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A Valentine’s Day Logic Puzzle – WALKTHROUGH

STEP ONE:

From the start, we see that Alan & Beth are celebrating Valentine’s Day together. This means that nobody else is celebrating with Alan, and nobody else is celebrating with Beth (since each individual is in only one relationship). Thus, we put X in the rest of Alan’s and Beth’s columns.

Whenever we successfully match a couple, we will need to do this for all remaining boxes in the matched row and column.

STEP TWO:

We also know that Sue & Steve and Mike & Megan cannot be together, since there is only one alliterative couple (which is Ryan’s). We put an X in those boxes. Also from the first clue, we see that Ryan is the only person seeing someone whose name begins with the same first letter. So Ryan is not dating Kate, Mike, or Sue (since they don’t begin with R).

Since Sue is heterosexual, she must be dating a guy. Thus she is not dating either Erica or Megan; Xs go in those boxes.

STEP THREE:

The only person left who Sue can be with is Leon. We put R in that box.

Since Leon is spending Valentine’s Day with Sue, no one else can be with Leon. Fill in Xs in the Leon’s remaining boxes.

STEP FOUR:

The end of this new clue states that Rick is in a same-sex relationship. Therefore Rick is not seeing either Erica or Megan (put Xs in those boxes).

STEP FIVE:

This is the toughest part of the puzzle.

Consider there being only one same-sex relationship:

- This would mean that Rick is in a relationship with another man.

- There would thus be 2 women and 1 man remaining in both the rows and the columns. With these options left, it is not possible to create 3 heterosexual couples (because the number of women in the rows, does not equal the number of men in the columns).

- Therefore, there MUST be 2 same-sex relationships.

- Using the same reasoning as above, the second same-sex relationship must be between 2 females (since Rick is in an all-male relationship).

Now that we know there must be a same-sex female couple, we re-examine the grid. Megan has only females remaining that she can be dating (Kate and Rachel). Therefore, Megan must be in the other same-sex relationship.

Thus, we know that Erica is NOT spending Valentine’s Day with a woman. This means Erica must be seeing Mike. We then put Xs in Mike’s and Erica’s remaining open boxes.

STEP SIX:

Steve is celebrating with a woman, so he is not seeing Rick. This means the only person left who Rick can spend Valentine’s with is Ryan. We put an X in Ryan’s remaining open box.

STEP SEVEN:

Only one of Kate and Rachel have a name that comes alphabetically before their partner’s.

- “Rachel” comes after “Megan”, but before “Steve”.

- “Kate” comes before both “Megan” and “Steve”.

Therefore, Kate must be the one who satisfies the condition in the newest clue. Consequently, Rachel must be dating Megan (because “Rachel” comes after only “Megan”), which means Kate is spending Valentine’s Day with Steve.

STEP EIGHT:

Happy Valentine’s Day!

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Name: Holly Minshull (minshkins)

Position: Associate Product Manager

Originally from: Seattle, Washington

Worked at Sporcle since: November 2012

About:  After studying Middle English at Western Washington University, Holly moved back to Seattle and joined the team at Sporcle. When she’s not Sporcling, Holly works with special needs kids, spends time with her friends, and plays in a rec soccer league. When SporcleHQ cast employees as Disney characters, Holly was chosen to be Belle.

Interests: Holly’s an avid fan of Seattle sports. (We recommend not asking her about the Sonics or the most recent Super Bowl.) She likes Indie Rock music, but she’s known around the office for being an enthusiastic Sam Smith fan. Lord of the Rings and Game of Thrones are two of her favorite series – both the books and on-screen. She also regularly attends Sporcle Live trivia nights.

Her favorite Sporcle quiz is Oldest to Youngest Minefield. But she holds a special place in her heart for quiz that introduced her to Sporcle–Countries of the World.

Strange Fact: When Holly was young, she had a large Tasmanian Devil collection. It included stuffed animals, an eraser, piggy bank, and other knick-knacks. She proudly shared them at her 3rd grade show-and-tell. 

Here at Sporcle: Holly is passionate about creating and promoting fun quizzes. She schedules content for the homepage, Facebook and other social media platforms. She works with our engineers to develop new products and improve the site. And she’s helping develop Sporcle’s newest quiz platform: Octoquiz.

Favorite thing(s) about working at Sporcle: Holly likes that every day is different at Sporcle. She likes the people and the community. And she loves the opportunity to be creative–whether coming up with a new product idea or creating a new quiz.

Sporcle teaches us tons of stuff every day. If you share our thirst for knowledge, check out these 10 cool facts.

1) Jellyfish don’t have hearts. Weird, but true. Don’t believe us? Check out this quiz for more strange heart related facts.

2) Matt Damon made a lot of movies in 2011. Surprisingly enough, none of them were action movies!

3) Tycho Brahe, an astronomer and rival of Copernicus, had a brass nose and a pet moose. Apparently the moose died after drinking a lot of beer and then falling down the stairs. We learned this one while researching the Famous Scientist Match-up  - check it out for more fantastic names!

4) Every single Friends episode, except three, is titled “The One With…” One of those special cases is in the first season, and it happens to share a common title with episodes from a lot of other TV shows. Can you guess what it is?

5) It’s really hard, but still possible, to match 30 lead singers to their bands in a minute. That being said, Mick Jagger should probably be singing “Time is not on your side…” for this one.

6) After they’ve been cut in half, baseballs and cigars look nearly the same. Also, CT scanners are strange looking whether or not they’re whole.

7) Leonard Cohen, Drake, and Avril Lavigne are more similar than you’d expect. It’s not just that they’re musicians… it’s that they all come from the same place.

8) “Don’t have a cow, man!” has been around since before The Simpsons, but Bart stole the expression and never gave it back.

9) Even though A Series of Unfortunate Events didn’t do so well in theaters, it still ended up being one of the highest grossing movies of 2004. Wait, Shrek 2 came out nearly 11 years ago? We feel old now.

10) Octopuses are able to open child-proof bottles. They’re extremely intelligent and curious, get bored if there isn’t anything interesting happening around them, and live very short lives. We learned all this while researching for yesterday’s #theresaquizforthat tweet. If you haven’t yet, head on over to the Sporcle Twitter feed for more daily facts!

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HIMYM Logic Puzzle: Find the Mother – WALKTHROUGH

CHARACTER NAMES:

• The Bartender – Carl
• The Waitress – Wendy
• Marshall’s Mom – Judy
• Lily’s Dad – Mickey
• Robin’s Co-worker/Barney’s Girlfriend – Nora
• Marvin’s Parents – Lily and Marshall
• Ted’s Daughter – Penny
• Law School Buddy – Brad
• Therapist – Kevin
• Baker – Victoria
• Barney’s Mom – Loretta
• Penny’s Brother – Luke
• Captain’s Ex-Wife – Zoey

STEP ONE:

Carl and Wendy can be plugged in right away.

STEP TWO:

Judy and Mickey can then be entered.

STEP THREE:

Based on the clues in step two, Barney is located in E1 and Robin is in A5.

STEP FOUR:

We can now see that C5 is both a co-worker of Robin and a serious girlfriend of Barney. The only person who fits these criteria is Nora (Patrice was never actually labeled at Barney’s girlfriend.)

STEP FIVE:

You then know the location of Marvin.

STEP SIX:

Based on the clues from Marvin and Wendy, it can be determined that Marshall is in B1 and Lily is in A2.

STEP SEVEN:

Penny is in B2, the only square adjacent to both Lily and Marshall.

STEP EIGHT:

Stella is adjacent to Nora, according to the clue from Lily. She cannot be located in B5 or D5, since she was never a co-worker of Robin’s or a girlfriend of Barney’s. She must therefore be in B4.

STEP NINE:

From Stella’s clue we see that somewhere in the puzzle must include Victoria, then Brad, then Kevin in a vertical line of three boxes. There are only two places remaining where this could happen (column D and column E). Column D wouldn’t work because that would put Victoria in D1, which contradicts Wendy’s clue. This means that Victoria is in E2, Brad is in E3, and Kevin in in E4.

STEP TEN:

Kevin’s clue shows that Zoey and Quinn are in D3 and D5. Quinn is the one who dated Barney (via Ranjit’s clue), so she goes in D5 and Zoey goes in E4.

STEP ELEVEN:

Brad’s clue shows that the Captain can only be located in A3 or B5, and he must be in A3 because of Robin’s clue.

STEP TWELVE:

Now there are five boxes remaining. The remaining characters are the Mother, Luke, Patrice, James, and Loretta. Patrice is the only one who worked with Robin (Robin’s clue), so she is located in B5.

STEP THIRTEEN:

James must be in D1 because D1 must be a male (Wendy’s clue), and he is the only remaining guy (besides Luke, who must be in either A3 or B3).

STEP FOURTEEN:

James’ clue states that the Mother and Luke occupy A3 and B3, leaving Loretta to D2.

STEP FIFTEEN:

Loretta’s clue leaves Luke in B3 and the Mother in A3. Congratulations!

Back to Quiz

Harry Potter Logic Puzzle – WALKTHROUGH

STEP ONE:

STEP TWO:

Mcgonagall is in D3 and Moody is in B5.

STEP THREE:

Ginny is in column A, and the 4 Weasleys in the quiz are all in the corners, so Ginny must be in A5.

STEP FOUR:

Ron’s brother who is 4 years older than him is Percy. Percy shares a diagonal with Ron, and since Percy is a Weasley he must be in a corner tile, so is in E5.

STEP FIVE:

Harry is in the same diagonal as Percy and Ron according to Ginny’s clue, so he is in either B2, C3 or D4. Since Luna is adjacent to both Harry and Ginny, she must be in B4 and Harry must be in C3. This means that Remus is in B2 as he must be adjacent to both Ron and Harry.

STEP SIX:

According to B2, Bellatrix is adjacent to Minerva, but not Remus. She is also adjacent to all 4 Triwizard champions, so must be adjacent to Harry, meaning Bellatrix is either in D2, D4 or C4. Column D is alphabetical, so Bellatrix cannot be in D4 as ‘B’ is before ‘M’. Hermione is also in column D, and so Bellatrix cannot be in D2 as then Hermione could not be in D in alphabetical order, so Bellatrix is in C4.

STEP SEVEN:

Bellatrix is adjacent to 4 empty boxes, which must include Fleur, Cedric and Viktor. Neither Fleur nor Cedric can go in either D4 or D5 as this would violate the clue that says column D is alphabetical, so Cedric and Fleur must occupy B3 and C5 in some order. C4 says that Hagrid is in column B, and the only square remaining in B is B1, so Hagrid goes there. He is adjacent to a female Dursley, which is Marge, since by C3, Lily’s sister (Petunia) is not in the quiz. Ginny is the only female in column A, and Hagrid’s clue implies that C2, D3 and E4 are all Hogwarts teachers, so Marge is in C1.

STEP EIGHT:

Neville is adjacent to Harry, but cannot be in B3 as we have already established that Fleur or Cedric is there. Neville also cannot be in C2 as Neville is not a professor from Harry’s education, and cannot be in D2 due to D being in alphabetical order. This leaves Neville in D4.

STEP NINE:

By Neville’s clue, Sirius and Voldemort are in row 3. B3 is still taken by either Fleur or Cedric. E1 has a Weasley, and since no Weasley has a name beginning with ‘S’ (not one mentioned in the books anyway) E2, E3 and E4 must all have names starting with ‘S’, as 3 in column E must have this according to Luna’s clue. This means Sirius is in E3 and Voldemort is in A3.

STEP TEN:

Cedric and Fleur are in B3 and C5 in some order which leaves Viktor in D5 as it is the only free square around Bellatrix. His clue states that Cedric is adjacent to Draco, and since Draco cannot be adjacent to C5, as their are no empty squares, Cedric must be in B3 and so Fleur is left in C5.

STEP ELEVEN:

According to D5, Kingsley is in this quiz, but is not adjacent to Cedric, so is not in A2, A4 or C2. Kingsley does not begin with ‘S’ and he is not a Weasley, so he cannot go anywhere in column E. This leaves either D1 or D2. By Harry’s clue in C3, Hermione is in column D, and since the column is in alphabetical order, Kingsley must be in D2 and Hermione in D1.

STEP TWELVE:

According to D2 there are 2 Dursleys in the quiz. Marge is already present, and we know from Harry’s clue that neither Petunia nor Dudley are in it. This leaves Vernon as the other Dursley, and he cannot be in column E (not being a Weasley or having a name starting with ‘S’). C5 states that if Vernon is in the quiz, which we now know he is, he is in a diagonal with Dumbledore. If Vernon was in A2, he couldn’t be in a diagonal with Dumbledore. This leaves Vernon in A4, and Dumbledore in C2.

STEP THIRTEEN:

By B3, we know that Draco shares a row with the human divination teacher (Trelawney), and we know Draco is adjacent to Cedric (clue D5). Therefore, Draco is in A2, and Sybill Trelawney is in E2.

STEP FOURTEEN:

By clue A3, we know that Snape is in the quiz, as he was a Hogwarts headmaster during the books. Snape is not a Weasley so must be in E4. The clue in E2 says that only 2 in column E survive the series. Percy and Trelawney both survive, so we now know that E1 is a Weasley who does not survive, leaving E1 being Fred. 

Congratulations, you’re finished!

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Sporcle teaches us tons of stuff every day. If you share our thirst for knowledge, check out these 10 cool facts.

1) Tom Cruise sat on the very top of the Burj Khalifa – the world’s tallest building – and lived to tell the tale. He also climbed it, ran along its outer walls, and jumped out one of its windows, all in the name of filming Mission Impossible: Ghost Protocol.

2) People have spent four times as much time watching PSY’s Gangam Style music video as they have building the Great Pyramid. It really makes us wonder how much – or how little – the ancient Egyptians would have done if YouTube had existed back then.

3) Most people can name the three most populous countries in order, but after that, it gets much harder. 

4) “Obstreperous” is an actual word in the English language. That being said, it’s not often that you’ll hear that particular word being used in an everyday conversation.

5) You don’t need to be a botany expert to be able to recognize fruit from pictures of their insides.

6) Only one UK Prime Minister was ever assassinated. Find out who by playing the UK History Grab Bag.

7) It’s possible to recognize great works of art, even after they’ve been pixelated. Now if only it were possible to recognize colors on the internet properly… we’re looking at you, black and blue dress.

8) Manute Bol, a basketball player from South Sudan, was 7’7″. He wasn’t the tallest man to ever live, but he was definitely up there!

9) You can get from Nigeria to Israel by land and only have to travel through three other countries to do so. Want to test your navigational skills? Try the Border Chain Match-Up!

10) Llamas can run at speeds of up to 35 mph. No wonder the Sun City police force had a really hard time catching two llamas that got loose yesterday! We learned this while researching for yesterday’s #theresaquizforthat tweet. If you haven’t yet, head on over to the Sporcle Twitter feed for more daily facts!

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 <— Back to Spy Vs. Spy Logic Puzzle

1. Clue A1 immediately tells us that A3, A5, E1, E3 and E5 are safe.
2. Clue E5 tells us that A4 was rigged by the black spy.
3. We know that there’s a pair of consecutive safe spots in column 2, thanks to A3’s clue. Let’s lay out all the possibilities: the two safe spots can be A2 and B2, B2 and C2, C2 and D2, D2 and E2, or E2 and F2.
   • But neither A2 nor E2 can be safe, because that would give us three consecutive safe spots in a row, which violates A4’s clue. So we can eliminate every possibility that includes either A2 or E2.
   • That leaves us with two possibilities: B2 and C2, or C2 and D2. However, consider the clue you got from E1. There’s one corner that’s adjacent to three rigged spaces (and no safe spaces). That corner has to be A1, since A6, F1 and F6 are each adjacent to at least one safe space.
   • That means that B1, A2 and B2 are all rigged. Since we know that B2 isn’t safe, the two consecutive safe spots in column 2 have to be C2 and D2.
   • Now let’s look at Column 4. Using the same logical process, we can immediately rule out any pairs that include A4 or E4. That leaves us with two possibilities: B4 and C4, or C4 and D4.
   • We don’t have the information necessary to narrow this down any further, but we do know that either way, C4 is safe.
4. There are three consecutive black traps in Row F, according to C4’s clue. Once again, let’s narrow down the possibilities.
   • A5′s clue told us that two corners contain white traps and two corners are safe. In other words, none of the corners contain black traps, which means that the three consecutive black spaces in Row F do not include F1 or F6.
   • That leaves us with two possibilities. The three consecutive black traps occupy either F2, F3 and F4 or F3, F4 and F5. Either way, F3 and F4 contain black traps.
5. We know that E4 cannot be safe, or else it would contradict A4’s clue. F4’s clue tells us that it can’t have been rigged by the white spy, either. Thus, E4 contains a black trap.
6. We’ve already identified two safe spaces adjacent to E4 (E3 and E5), so none of the other spaces surrounding it are safe. This includes F5, which we know cannot contain a white trap either (F4’s clue). Thus, F5 was rigged by the black spy.
7. Thus far, you’ve identified nine safe spaces (A1, A3, A5, E1, E3, E5, C2, D2 and C4). F5’s clue tells us that there are 12 safe spaces in total, so there are three more you need to find. From A3’s clue, we can conclude that two of those are in Column 6, and the last remaining one is either B4 or D4. From this point forward, we can assume that every remaining space outside Column 4 and 6 is rigged.
   • Therefore, we know that F2 was rigged, but there are exactly three consecutive black traps in Row F (and we found them all), so F2 cannot contain a black trap. F2 was rigged by the white spy.
   • We also know that F1 was rigged, but it couldn’t have been rigged by the black spy either, since none of the corners contain black traps (A5’s clue). F1 was rigged by the white spy.
8. 8. Since Row F must contain at least one safe space (F1’s clue), we can conclude that F6 is safe.
   • That leaves us with two remaining safe spots; one in Column 6, and one in Column 4. But the two safe spots in Column 6 have to be consecutive (A3’s clue), so E6 is safe.
   • And now that we’ve identified two safe corners (A1 and F6), we can conclude that the other two corners contain white traps (A5’s clue). A6 contains a white trap.
9. A6′s clue helps us figure out D4’s status. D4 cannot contain a black trap, since that would give us three consecutive black traps in Column 4. However, we know that it cannot be safe either, since E4 is only adjacent to two safe spaces (E4’s clue). D4 contains a white trap.
   • Now that we know that D4 isn’t safe, we can conclude that B4 is (A3’s clue).
10. C2′s clue told us that one column contains four white traps. Let’s figure out which one it is.
    • In order for Column 1 to have four white traps, B1, C1 and D1 would all have to contain white traps, which would contradict A6’s clue.
    • Column 3 only contains one white trap (B4’s clue).
    • We’ve already identified every space in Column 4, and we only found one white trap.
    • We haven’t found any white traps in Column 5 so far, and there are only 3 spaces remaining.
    • In order for column 6 to have four white traps, F2, F3 and F4 would all have to contain white traps, which, again, would contradict A6’s clue.
    • The only remaining column is Column 2. We’ve identified one white trap in that column so far, and there are only three more spaces remaining, so A2, B2 and E2 were rigged by the white spy.
11. B2′s clue tells us that one corner is adjacent to three spaces that all have the same status. We know that this can’t apply to F1 (adjacent to two safe spaces and one white trap) or F6 (adjacent to two safe spaces and one black trap) so that leaves us with A1 and A6.
    • A6 is adjacent to one safe space. In order for it to fit B2’s clue, both B5 and B6 would have to be safe as well. However, we have already identified all 12 safe spaces, so this cannot be true.
    • Therefore, A1 must fit B2’s clue, which means that it must be surrounded by white traps. B1 contains a white trap.
12. B4’s clue tells us that there is only one white trap in Column 3, and as we have established, none of the remaining spaces are safe. So if B3 contains a white trap, both C3 and D3 would have to contain black traps.
    • But this can’t be true. Both C3 and D3 lie on main diagonals, so according to B1’s clue, they can’t both contain black traps. B3 contains a black trap.
    • Let’s look at B4’s clue some more. Of the three remaining spaces that lie on main diagonals (C3, D3 and B5), exactly one contains a black trap.
    • This means that if B5 contains a black trap, then neither C3 nor D3 can. They’d both have to contain white traps, but that would contradict B4’s clue.
    • B5 contains a white trap.
13. E4 is adjacent to eight spaces in total, so B5’s clue leaves us with two possibilities. E4 is either adjacent to one white trap and two black traps, or two white traps and four black traps. But we’ve already identified three black traps adjacent to E4 (F3, F4 and F5), so it must be the latter. That means that of the two remaining spaces surrounding E4 (D3 and D5), one contains a white trap and one contains a black trap.
    • Look back at the clues in F3 and F5. There are 12 safe spaces in total, and the remaining 24 spaces are divided equally between the black spy and the white spy. This means that each spy rigged 12 spaces.
    • So far we’ve identified 9 spaces that were rigged by the white spy (B1, F1, A2, B2, E2, F2, D4, B5 and A6). This means that there are 3 remaining; two are in Row C (B3’s clue), and the last remaining one is either D3 or D5. Thus, we can safely assume that all other spaces contain black traps.
    • B6, D1 and D6 contain black traps.
    • Now that we know the status of B6 and D6, we can tell that C6 contains a white trap (A6’s clue).
14. C6′s clue tells us that there are two columns that contain an odd number of black traps. We already know that one of them is Column 4 (A4, E4 and F4), but what about the other one?
    • Let’s look at Column 3. There are two remaining spaces in that column that we haven’t identified, and they both lie on main diagonals. Thanks to B1’s clue, we know that exactly one of them contains a black trap.
    • We’ve already identified two other black traps in the column (B3 and F3), so that makes three in total. Columns 3 and 4 each contain an odd number of black traps.
    • This means that each of the other columns either contains an even number of black traps, or none at all. C1 contains a black trap.
15. C1′s clue tells us that C5 contains a white trap, meaning that D5 has to contain a black trap (A6’s clue). This, in turn, means that D3 has to contain a white trap (B5’s clue), which means that C3 contains a black trap (B1’s clue).

<– Back to Lively Letters Logic Puzzle II

1. A1 tell you “C4 is not a letter or a number”, which means it has to be an X

2. C4 tells us that “All letters are in cells that have BOTH a prime cell number and a cell letter whose position in the alphabet is also a prime number. This means that the letter can only be in cells that have both a cell number of either, 2, 3 or 5 (1 and 4 are not prime numbers) and a cell letter of either B, C or E. Therefore all other cells are either a number or an X. A2 is one of these other cells since it does have a cell letter of B, C or E. We are also told that A2 is not a number. Therefore A2 is an X.

3. A2 tells us that “There are no numbers in the corners, but not all corners are Xs”. Since C4 tells us that A5 and E1 can’t be letters, they have to be Xs. This also means that E5 is a letter since they are not all Xs.

4. A2 also tells us that “If D2 is not an X, then B2 has no letters in its row and no letters in its column”. So if B2 had no letters in its row and no letters in its column, that would only leave 4 cells as possibilities for letters based on the criteria laid out in the C4 clue: C3, C5, E3 and E5. However, since we know that there are 5 letters (A-E) this is not possible. Therefore D2 has to be an X.

5. D2 tells us that “D4 is the only number in this row” (row D). So, since C4 tells us that there are no letters in this row, we know that all of the cells in the row other than D4 or Xs (D1, D3, and D5)

6. A5 tells us that “C5 is a number that is either its cell letter’s position in the alphabet or its cell number”. This tells us that C5 is either a 3 or a 5, meanwhile E1 tells us that “No numbers that are in columns left of a number are greater than that number”. This tells us that C5 can’t be a 3 because if it were, there would have to be a number greater than 3 in a leftward column (and there is not enough room in column 5 for 3, 4 and 5 since we already know that E5 is not a number) Therefore C5 is 5.

7. D3 tells us that “Every column has exactly 3 Xs”. This tells us that B5 is an X since it is the only place that the third X in column 5 can be (again, E5 is a letter).

8. D3 also tells us that B1 and C1 have to be numbers since there are already 3 Xs in the column and the cells don’t match the C4 requirement for letters, Then B5 tells us “There are 2 numbers in this row” (row B). This has accounted for the last of the 5 numbers. We know how that B1, C1, D4 and C5 and somewhere B2-B4 are the 5 numbers. Therefore all other cells that don’t meet C4’s criteria for letters now have to be Xs. So A3, A5 and E5 are all Xs.

9. E4 tells us that “There are two rows with only one X.” Since we already have multiple Xs for rows, A, D and E, we know this means rows B and C have to be these rows with only one X. This tells us C2 and C3 have to be letters, since we already know they can’t be numbers. As for row B, B4 doe not meet the C4 criteria, which means B4 has to be the last number (and B2 and B3 have to be the last letters). So now that all numbers and letters are accounted for, all other cells have to be an X. The only cells left are E2, and E3 so they are X’s

10. E3 tells us that “Three numbers are the same as their cell number and/or their cell letter’s position in the alphabet”. We already know that 5 is one of these numbers since it has a cell number of 5. 4 ha to be one of these numbers since it has to be column 4 (because of the clue in cell E1). 3 can’t be one of these numbers since it is in column 4 and row B or D. Therefore 1 or 2 has to be the other one, but not both. This means B1 has to be 1 and C1 has to be 2 because if it were reversed they would both satisfy the condition and we’ve have four numbers that do so instead of three.

11. Since A3 tells us that “Be and C are adjacent” that means they have to be in two of the four cells of B2, B3, C2 and C3. D5 tells us that “A is adjacent to B”. So A must be also in one of these four cells. Then B1 tells us that “B is not diagonally adjacent to C”. Then we combine this with D1’s condition: “All letters except for one are both different from their cell letter and have a position in the alphabet that is different from their cell number”. B1’s clue guarantees that is either B or C that is the exception for this D1 condition. This is because the only possibility that both B and C are “different from their cell letter AND have a position in the alphabet that is different from their cell number” is with B and C diagonally adjacent (B2 being C and C3 being B). So, all of this adds up to telling us that E5 can’t be E (because there is only one letter that is the exception for the D1 condition). And since we’ve already said, A B, and C are in one of the other four cells, the only letter left is D. So E5 is D.

12. C1 tells us that “A is not adjacent to 4” and E5 tells us “E is not adjacent to 4”. This means that cell C3 is neither A nor E since cell C3 has to be adjacent to 4. It also cannot be C because if it were then both C and B would be exceptions for the D1 condition, but we know that there is only one that is. Therefore, C3 has to be B.

13. C3 tells us that “E and 4 are in the same row”. This means B4 has to be 4 because it is impossible for D4 to share a row with E. That also makes D4 have to be 3 as it is the last number. Then we know that “E is not adjacent to 4”, so E has to be in cell B2 in order for it to share the same row as 4. Then we also know that “A is not adjacent to 4”, so A has to be in the only non-adjacent cell left: C2. This leaves us with the last cell having to be C (also note that this satisfies the D1 condition).

It’s St. Patrick’s Day – don’t forget to earn your Luck of the Irish badge, then come right on back and get these badges too.

Joke’s On You: Things are about to get ridiculous in here. Prove your dedication to silliness on April Fool’s Day by getting 100% on Countries of Antarctica, The Impossible Quiz, and Click the Correctly Spelled ‘K’ountry. Just don’t take things too seriously!

Map Schmap: A continuation in the series, all you need to do to earn this badge is play 1,000 Map quizzes. They don’t have to be geography, either – cast photo quizzes and lyrics maps are fair game, too!

Go Green: It’s not always easy being green, but this badge might sweeten the deal a little. Get 50% or better on Green Characters, Foods That Are Green, Green Movies, and Going Green to earn it. Bonus points if you hum “The Rainbow Connection” while you’re at it.

As always, check the badge page for the latest updates.

<— Back to Mascot Murder Mystery quiz

1) From Ronald McDonald’s clue, we can tell that Captain Morgan, Aunt Jemima and Charlie Tuna are safe.

2) Charlie Tuna tells us that Marlboro Man is safe.

3) Marlboro Man tells us that the only victim in his row is Mr. Peanut.

4) Mr. Peanut tells us that there are 11 victims in total. Aunt Jemima told us that seven of those victims are adjacent to corners, so we can deduce that four of them are not adjacent to corners (including Mr. Peanut). The only remaining spaces that are not adjacent to corners are Michelin Man, Mr. Clean, the Trix Rabbit, the Geico Gecko, Chester Cheetah, the Noid and the Energizer Bunny. However, as revealed by the Marlboro Man, there are no victims in his row other than Mr. Peanut, so Michelin Man, Mr. Clean and the Trix Rabbit can be crossed off that list.

This leaves us with the other mascots in Column 3: the Geico Gecko, Chester Cheetah, the Noid, and the Energizer Bunny. Of these four, three are victims and one is safe. Mr. Peanut helpfully told us that he is vertically adjacent to one safe mascot, so we know that the one safe mascot in Column 3 is either Chester Cheetah or the Noid. Therefore, both the Geico Gecko and the Energizer Bunny must be victims.

5) The Energizer Bunny cannot see the murderer. Both the Michelin Man and the Trix Rabbit are in his line of sight, so neither of them can be the murderer. Additionally, the Marlboro Man’s clue told us that they’re not victims, either. So both Michelin Man and the Trix Rabbit are safe.

6) Look back at the Geico Gecko’s clue; one row is filled with nothing but victims. Rows 1, 3 and 5 each have at least one safe mascot, so that leaves us with either 2 or 4.

But the Trix Rabbit’s clue just told us that the Energizer Bunny is only adjacent to one victim, and he’s adjacent to three boxes in Row 4. Those three boxes can’t all be victims, so that only leaves us with Row 2.

Pillsbury Doughboy, Joe Camel, Chester Cheetah, the AFLAC Duck and Little Debbie are all victims.

And now that we know that Chester Cheetah is a victim, we can tell that the Noid is safe thanks to Captain Morgan’s clue.

7) Little Debbie’s clue tells us that Flo is safe.

8) According to the Pillsbury Doughboy’s clue, if Colonel Sanders is a victim, then the Kool-Aid man has to be one too. However, as of this moment we have identified four victims adjacent to corners; two adjacent to Ronald McDonald and two adjacent to Captain Morgan. As Aunt Jemima told us, there are a total of seven victims adjacent to corners. If there are two victims adjacent to Aunt Jemima, then the last remaining corner victim would have to be adjacent to Charlie Tuna, since Michelin Man told us that every corner is adjacent to at least one victim. That would leave…

2 victims adjacent to Ronald
2 victims adjacent to Morgan
2 victims adjacent to Jemima
1 victim adjacent to Charlie

However, Michelin Man’s clue ALSO told us that one corner is adjacent to three victims, so this can’t be true.

Since Colonel Sanders is a food mascot, he cannot be the murderer according to Flo’s clue. So Colonel Sanders is safe.

Since Aunt Jemima has to be adjacent to one victim, we can deduce that Kool-Aid Man must be a victim.

Kool-Aid Man is adjacent to the Energizer Bunny, so all the other spaces adjacent to him must be safe, including Smokey the Bear and Tony the Tiger (neither of them are in Chester Cheetah’s line of sight, so neither of them can be the murderer).

8) Tony’s clue tells us that Cap’n Crunch is safe.

9) Captain Morgan, Aunt Jemima and Charlie Tuna are each adjacent to at least one safe mascot. This means that Ronald McDonald is the only corner remaining who can be adjacent to three, so according to Michelin Man’s clue, the Jolly Green Giant must be a victim.

10) That leaves us with Mr. Clean and Snuggle Bear as the possible murderers. However, Charlie Tuna has to be adjacent to one victim according to Michelin Man’s clue, and both Smokey and Tony are safe, so Snuggle Bear must be a victim. That leaves us with only one unidentified space… the murderer, Mr. Clean.

<— Back to Nimble Number Logic Puzzle V

It’s been waaaaaay too long since we’ve had a quiz contest. So we thought, let’s change that! This contest is different from those in the past, because it’s an OctoQuiz Contest.

What’s OctoQuiz? It’s a new site we’re developing, that features 8-question quizzes. (You can read more about it in this blog post.) The Sporcle Staff have been having an OctoQuizmaking contest for the whole month of March. And we’re inviting you to join us for the final week.

The Rules: 

1. Go to the Create section, and click the ‘Create an Octoquiz’ button. (Or go directly to OctoQuiz here.)

2. Login with your Sporcle credentials.

3. Click ‘Create a Quiz’ and choose whether you want to make a Multiple Choice or Personality style quiz.

4. Experiment with the quiz format and add some images. When it’s ready to go, click the green ‘Publish Quiz’ button.

5. Come back to this blogpost and post your quiz in the comments.

Multiple entries are accepted, and you have until the end of March to submit them.

We’ll review all the quizzes, and the winning one will be promoted on the Sporcle homepage and the OctoQuiz Facebook page. Even if your quiz doesn’t win, there’s a good chance it will be published on OctoQuiz in the coming days and months.

And the most fun part is sharing these quizzes with family and friends. Share your quiz on Facebook or tweet it at someone you think will really enjoy it.

Spring is an important time of the year for wrestling fans. It’s when WWE brings us WrestleMania. WrestleMania 31 will happen in Santa Clara, California, the first time the San Francisco Bay Area has hosted the event.

I’m nscox, and I am both a wrestling fan and for nearly five years the curator of Sporcle’s WWE subcategory (as well as Canada and the Olympics).

Wrestling Explained

The WWE has several weekly shows, most notably Raw. These programs help build up feuds which are then settled at big events and watched on pay-per-view (ppv). Historically, the WWE and other companies made a lot of their money from these ppvs, but a lot of this has changed in the last year thanks to the introduction of the WWE Network.

There are 12 ppvs a year, and the biggest and most important is WrestleMania (followed by the Royal Rumble and SummerSlam). What makes WrestleMania different? Most ppvs are held in arenas, while WrestleMania gets the full stadium treatment. The WWE pulls out all the stops, featuring the biggest stars, presenting the most anticipated matches, bringing in outside celebrities and having a different set every year.

Yes, wrestling is “fake” (better descriptions include “staged” and “predetermined”) but it takes skill and work to do safely. The foundation of training is learning to perform these moves with minimal risk, but even then accidents happen or damage adds up over time.

WrestleMania 31: This Time, It’s Personal

This year there will likely be a Tag Team Championship match.  The women of the WWE are represented in a tag team matchup. The second annual Andre the Giant Battle Royal features a weaker lineup than last year (which was one of the best battle royals in years) but should still be good. And finally the undercard is topped off by a seven-man ladder match for the Intercontinental Championship.

The Undertaker will face (and probably beat) Bray Wyatt. Rising star (and recent Daily Show foil) Seth Rollins will face Randy Orton. WCW legend Sting will have his first (and probably only) WWE match, facing Triple H.

The United States Championship will be defended at ‘Mania for the first time in a while with Rusev defending against John Cena. Rusev should win, but it’s John Cena so he’ll win. Finally in the main event, Brock Lesnar defends the WWE Title against Roman Reigns.

10 Things You Should Know About WrestleMania

1) A Bit of History

The first WrestleMania was in 1985 at Madison Square Garden. In the main event, Hulk Hogan teamed with Mr. T to defeat “Rowdy” Roddy Piper and “Mr. Wonderful” Paul Orndorff. This is the sixth WrestleMania to emanate from California (the previous ones were 2, 7 & 21 in L.A. and 12 & 16 in Anaheim) but the first in San Francisco.

2) 9.99: The Anti-3:16

Last year WWE launched their own TV network! Response to the network has been underwhelming so far, but the advantage is that pay-per-views, which used to cost around $60 each, can now be bought for $9.99 (US) per month (it’s more in just about every other country). Amazingly, the WWE created a villain out of “9.99” promoting the number so much fans became sick of it. Not surprisingly, the “Austin 3:16” craze was not recreated.

3) Wrestling With Superlatives

The champion heading into this year’s show is Brock Lesnar, who won the title back in August, but has defended it only a handful of times. Bruno Sammartino is the longest reigning WWF Champion, holding the belt twice for a combined 4040 days (over 11 years) including one reign that lasted for more than 8 years.  Lou Thesz held the NWA Championship for a combined 3,749 days. And the record for longest reigning world champion is held by Verne Gagne, who held the AWA Title 10 times for a combined 4,677 days.

4) Hall of Fame

A big part of the WrestleMania festivities is the WWE Hall of Fame. This year will be headlined by the legendary “Macho Man” Randy Savage, as well as Rikishi, The Bushwhackers, Alundra Blayze, Tatsumi Fujinami and Diesel. Rounding out the group is the Kyrgyzstan of wrestlers: Larry Zbyszko. Also going in is Arnold Schwarzenegger. The WWE typically inducts one celebrity per year.

5) Family Business

There are a lot of families in the wrestling business: Hart. McMahon. Von Erich. Guerrero. But the most successful is the Anoa’i Family. Originating in American Samoa, this group includes The Wild Samoans Afa & Sika, WWF Champion Yokozuna and brothers Rikishi, Umaga and the Tonga Kid. Currently in the WWE you have the Uso Brothers (sons of Rikishi) and Roman Reigns (son of Sika). The patriarch had a blood brother named Peter Maivia. Though not actually related, the current crop considers his grandson their cousin. His name? Dwayne “The Rock” Johnson.

6) Russians and Iranians and Canadians, Oh My!

One of the main matches this year is John Cena vs. That dastardly evil Russian Alexander Rusev. At one point a common wrestling villain was the evil foreigner who hated America. The business has changed and today there are wrestlers from over a dozen different countries. But before 2000, most of the evil stereotypical foreigners were not from the countries they claimed to be from. Nikolai Volkoff? Croatian. Nikita Koloff? American. Abdullah the Butcher? Canadian. Every now and then, they were actually citizens of the country they claimed to be from. The Iron Sheik? Iranian.

7) Women in Tights

The Divas Division is in an odd place with the rise of the NXT women. The problem is simple: For years women’s matches in the WWE were treated as filler. Most have no wrestling background and were hired for their looks, so the matches tended to be short and bad. Some fans are starting to want change, thanks partially to the rise of Ronda Rousey in UFC, and the “Give Divas a Chance” hashtag is starting to catch on.

8) Celebrity Magic

One of the main draws of the first WrestleMania was the celebrity presence, which included Mr. T, Muhammad Ali, Cyndi Lauper and Liberace. This trend has continued and throughout the years a lot of celebrities have appeared in WWE or WCW. Some have participated in matches, including Mr. T, Snooki, Drew Carey, Jay Leno, and Dennis Rodman. Kevin Federline once faced John Cena, the WWE Champion at the time, and won. And David Arquette won the WCW World Heavyweight Championship (in 1999).

9) The Dead Man Goes Streaking

Everyone knows that wrestling has a silly side. Wacky characters include a wrestling plumber, an evil clown and a guy in a turkey suit. But the amazing part of wrestling is that an idea that seems dumb on paper can work. Take Mark Calaway for example. In 1990, he was given the gimmick of an undead wrestler who was controlled by an urn. Calaway made it work and 25 years later The Undertaker is still around. He made his Mania debut in 1991 against Jimmy Snuka, and began “The Streak”, winning 21 straight WrestleMania matches. Last year The Streak was ended by Brock Lesnar in a legitimately shocking moment.

10) A Sizeable Detour

The tallest person who ever engaged in professional wrestling was Edouard Beaupré (1881-1904) a French-Canadian who was 8 ft. 3 when he died. The tallest WWE wrestler was Giant Gonzalez, who was 7’7 but billed at 8 feet tall. Andre the Giant has many legends about his drinking ability. He is said to have drank 156 bottles of beer in one sitting, as well as often drinking several bottles of wine before matches.

Heaviest wrestler is a tricky because weights can change. According to WWE, Happy Humphrey (1926-1989) was the heaviest, averaging 750 pounds over his career, and at one point weighing in at 900. Amazingly, after his retirement he entered a program at a medical clinic and lost 570 pounds in 2 years. There’s also Haystacks Calhoun, possibly the heaviest WWF wrestler at a billed 630 and Yokozuna, the heaviest WWF champion at around 500 pounds.

Final Thoughts

And there you have it. Thanks for reading and hopefully some of the things you learned will help you in your future Sporcle endeavors.

If you want more WrestleMania quizzes, here’s a listing of every one with an appropriate tag. Check out the Sporcle WWE Facebook page for more great quizzes, and you can always email me (nscox22@hotmail.com) for feedback or suggestions for curator picks!

nscox

​<— Back to Sink or Swim Logic Puzzle

Step 1: Combining clues from A1, E2, and B4 we can figure out what B2 and E4 are. Because every swimmer is underwater at least 2 times consecutively, and swimmers can’t go from being out of the pool to underwater directly, Swimmers 2 and 4 only have a few options of when they can can under water. Swimmer #2 can be underwater in A2, B2, and C2. B2 must be underwater or it is not possible for Swimmer 2 to be underwater two consecutive times. Similarly, Swimmer 4 must be underwater in E4.

Step 2: B2′s clue says, ‘Exactly one swimmer was in the pool for three consecutive time slots.’ The only possible swimmers that could do this are #3 and #5. F5′s clue says, ‘Exactly one swimmer was underwater four times.’ The only possible swimmers that could do this are Swimmers 3 and 5. The only order in which a swimmer can be underwater four times is U, U, U, U, I, O (or reversed). The only order in which a swimmer can be in the pool for three consecutive slots is U, U, I, I, O (or reversed). Notice that 4 out of the 6 answers are the same for these two columns. Because Swimmer 5 is already out of the pool for the sixth time slot, E5 must be in the pool, and A5 & B5 must be underwater. C5 and D5 we leave blank because they are either both underwater or both in the pool.

Step 3: E4′s clue says, ‘Exactly one swimmer was out of the pool for three consecutive time slots.’ This swimmer cannot be Swimmer 3 or 5 because those are the swimmers that fulfill B2 and F5′s clues. This swimmer cannot be Swimmer 1 because it isn’t possible for Swimmer 1 to be out of the pool 3 consecutive times. This means that Swimmer 2 or Swimmer 4 is the swimmer that is out of the pool three consecutive times.

Now let’s look at cell A3. A3 is either underwater or out of the pool, because we already know that Swimmer 3 has to be U, U, _, _, I, O or O, I, _, _, U, U. Let’s see if A3 can be out of the pool. A5′s clue says, ‘At any given time slot at least three swimmers are doing the same thing.’ If A3 is out of the pool this would mean both A2 and A4 would have to be in the pool. There is no other way to have three swimmers doing the same thing in the first time slot if A3 is out of the pool. If both A2 and A4 are in the pool, then it’s not possible for either Swimmer 2 or 4 to be out of the pool for three consecutive time slots. This means that A3 cannot be out of the pool and must be underwater. Knowing A3 is underwater also tells us that B3 is underwater, E3 is in the pool, and F3 is out of the pool.

Step 4: A5′s clue helps us fill in a few more spaces. We already determined that A2 and A4 cannot both be in the pool. This means that at least three swimmers were underwater in the first time slot. A4 cannot be underwater so the third underwater in the first row must be A2. In the fifth time slot (row 5), E1 must be in the pool to fulfill A5′s clue. With E1 being in the pool, F1 must be out of the pool, because there’s no other time in which Swimmer 1 could be out of the pool.

Step 5: For Swimmer 1 to be underwater for at least two consecutive time slots, C1 must be underwater. F1′s clue says, ‘Swimmers 4 and 4 were never doing the same thing at the same time.’ We know that at least one of C4 or D4 must be in the pool because B4 is out of the pool and E4 is underwater. We know that C5 and D5 are either both in the pool or both underwater. Because of F1′s clue C5 and D5 cannot both be in the pool so C5 and D5 must both be underwater. This means that Swimmer 3 is the simmer that was in the pool for three consecutive time slots, so C3 and D3 are both in the pool. D4 cannot be out of the pool (B4′s clue) and cannot be underwater (F1′s clue), so D4 must be in the pool. This means that F4 must be underwater (E2′s clue).

Step 6: D4′s clue says, ‘Swimmers 1 and 2 were underwater for a different number of time slots.’ This means that out of Swimmers 1 and 2, one was underwater three times and one was underwater two times. Let’s try Swimmer 2 being underwater three times. This would mean that C2 is underwater and D2 is in the pool. This would mean Swimmer 4 is the swimmer that is out of the pool for three consecutive time slots. So if Swimmers 1 and 2 are underwater for a different number of time slots, which is what D4 tells us, Swimmer 4 has to be the swimmer out of the pool for three consecutive time slots. This means A4 and C4 are out of the pool.

Step 7: C2 must be underwater to fulfill A5′s clue. D2 must be in the pool because of B4′s clue. C2′s clue says, ‘Swimmers 1 and 2 did the same thing at the same time as many times as Swimmers 2 and 3 did the same thing at the same time.’ Without knowing F2, Swimmers 2 and 3 are doing the same thing at three times. We know that B1 and D1 are in the pool and underwater, we just don’t know which one’s which. If B1 is in the pool and D1 is underwater, then Swimmers 1 and 2 are doing the same thing at the same time once. That wouldn’t fulfill C2′s clue so B1 must be underwater and D1 must be in the pool. D1′s clue tells us that no swimmer finished in the pool. This means that F2 cannot be in the pool. F2 also cannot be underwater so F2 myst be out of the pool.

​<— Back to Letter Grid Logic Puzzle (A-Y) Solution

From the clue in cell 8, we know that cell 18 is c.

We also know that the corner cells contain E, I, O and U. The letter E must be in the fifth column, since there is only one step between C and E (C-D-E and sporC-L-E.) Therefore, either cell 21 or cell 25 is E, and the two cells in between C and E are D and L (cells 17 and 22 or cells 19 and 24.)

If F has to be adjacent to E, it can only be in cell 16 or 20. Now, we know that the other corner of the fifth row is O (spO-R-Cle), since there is only one step between O and C. If we take a look at the letter I: there are two steps between the F and the I (F-G-H-I). F is either in cell 16 or 20. If F is in the first row (cell 16), I must be in the first row as well (cell 1). The same goes for the fifth row, if F is in cell 20, I must be in cell 5, otherwise there would be no possibility to position G and H.

This indicates that E and I are in the same row, and therefore O and U are in the same row as well. Now, we know that T is not in row 1 or 2, so U cannot be in row 1, since T and U are adjacent. Therefore, cell 5 is U, cell 25 is O, cell 1 is I and cell 21 is E.

Logically, cell 16 is F. Now we know that cells 17 and 22 are D and L, we can try to position those letters. There are 2 cells between I and L (I-J-K-L), so L cannot be in the fifth column. Cell 17 is L and cell 22 is D. We also know that G must be in cell 11 or 12 (adjacent to F), H in cell 6 or 7 (adjacent to G and I), J in cell 6 or 7 (adjacent to I and K) and K in cell 11, 12, or 13 (adjacent to J and L). L is in the second row, and O in the fifth. Therefore, M must be in the third row and N in the fourth row. N must be in cell 19 or 24, just like R (adjacent to C and O). If M would be in cell 13, then cell 23 would be completely isolated, and no letter would be able to be put there. Therefore, cell 23 is M.

P is adjacent to O, and not in cell 19 or 24, so cell 20 is P. If cell 19 would be N, and cell 24 R, R would not be adjacent to Q and S, which is impossible. Therefore, cell 19 is R and cell 24 is N.

Let’s take a look at cell 2. Cells 6 and 7 are H and J, so cell 2 has only one neighbor which has to be determined. Therefore, cell 2 must be Y, and cell 3 must be X.

With the clue from cell 3, we know that K and W are adjacent. W can only be in cell 4 or 9 (adjacent to X), and K in cell 11, 12, or 13. There is only one possibility: cell 9 is W and cell 13 is K. The only possibilities for cells 11 and 12 are B and G. B must be adjacent to A and C, so cell 12 is B and cell 11 is G. J must be adjacent to K, so cell 7 is J and cell 6 is H. K is not adjacent to Q, but Q is adjacent to R. Therefore, cell 15 must be Q. The letter S must be adjacent to P and R, so cell 14 is S. T is adjacent to S, so cell 10 is T. Only V remains, so cell 4 is V.

These badges are just too hot, so you should listen to Uptown Funk while you’re earning them.

Throwback Thursday: It’s throwback time, Sporcle style. Ditch the old photos and nostalgic filters, mosey your way on over to the History section, and play 15 quizzes on a Thursday. You won’t even need to use a #tbt hashtag!

Stone Cold: Want to be King of the Ring? Think you can hold your own with a crowd of professional wrestlers? Has your time come? Then build up a little courage and play 316 WWE quizzes… because Stone Cold said so.

Sort-A-Badge: To get this badge, you’ll need to play the original sorting quizzes: Country Sorting Blitz, World Language Sorting Blitz, Food, Drink, and Poison Sorting Blitz, and Animal Sorting Blitz. Get ready to make a lot of snap decisions – if you score 100% on all four quizzes, you’ll earn this badge.

As always, check the badge page for the latest updates.

<— Back to ‘A Pokémon Logic Puzzle’