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99+ Problems & They're All Lenny Conundrums (Part 1)

by minnesotan


Also by sacados

In honor of the upcoming 500th Lenny Conundrum, we will walk you through the solutions to all of the puzzles from the past two years. Some are simple, some are quite in-depth, and a few are just a matter of guessing! We've analyzed every LC solution from round 400 to round 495, given them each a category and difficulty, and explained their solutions. In order to keep the reading manageable, we've split up the puzzles into two sections. In this first part, we'll take a look at the conundrums from rounds 400-451:

To see each week's question and answer, put the number of the round in place of ### in the following link:

Are you ready? Let's get started!

Round 400:

Category: Patterns

Difficulty: Extremely Difficult

Solution: This was a very special round, being a 100th round. The solution is given on the answer page so I will not explain it here. This round was only correctly answered by 10 people and the prize was a rather valuable Lenny Conundrum Negg.

Round 401:

Category: Guessing

Difficulty: Easy

Solution: The last round was #400, so lots of people guessed that as the answer. This happened with round 300 as well. Huh, I guess history really does repeat itself.

Round 402:

Category: Patterns

Difficulty: Medium

Solution: The answer page mentions the items are taken from shopkeeper images and explains that the number is the sum of the shop IDs. Here are the shops each item came from and the shop ID (in parentheses), in order:

The Bakery (15) and Faerie Furniture (75) = 15+75 = 90

Collectable Card Shop (8), Haunted Weaponry (59) = 8+59 = 67

Tropical Food Shop (20), Molten Morsels (112), Faerieland Petpets (40), Cog's Togs (111), Space Weaponry (23), Grundos Cafe (22), Spooky Food (30), Roo Island Souvenirs (76), and Tyrannian Furniture (43) = 20+112+40+111+23+22+30+76+43 = 477

Round 403:

Category: Math

Difficulty: Easy

Solution: All we have to do is go to the map of Moltara City, count the lamps and fans, and do some arithmetic. On the map we can see seven hanging lanterns, four spherical lanterns, and five fans. Thus, the wattage is 7*79+4*47+5*113 = 1306.

Round 404:

Category: Problem-solving

Difficulty: Easy

Solution: For some reason, this conundrum is nowhere to be found. It's like the page is missing or something. If you were somehow able to find a puzzle here, open it in MS Paint (or another graphics program of your choice), and flood fill (bucket tool) it with any color other than white, you see the question, "If you try to load a Neopets page that doesn't exist and you get a '404' error that displays the silhouettes of three Neopets, including a Grundo, what Neopet species is depicted on the right side?" If we open a nonexistent page and refresh until finding a Grundo, we'll see a Moehog on the right side.

Round 405:

Category: Patterns

Difficulty: Easy**

Solution: This Lenny Conundrum began on the first day of Relaxing in Y13 during Altador Cup VI. If one visited the main Altador Cup page on this day, they would have seen the following news coverage:

The Altador Cup has narrowly averted disaster, as the Yooyus from this year's tournament that escaped from their pen a few days ago have been found. It seems the Yooyus got loose when a member of their training team left their pen unlatched before going home, allowing the Petpets to get out. A search began the following morning, followed by efforts to prepare a team of replacements. Thankfully, a farmer found the Yooyus asleep in his barn this morning. "We had some rain last night, so I guess they decided to take cover," the farmer said. "You can imagine my surprise when I found them curled up in there. Like everyone, I'm just glad they're safe and the tournament con go on as planned."

See the pattern? Look at the last word of each sentence: found, out, replacements, morning, said, there, planned. The words in the conundrum are this list in reverse order. The missing word must be "replacements".

** = This conundrum was a lot easier to solve when it was live on the site. Now it's quite difficult to find this paragraph and nearly impossible to connect it to the conundrum.

Round 406:

Category: Math

Difficulty: Tedious but easy

Solution: To solve this now, we have to note the month it was referring to is the Month of Relaxing in Y13. If we look at a calendar we'll see this month had five Wednesdays, five Thursdays, and four of every other day. The fastest way to solve this puzzle was using a spreadsheet, though for most people it was easier to do by hand. I won't explain any formulas here as they're slightly advanced. I will simply list the number of jellies for each day in the month and the day of the week (M = Monday, T = Tuesday, W = Wednesday, Th = Thursday, F = Friday, S = Saturday, and Su = Sunday):

3 (W), 3 (Th), 11 (F), 42 (S), 9 (Su), 18 (M), 5 (T), 3 (W), 94 (Th), 22 (F), 348 (S), 72 (Su), 144 (M), 9 (T), 3 (W), 786 (Th), 33 (F), 2457 (S), 501 (Su), 1002 (M), 13 (T), 3 (W), 5581 (Th), 44 (F), 16875 (S), 3432 (Su), 6864 (M), 17 (T), 3 (W), and 38397 (Th).

Adding these all together we get 76794.

Note: The puzzle didn't make it clear whether the pattern for Fridays was counting by elevens (11, 22, 33, 44) or doubling each week (11, 22, 44 ,88). As you can see in the list above, the correct choice was the former.

Round 407:

Category: Patterns

Difficulty: Medium

Solution: The key to this puzzle is Nimmo Day. It's a logical assumption that this Nimmo went shopping for Nimmo items. It turns out those items were Battledome weapons. The items are listed in the proper order in Neopets' Battlepedia. (link "Neopets' Battlepedia" to We can see the first item is Nimmo Apple Bomb with an estimated value of 2811. The second item is Nimmo Elixir of Healing with an estimated value of 13520. The next item is Nimmo Helmet with an estimated value of 2156. The question asks for the sum of the estimated values of the next two items (Nimmo Power Staff and Nimmo Throwing Star) which have estimated values of 3022 and 2054. 3022+2054 = 5076.

Round 408:

Category: Guessing

Difficulty: Extremely difficult

Solution: Only eleven very lucky people guessed in the correct range. The additional prize mentioned ended up being a Midnight Jelly Stamp.

Round 409:

Category: Patterns

Difficulty: Easy

Solution: This conundrum was also released during and related to Altador Cup VI. The species were the species corresponding to the captain of each team in alphabetical order. (AL, BV, DC, FL, HW, KL, KI, KD, LD, MQ, MD, MT, MI, RI, SK, TM, TY, VP)

During ACVI, the captains were:

Altador: "Trapper" Remis (Poogle)

Brightvale: "Squeaky" Tressif (Lupe)

Darigan Citadel: Layton Vickles (Hissi)

Faerieland: Kakoni Worrill (Bruce)

Haunted Woods: Krell Vitor (Kyrii)

Kiko Lake: "Poke" Cellers (Kiko)

Krawk Island: Garven Hale (Bori)

Kreludor: Derlyn Fonnet (Gnorbu)

Lost Desert: Leera Heggle (Kau)

Maraqua: Elon "The Black Hole" Hughlis (Acara)

Meridell: "Wizard" Windelle (Techo)

Moltara: Aldric Beign (Kougra)

Mystery Island: Volgoth (Mynci)

Roo Island: Lilo Blumario (Blumaroo)

The answer was therefore the species of Shenkuu's captain, Mirsha Grelinek the Gnorbu.

Round 410:

Category: Math

Difficulty: Easy

Solution: In Double or Nothing our Neopoints double each time the coin lands on heads. Thus, after N heads we (theoretically**) have 10*2^N Neopoints. A Portable Kiln costs 3158000 NP in the Hidden Tower. We can set 3158000 equal to 10*2^N and solve for N. Solving for N we get approximately 18.3. Because 10*2^N must be more than the cost of the Portable Kiln to be able to afford it, N must be greater than 18.3. The smallest integer above 18.3 is 19.

**Note: Due to the actual mechanics of Double or Nothing, 20 would have been a more suitable answer.

Round 411:

Category: Problem-solving

Difficulty: Easy

Solution: At first you may want to try finding the letters of individual species names. That would be somewhat difficult though, and there's a better way. Count all the letters in the names of the 54 species, and count the letters in the word-search. In counting, we see the puzzle has only 24 I's, 16 K's, 24 R's, and 20 U's whereas the species have 25, 17, 25, and 21 of each of those letters respectively. All the other letters check out, so the missing species uses the letters IKRU. The species must be RUKI.

Round 412:

Category: Pop Culture

Difficulty: Easy

Solution: Searching the web for the phrase "we like tha moon" brings you to a certain video. Watching this video reveals some lyrics. The first line of the song is, "We like tha moon coz it is close to us." The answer is simply the second half of this line. Interestingly, the name listed on dirigibles' user lookup are also lyrics from this song and he hasn't been seen since this conundrum. Coincidence? Probably not. This conundrum is believed by some Neopians to be a farewell to Dirigibles.

Round 413:

Category: Math

Difficulty: Easy

Solution: He buys it for 5112 and sells it for 7555, which is a 2443 NP profit. However, he lost 446 between the time of the first purchase and the time of the final sale, bringing the profit down to 1997.

Round 414:

Category: Math

Difficulty: Very Easy

Solution: Three places in front of last is also two ahead of seventh, which is fifth. Because fifth place is three places in front of last there must be eight participants.

Round 415:

Category: Math

Difficulty: Medium

Solution: Let's start by calling the three digit number ABC where A, B, and C each represent a digit of the answer. The first condition says that A+BC is a perfect square. The second says that A*BC is a perfect square. Finally, the third says that A+B+C is also a perfect square. Let's start with the third condition. Because A, B and C are all different, A+B+C must be (inclusively) between 6 (1+2+3) and 24 (9+8+7). Thus A+B+C is 9 or 16. The ways we can get A+B+C=9 with three different digits are 1+2+6, 1+3+5, and 2+3+4. The numbers possible from this that satisfy the first condition are 135, 531, 234, and 432. However none of these numbers satisfy the third condition so now we know A+B+C=16. This is the point at which seasoned Kakuro solvers will leap with joy and everyone else will groan. The ways this is possible are 1+6+9, 1+7+8, 2+5+9, 2+6+8, 3+4+9, 3+5+8, 3+6+7, and 4+5+7. The only three digit numbers that are possible from this which fulfill the first condition are 916, 619, 817, and 718 of which only 916 fulfills the second condition. The only three digit number that satisfies all three conditions is 916.

Round 416:

Category: Math

Difficulty: Easy

Solution: Let's make the assumption that the new palindrome is the smallest palindromic number greater than 14941. The next time the odometer is palindromic is at 15051 billion kilometers, therefore he travels 110 billion kilometers in the next two hours. At this point we require a very basic physics equation: average speed = distance/time. 110 billion km / 2 h = 55 billion km/h.

Round 417:

Category: Word Puzzle

Difficulty: Easy

Solution: If we actually go through this step by step we'll find not one, but three solutions:

S (Owing), O (Wing), G (Win), W (In), N (I) = SOWGN

O (Swing) S (Wing), G (Win), W (In), N (I) = OSGWN

and of course the official answer:

O (Swing), W (Sing), G (Sin), S (In), N (I) = OWGSN

We could decide that the first answer was unlikely as owing is a very uncommon word. Deciding between the other two answers wasn't as easy, however, leaving us with a 50/50 shot of choosing the right one.

Round 418:

Category: Problem-solving

Difficulty: Easy

Solution: One method would be to label the dots as such:




We can make these six squares: ABED, BCFE, DEHG, EFIH, ACIG, and BFHD.

Round 419:

Category: Logic

Difficulty: Easy

Solution: There are three cases.

1) He pulls out two Deluxe Angel Usukis and throws one Devilish Usuki back in

2) He pulls out two Devilish Usukis and throws one back in

3) He pulls out one of each Usuki and throws the Deluxe Angel Usuki back in.

All three cases leave the parity of the number of Deluxe Angel Usukis in the box unchanged. Because Lawyerbot began the process with an odd number of Deluxe Angel Usukis, he can never have zero of them. Therefore if he has only one Usuki left in the box it must be a Deluxe Angel Usuki.

Round 420:

Category: Word Puzzle

Difficulty: Easy

Solution: We solve this conundrum by removing the ten letters in the words "six letters" rather than removing only six letters.

Round 421:

Category: Math

Difficulty: Somewhat Difficult

Solution: This is a geometry problem involving some trigonometry that's somewhat tedious but not terribly complex. Of course it could also have been solved with some weird integrals, but who would want to put themselves through THAT? Certainly not you or me. No, not us. We'll stick to the law of sines and Heron's formula, right? Right. The farthest points the Babaa can reach are on the Southwest wall of the pen and on the Northeast wall of the pen. It's easy to see that the area the Babaa can graze is composed of two congruent triangles and a sector between them. To find the area of the sector we need to figure out how large the angle is. In order to do that we must first find the angle measures in the triangles. Each triangle is a full regular hexagon angle (interior) opposite a side length of 50 yards and adjacent to a side length of 40 yards. The measure of the interior angles of a regular polygon is 180-360/n degrees. The interior angles of a regular hexagon measure 120 degrees. We can now use the law of sines to find the measure of the angle (call it B) opposite the side length (call it b) of 40:

sin(A)/a = sin(B)/b

sin(120)/50 = sin(B)/40

Solving for B we get 43.8538 or 136.1462

The triangle already has angle A = 120 degrees, so only the former is a possible measure for angle B because the sum of 120 and 136 exceeds the 180 degrees in a triangle. We now have an unambiguous measure for angle B. Since A+B+C = 180 and A+B = 163.8538, the remaining angle (C) is 16.1462 degrees. Using the law of sines again with this newly found angle measure, we can find the length of the remaining side to be 16.0555 yards long. Going back to angle C, we can now finally solve for the angle measure of the sector: 120 - 2C = 120 - 32.2924 = 87.7076 degrees.

Thus the area of the sector is (87.7076 / 360)*(Pi * 50 * 50) = 1913.4830 square yards.

Finding the area of the triangles is a little trickier. Here we will use Heron's formula for its ease. Also, Heron's formula is cool. The area of each triangle by Heron's formula is:

sqrt(s * (s-a) * (s-b) * (s-c)) where s is the semiperimeter (40+50+16.0555)/2 = 53.0278

Plugging this and the side lengths into Heron's formula we have:

Area = sqrt(53.0278 * (53.0278-50) * (53.0278-40) * (53.0278-16.0555)) = 278.0924 square yards

The Babaa's total grazing area is therefore 1913.4830 + 278.0924 + 278.0924 = 2469.6678 square yards.

We are almost done; now we just have to find how long it will take the Babaa to each this much grass.

At 13 minutes per square yard, it will take 13 * 2469.6678 = 32105.6814 minutes for the Babaa to eat all this grass. This is 32105.6814 / 60 = 535.0947 hours. Dividing by 24 gives us an answer of 22.2956 days.

Now I know that you're thinking I wasted 10 minutes getting the wrong answer, but my answer is quite correct. The problem says to round to the nearest day, but it really wanted us to round UP to the nearest day. It is very unfortunate that the puzzle left out such an important word. :(

Round 422:

Category: Logic/Puzzle

Difficulty: Easy

Solution: Move Dubloons 1 and 4 to the left of Dubloon 8 and to the right of Dubloon 9, then move Dubloon 10 above Dubloons 2 and 3.

Round 423:

Category: Math

Difficulty: Very Easy

Solution: In Ultimate Bullseye a Fire Hoop doubles the score of the shot. The most points we can get from a shot with the Fire Hoop is 2*40 = 80 if we hit the bullseye. A second bullseye would give us the 40 more points we need to our score up to the desired 120.

Round 424:

Category: Math

Difficulty: Who knows? The answer is wrong.

Solution: Ask anyone on the Neoboards, and they'll probably agree that 64 is incorrect. Doing the geometry (or trigonometry) gets you an area of 66.6667 which rounds up to 67. The easiest way to get that is to solve for the horizontal width of the path and multiply it by 40. If we call that width x, we just have to solve the following quadratic: (55-x)^2/1600 + 1 = x^2. The only positive solution is x=5/3, so the area of the path is (5/3)*40 = 66⅔. Arty will be one rather upset gardener when he finds out he's nearly three square yards short.

Round 425:

Category: Word problem

Difficulty: Easy

Solution: His birthday must be on the last day of the year. If we assume that yesterday was his birthday and that today is the first day of the new year, he was 3 years old the day before yesterday and is currently 4 years old. On the last day of the year he will turn 5, and on the last day of the following year he will turn 6. The solution should read "Month of Celebrating, Day 31" after converting it to the Neopian date system.

Round 426:

Category: Math

Difficulty: Hard

Solution: This type of problem is called an alphametic or verbal arithmetic puzzle. The requirements for the solution are that each letter represents a distinct digit 0 to 9 and none of the leading letters/digits are 0. Solving this type of puzzle can get rather tedious and require overwhelming amounts of logic and arithmetic. Without walking you through the long process of solving this I will tell you that TWELVE = 130760, THIRTY = 194215, and NINETY = 848015. Solving this one fully makes a good exercise for the reader, yes? ;)

Round 427:

Category: Logic

Difficulty: Very easy

Solution: Is it possible? Absolutely.

Round 428:

Category: Patterns

Difficulty: Easy

Solution: This pattern is known as the look-and-say sequence. See why? Try reading the number out loud, digit by digit. The first number is 1. The second number is 11, or "one 1." The third number is 21, or "two 1s." The fourth number is 1211, or "one 2, one 1." The pattern continues like this until 31131211131221. The next line should therefore be "one 3, two 1s, one 3, one 1, one 2, three 1s, one 3, one 1, two 2s, one 1" or 13211311123113112211.

Round 429:

Category: Patterns

Difficulty: Easy

Solution: Each letter in the sequence corresponds to the first letter of a word inside the box. The missing letters are E from "every," L from "live," I from "in," and V from "very."

Round 430:

Category: Word problem

Difficulty: Easy

Solution: Austin is better than Doug and Zorblox is better than Doug. Therefore Zorblox is better than Doug and also far better than Austin, so the answer is C.

Round 431:

Category: Problem-solving

Difficulty: Medium

Solution: Solution is given on answer page and will not be provided here.

Round 432:

Category: Math

Difficulty: Easy

Solution: We can take note of the fact that to get a 5 as the 10s digit from multiplying a number by two, we'll have to get a 4 and carry over a 1. Thus, the 10s digit in the first number must be a 2 or a 7 and the ones digit must be greater than 5. Now that we have significantly narrowed down the possibilities for the first number, let's try out some numbers.

126-129 range: This range is not possible because multiplying these numbers by two results in a 2 in the hundreds place duplicating the 2.

176-179 range: This range is not possible because multiplying these numbers by three results in a 5 in the hundreds place duplicating the 5.

226-229 range: This is a little more work, but we can easily find no numbers in this range work without duplicating any numbers.

276-279 range: This range is not possible as multiplying these numbers by two results in a 5 in the hundreds place duplicating the 5.

326-329 range: Once we get here and try the numbers, we find out that 327 actually works. Therefore the answer must be to have 327 in the first row, 654 in the second row, and 981 in the third row.

Round 433:

Category: Logic

Difficulty: Medium

Solution: This is a logic puzzle. As such, it is of a slightly tedious and subjective nature and I will not give a step-by-step solution. However, I will tell you how everything works out and I urge you to try to come to the same conclusions on your own.

Brilch: Stuffing department; First floor; 53 day record

Embroal: Warehouse; Third floor; 28 day record

Lyrian: Store; Second floor; one day record

Phabrick: Design department; First floor; 348 day record

Trilnin: Engineering department; Third floor; 17 day record

Round 434:

Category: Problem-solving/Site knowledge

Difficulty: Easy

Solution: Solution is given on answer page; a solution will not be included here.

Round 435:

Category: Word problem

Difficulty: Easy

Solution: We'll work this out by reading it backwards. First, the answer to 2+4*5 = 2+20 = 22 = TWENTY TWO. The sixth letter of TWENTY TWO is Y. Y is the 25th letter. 25 has three straight lines. Thus the answer is 3.

Round 436:

Category: Problem-solving/Simple math

Difficulty: Medium

Solution: We have to go through the species and add all the toes in their footprints:

Acara: Four feet with three toes each = 12

Aisha: No toes = 0

Blumaroo: Two feet with two toes each = 4

Bori: Four feet with three toes each = 12

Bruce: No toes = 0

Buzz: Two feet with three toes each = 6

Chia: No toes = 0

Chomby: No toes = 0

Cybunny: Four feet with three toes each = 12

Draik: Two feet with three toes each = 6

Elephante: Two feet with three toes each = 6

Eyrie: Four feet with three toes each = 12

Flotsam: No feet/toes = 0

Gelert: Four feet with three toes each = 12

Gnorbu: No toes = 0

Grarrl: Two feet with three toes each = 6

Grundo: Two feet with two toes each = 4

Hissi: No feet/toes = 0

Ixi: No toes = 0

Jetsam: No toes = 0

JubJub: Two feet with three toes each = 6

Kacheek: Two feet with two toes each = 4

Kau: No toes = 0

Kiko: No feet/toes = 0

Koi: No feet/toes = 0

Korbat: Two feet with two toes each = 4

Kougra: Four feet with three toes each = 12

Krawk: Two feet with three toes each = 6

Kyrii: Two feet with two toes each = 4

Lenny: Two feet with two toes each = 4

Lupe: Four feet with three toes each = 12

Lutari: Two feet with three toes each = 6

Meerca: Two feet with two toes each = 4

Moehog: No toes = 0

Mynci: Two feet with three toes each = 6

Nimmo: Two feet with three toes each = 6

Ogrin: Four feet with three toes each = 12

Peophin: No toes = 0

Poogle: No toes = 0

Pteri: Two feet with two toes each = 4

Quiggle: Two feet with three toes each = 6

Ruki: No toes = 0

Scorchio: Two feet with three toes each = 6

Shoyru: No toes = 0

Skeith: Two feet with three toes each = 6

Techo: Two feet with three toes each = 6

Tonu: Four feet with three toes each = 12

Tuskaninny: No feet = 0

Uni: No toes = 0

Usul: No toes = 0

Wocky: No toes = 0

Xweetok: Four feet with three toes each = 12

Yurble: Two feet with three toes each = 6

Zafara: Two feet with three toes each = 6

If we count up all these toes collectively we'll get the answer, 242.

Round 437:

Category: Word Puzzle

Difficulty: Moderately challenging

Solution: This round was a particularly difficult word puzzle. To complete it, you had to find the correct pairings and the correct anagram of each of those pairings. It turned out that all of the anagram pairs formed the names of petpets, but this was not mentioned in the puzzle, making it significantly harder to solve. The correct pairings were as follows:

HERD + SNOW = Werhonds

ROOT + SCAN = Octornas

CLIP + MONK = Clompkin

KITS + HOOK = Kookiths

CITY + SOAK = Yackitos

Taking the first letter of each petpet name above gives the answer: Wocky.

Round 438:

Category: Word Puzzle

Difficulty: Medium

Puzzle (does not appear on-site):

hh, Maraqua. Land of aquatic fun, fine food, fishing antics... but a really bad place for books. Like, really bad. The Lenny Library has unfortunately found a very rare copy of A Brief History of Neopian Turnips in an underwater abode, and needs to restore it. As you can see, many of the letters are smudged beyond recognition. The Neopian Center of Unimaginable Knowledge knows the answer though, but have decided to challenge the great minds of Neopia to see who is worthy to join their mighty ranks. So they have posed this question:

You should be able to complete the Neopian words below using all of the letters A to Z, each once only. Once completed, three leftover letters should remain.










What are the three leftover letters? Please list the remaining three letters in alphabetical order.

Solution: This was just a matter of solving obvious words first and harder ones later. From top to bottom we have:

Skarl (uses S, K, and A)

Jacques (uses J and Q)

Zomutt (uses Z and U)

Negg (uses N and G)

Dubloon (uses D, B, and O)

Chia (uses C and H)

Eyrie (uses Y and E)

Vex (uses V and X)

Meepit (uses M, P, and T)

Faerie (uses F and I)

Looking at the letters we used we can see the three leftovers are L, R, and W. The answer is LRW.

Round 439:

Category: Cryptology

Difficulty: Fairly easy

Puzzle: (like 438, does not appear on-site)

Heermeedjet came down with Neggitus one evening and was unable to finish his rounds. His goal was to steal an elusive golden doughnut from the Bakery, so Merouladen volunteered to help him complete the task. Merouladen successfully broke into the bakery and stole the golden doughnut but was caught just after he managed to hide it! Stuck in a jail cell, Merouladen worried that the golden doughnut would be eaten by a wandering Grarrl.

When Heermeedjet recovered, he received a coded message from Merouladen informing him of the hiding place so he could retrieve the golden doughnut as soon as possible:


Where did Merouladen hide the golden doughnut?

Please list the decoded message only, with the appropriate spaces between the words to show a proper sentence.

Solution: The hint here was the word bakery. Although both this keyword and the step that immediately follows were completely unnecessary, they help a little bit. The first step is to look at every other letter in the ciphertext: ROLLBREADMUFFINBAGUETTE. Notice the link to the bakery? Look again and you'll see the words roll, bread, muffin, and baguette. Remove these letters and get: DNALEIREAFNIEERTEHTREDNU. If we reverse this text and add spaces in the appropriate locations we get the answer: Under the tree in Faerieland.

Round 440:

Category: Math (counting)

Difficulty: Moderate**

Solution: The puzzle unfortunately omits a key detail: all four dice must show unique numbers. Your first instinct may tell you to do just as King Roo did and actually find and add together all combinations, but I assure you it is not needed. Let's think about the puzzle in terms of sums. First, assume the first die is not touched. The last three dice can be arranged 120 ways (6!/3! or 6*5*4). Now there are seven ways this can happen, one for each number possible on the first die. We thus have 120 combinations where the first die shows a one, 120 combinations where the first die shows a two, and so on. Because it can show 1, 2, 3, 4, 5, 6, or 9 and 1+2+...+9 = 30, the first die contributes 120*30*1000 (360000) to the sum because it's in the thousands place. If we repeat this for the other three dice, a mathematical expression for the sum is (1000*120*30)+(100*120*30)+(10*120*30)+(1*120*30). The value of this expression is our answer: 3999600.

** = The difficulty for this round assumes you applied the uniqueness restriction when solving it. I myself arrived at a wrong answer (30*343*1111 = 11,432,190) by not imposing uniqueness on the sample space.

Round 441:

Category: Math

Difficulty: Easy

Solution: This was really just a bit of trial and error (and cleverness) and a tiny bit of guesswork in what would be counted as correct. 8/(3-8/3) = 8/(1/3) = 3*8 = 24. Therefore the answer is the concise expression 8/(3-8/3)=24. Although correct, the answer 8/(3-(8/3))=24 was not accepted because the extra parentheses are not needed, as division has precedence over subtraction in the good ol' order of operations.

Round 442:

Category: Word Puzzle

Difficulty: Moderately challenging

Solution: Ahh, a trip down memory lane for the avid and loyal conundrum solvers. This conundrum was just like rounds 257 and 259. The best way to go about solving this was to try beginning with the common question words who, what, when, where, why, and how. After much experimentation you'd arrive at the question "How many Neopet species names end in a vowel or Y?" pictured below.

Without listing them, I can tell you 38 out of the 54 species have a name that in a vowel or the letter Y. Now I challenge you to try working out round 259. ;)

Round 443:

Category: Math (counting)

Difficulty: Quite simple

Solution: This was a repeat of Round 83 with a larger combination lock and a faster thief. We just do some simple arithmetic to count the number of possible combinations and some more arithmetic to calculate how many days it will take. There are 76*76*76 + 76*76*76*76 (76 because we can't forget the zero!) combinations to try, times two to account for the starting direction. Multiply by 8 to get the time in seconds. Divide by 60*60*24 to convert to days. (76^3+76^4)*2*8/(60*60*24) = 540818432/86400 = 6259.473

Rounding to the nearest day gives us 6259

Round 444:

Category: Math (patterns)

Difficulty: Easy

Solution: This puzzle began on Pi Day (the 14th day of Running) in Year 14. This was the first clue that there was some oh so tasty Pi involved. On the first day Clodbottle has three petpetpets. Day two brings Milton three more petpetpets. On day three the professor sees one new petpetpet. On day four, he had yet another extra petpetpet. The pattern in the day-to-day growth that's emerging is 3.141.... Continuing to day fifteen, we need 14 digits of Pi: 3.1415926525897. We add the digits to get 68 and add that to the three petpetpets he started with on day one giving us 71 petpetpets total on the final day.

Note: Some people got an incorrect answer by using the observation that petpetpets pass away after a week, and subtracted the first eight days. This was incorrect because the digits of Pi measure daily net growth.

Round 445:

Category: Math/Logic

Difficulty: Extremely Easy

Solution: The last part of this question asks you to multiply by the number of Lupe villains in the Gallery of Heroes. There are clearly no villains in the Gallery of Heroes. Multiplying by zero gives us a final answer of 0.

Round 446:

Category: Cryptology

Difficulty: Easy

Solution: One of the first things you'll notice while reading this story is the abundance of spelling errors. Throughout the puzzle, you'll find eight words that have a letter missing. These eight letters form a Neopian (Gadgadsbogen themed) word when rearranged. The words and the corresponding letters are:

touists (tourists missing R)

ben (been missing E)

Uchini (Puchini missing P)

Salider's (Salinder's missing N)

Banngos (Banangos missing A)

Honeylumes (Honeyplumes missing P)

smal (small missing L)

serching (searching missing A)

The missing letters (REPNAPLA) are an anagram of the item Arnapple.

Round 447:

Category: Pattern

Difficulty: Easy**

Solution: The words in the sequence were taken from the New Features page (link "New Features page" to from the 5th and 6th days of Eating in Year 14. The words are the first words of every other bullet point. Therefore the next word is "Ooh."

** = This round began on the 5th day of Eating in Y14. This meant the answer was from the current week's news. It was much easier to find this pattern during the week this round was live on the site.

Round 448:

Category: Word Puzzles

Difficulty: Medium

Solution: The numbers in the puzzle (first, second, third, etc.) referred to the letters in the answer. "Is in ___, but never in ___" refers to the letters in the two words being compared. Thus, we have the following choices:

First letter: C, L, P, U

Second letter: A, H, I, P, T, W

Third letter: E, G, I, K, T

Fourth letter: F, L, O, P, R, W

Fifth letter: B, L, S

To solve this, we have to find the correct letter for each position. The last line of this poem tells us we are looking for something from the Lost Desert. We can see that the correct letters are C, H, E, O, P, and S forming the word Cheops, a Desert Food.

Round 449:

Category: Math

Difficulty: Easy

Solution: Only the numbers are relevant. Upon inspection, it's not hard to recognize that 18 is three times 6 and 72 is four times 18. From this you may be able to correctly surmise the pattern: that each number is a multiple of the last and the multiplier increases by one each time.

6*3 is 18, 18*4 is 72, 72*5 is 360, and 360*6 is 2160. Therefore the missing number must be 360.

Round 450:

Category: Math

Difficulty: Easy

Solution: Let's start at the beginning with his shopping spree.

Tropical Food: 20000 – 15% = 20000*.85 = 17000

Tombola: 17000 + 15% = 17000*1.15 = 19550

Jelly Sponge Cake: 19550 – 8% = 19550*.92 = 17986

Since he has between 10000 and 25000 NP, he is eligible for a Super Gold Plus account which will acquire 7% annual interest compounded daily. First let's do a quick check to see if his interest will be constant or if it will go up during the ten day period in which we are currently investing our interest (see what I did there?). Currently, at 17986 NP, he will get 4 NP per day in interest ((.07*17986)/365, rounded UP). If he got only 4 NP per day for ten days, he would have 17986+40 NP or 18026 NP. In order to get 5 NP per day at 7%, he would need no less than 20858 NP. Therefore his interest did not go up during those ten days and our answer does not need to be changed. Our final answer is 18,026. Don't forget the comma!

Round 451:

Category: Guessing/Luck

Difficulty: None

Solution: D'oh! I just told you not to forget the comma. Don't worry, you weren't alone. You along with approximately 165 others forgot to insert the comma.

Are you getting the hang of how most Lenny Conundrums are solved? Keep your eye out for the solutions to rounds 451-495 in next week's editions. We've also included 5 special puzzles that have never been seen before so you can practice on your own! Make sure to check it out!

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