote, you know it’s an I or an A.

Numerous anagram puzzles involve adding a single letter to a word, anagramming the result, and getting something new and unexpected.

That sort of letter addition reminds me of a brain teaser:

There’s a word in the English language in which the first 2 letters signify a male, the first 3 signify a female, the first 4 signify a great man, and the whole word signifies a great woman. What is that word?

In all of these examples, single letters are part of a greater puzzle. But what about puzzles composed entirely of single letters?

Easy. I can think of two.

In the first, single letters are the first letters of words forming some sort of pattern. Can you deduce the pattern AND provide the next entry in the series?

  • O, T, T, F, F, S, S, E, ?
  • J, F, M, A, M, J, J, A, S, O, N, ?
  • S, M, H, D, W, M, ?
  • M, V, E, M, J, S, U, ?
  • D, K, P, C, O, F, G, ?

In the second, the single letters are still the first letters of words, but we’ve added numbers and it’s up to you to deduce what the letters represent.

  • 3 F in a Y
  • 366 D in a LY
  • 12 S in the Z
  • 4 Q in a D
  • 13 C in a S

All of this, sprung from a single letter. Pretty impressive, isn’t it? No wonder we can accomplish so much with words, given building blocks like these.

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It’s Follow-Up Friday: Mother’s Day Answers 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’ll be posting the answers to our Mother’s Day Unscramblers puzzle!

On Sunday, in honor of the many puzzle-loving moms in the audience, I created a twofold challenge. First, you had to unscramble the names of 12 famous sitcom mothers, and then you had to match them with their respective TV shows.

So, without further ado and hullabaloo, here are the answers!

1. Marion Cunningham — H. Happy Days
2. Samantha Stephens — D. Bewitched
3. Morticia Addams — B. The Addams Family
4. Jill Taylor — L. Home Improvement
5. Peggy Bundy — G. Married… With Children
6. Sophia Petrillo — K. The Golden Girls
7. Claire Dunphy — J. Modern Family
8. Florida Evans — E. Good Times
9. Maggie Seaver — F. Growing Pains
10. Thelma Harper — A. Mama’s Family
11. Harriette Winslow — I. Family Matters
12. Jane Jetson — C. The Jetsons

How did you do? Are you a Mother’s Day puzzle master, or did you sitcom this one out?

Thanks for visiting PuzzleNation Blog today! You can share your pictures with us on Instagram, friend us on Facebook, check us out on Twitter, Pinterest, and Tumblr, and be sure to check out the growing library of PuzzleNation apps and games!

Happy Mother’s Day!

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Today is Mother’s Day, and as always, I’d like to celebrate with a puzzle! So, in honor of the day and mother’s everywhere, I’ve conjured up a Penny/Dell Puzzles-style Unscramblers puzzle for you!

Rearrange the pairs of letters in the left-hand column to form the names of 12 TV sitcom moms. Then match them with their TV shows in the right-hand column!

Enjoy! And Happy Mother’s Day to all the marvelous, wonderful, inspiring, hard-working moms out there!

Thanks for visiting PuzzleNation Blog today! You can share your pictures with us on Instagram, friend us on Facebook, check us out on Twitter, Pinterest, and Tumblr, and be sure to check out the growing library of PuzzleNation apps and games!

It’s Follow-Up Friday: Birthday Puzzle 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 return to the subject of birthday brain teasers!

Working on the Cheryl’s Birthday brain teaser a few days ago reminded me of another birthday-fueled puzzle that’s been around forever.

How many people do you need to enter a room before the probability of any 2 or more people sharing a birthday (day and month only, not year) is greater than 50%?

Assume for the sake of the puzzle that birthdays in the population at large are equally spread over a 365 day year.

Now, given that there are 365 days in the year, you’d assume the number of people necessary to get that probability of a shared birthday above 50% would be more than half of 365, or 183 people.

But it turns out that, statistically speaking, you don’t need anywhere near that many people.

Let’s break it down. Person A has a birthday. Person B has a birthday. There’s only one possible pairing, A-B. Person C has a birthday, but creates three possible birthday pairings: A-B, A-C, and B-C.

Person D could have a different birthday, but the introduction of Person D begins escalating the number of POSSIBLE shared birthdays. With these four people, we have SIX possible pairings: A-B, A-C, A-D, B-C, B-D, and C-D.

Our fifth person, Person E, allows for TEN possible pairings: A-B, A-C, A-D, A-E, B-C, B-D, B-E, C-D, C-E, and D-E. The probability of a shared birthday is increasing much faster with each new person.

As it turns out, it only takes 23 people to give us a 51% probability of a shared birthday.

And that would certainly save on catering.

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Are you smarter than a Singaporean student?

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We love brain teasers here at PuzzleNation Blog. Whether we’re dealing with curious parking spaces, men in hats, the crew of the Enterprise playing games, or the seesaw-based conundrum that so baffled Captain Holt on Brooklyn Nine-Nine, we thoroughly enjoy tackling these often diabolical and curious logic problems.

And one has been making the rounds on Facebook, Twitter, and other social media platforms recently. This one comes from a Singaporean classroom, and has made headlines all over the Internet.

Hmmm, doesn’t seem like a lot of information, does it?

So, we have ten possible dates.

Let’s put them in a chart to organize them as best as we can.

Now, let’s analyze each statement in order, since the progression is the key to solving this brain teaser.

Albert says: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Since Albert is told the month, and there are multiple options for each month, there is no way he could know. At first. But he does know Bernard doesn’t know either. How?

Deduction. If Cheryl told Bernard 18 or 19 (the only days that appear once), Bernard WOULD know Cheryl’s birthday. So Albert can eliminate those two options.

And for Albert to KNOW that, Cheryl cannot have told him May or June, since those were the only months with days that appear once.

A lot of information in a single sentence. Let’s move on to the next sentence.

Bernard says: At first I don’t know when Cheryl’s birthday is, but I know now.

Bernard is on the same track as Albert. He’s eliminated May and June. And he says he knows Cheryl’s birthday. If you look at our chart now, there are three singlet dates (15 for August 15, 16 for July 16, and 17 for August 17). If he was told 14, he wouldn’t know if it were July or August, so we can eliminate those.

From ten possible days, we’re down to three. And Albert’s final sentence finishes the job.

Albert says: Then I also know when Cheryl’s birthday is.

Since Albert is only told the month, it has to be July, because there are two possible dates left in August.

Therefore, in impressively brisk fashion, both Albert and Bernard have deduced that Cheryl’s birthday is July 16. And so have we!

We’ve also deduced that Cheryl is sort of a pain in the ass. But I suspect that wasn’t much of a brain teaser.

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A Puzzle in Honor of Leonard Nimoy

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Late last week, news went out that actor, sci-fi luminary, and respected advocate for science Leonard Nimoy had passed away. As a fan of much of his work — not just Star Trek, but Fringe, In Search Of…, and numerous other shows made better by his presence — I was sad to say goodbye to such a terrific and influential figure.

And so, I thought it was only appropriate to honor him in true PuzzleNation fashion: with a Star Trek-themed brain teaser.

This puzzle was created by a University of Kentucky mathematician and puzzler named Raphael Finkel, and came to my attention thanks to a post on IO9.com. It’s a logic puzzle that will really test your deduction skills, perfect to honor pop culture’s most famous Vulcan and their great tradition of logic and reason.

Six Fearsome Heroes

Grobly Grizik is planning to write a novel fashioned after Star Trek: The Next Generation. In this novel, six of the crew members compete both at Fizzbin and at Tri-Dimensional chess. Each crew member gets two independent rankings for proficiency at these games (one ranking for each game), with “1” ranked lowest and “6” highest.

Every crew member has a personal hero among the crew, and every crew member is afraid of some crew member. Everyone is the hero of somebody, and everyone is feared by somebody. Nobody either fears him/herself nor counts him/herself as a hero. Nobody fears his/her own hero.

From the given clues, discover every crew member’s ranking at Fizzbin and at Tri-D chess, as well as whom he/she fears and whom he/she counts as a hero:

  • Geordi ranks 2 at Tri-D Chess.
  • Picard ranks two positions behind Troi at Fizzbin.
  • Troi is feared by the person Geordi fears.
  • Worf’s hero ranks 3 times lower at Tri-D Chess than the crew member who is best at Fizzbin.
  • Picard’s hero fears Geordi.
  • Data’s hero is not Geordi.
  • Data is the hero of Riker’s hero.
  • The person who is worst at Fizzbin is better than Troi at Tri-D Chess.
  • The person ranked number 3 at Tri-D Chess is ranked 4 positions higher than Data at Fizzbin.
  • Riker is feared by the person Picard fears and is the hero of Worf’s hero.
  • Riker is ranked 2 lower at Tri-D Chess than the crew member ranked 2 at Fizzbin.

Let us know if you solve this diabolical puzzle. And please share your favorite Nimoy roles and memories. He was a remarkable talent and will be missed.

Thanks for visiting PuzzleNation Blog today! You can share your pictures with us on Instagram, friend us on Facebook, check us out on Twitter, Pinterest, and Tumblr, and be sure to check out the growing library of PuzzleNation apps and games!