99+ Problems & They're All Lenny Conundrums (Part 2) by minnesotan

Also by sacadosLast week we looked at the solutions from Lenny Conundrum rounds 400451. This week, we've explained the solutions from rounds 452495: As last time, to see each week's question and answer put the number of the round in the following link: http://www.neopets.com/games/conundrum_feature.phtml?round=### Round 452:
Category: Word Puzzles Difficulty: Somewhat difficult Solution: The majority of the work done in solving this puzzle is finding the answers to all five Neopian trivia questions. The answers (in order) are: Ogrin, Scado, Heart, Linae, and Talek. Looking at the first letters we see they are an anagram of Sloth. So to get the answer we need to take the last letter of "Scado," the fourth letter of "Linae," the third letter of "Ogrin," the second letter of "Talek," and the first letter of "Heart." Doing this we get OARAH. Round 453:
Category: Silly/Funny Difficulty: Very Easy Solution: If you read carefully, you'll notice Charles is the only Ogrin mentioned. Therefore the answer can only be one. Round 454:
Category: Math Difficulty: Easy Solution: Ahhh, it's PLUMPY!!!!! The first important statement in solving the puzzle is: "If you took 20 and then added half his weight, that's how much he actually weighs." If 20 plus half his weight is his weight, then 20 must also be half his weight (a half plus a half is a whole!) and Plumpy is therefore one very heavy Angelpuss at a whopping 40 pounds. We aren't done yet, though. The question asked us how much the scale said Plumpy weighed. The next key statement in solving this is: "the scale was off by 12 pounds but he hadn't actually lost any weight." This tells us the scale read 12 pounds lower than Plumpy's actual weight. It read 4012, or 28 pounds. Round 455:
Category: Math Difficulty: Slightly hard Solution: Yay, more Plumpy! Before we can begin any calculations, we must remember that in the last Lenny Conundrum we determined Plumpy weighs 40 pounds. Do NOT use the 28 pounds from the broken scale. This means Plumpy will eat: 1.6*40 = 64 pounds of food on Sunday 40 pounds of food on Monday .8*40*2 = 64 pounds of food on Tuesday 40 pounds of food on Wednesday .8*40*2 = 64 pounds of food on Thursday .3*40*2 = 24 pounds of food on Friday 1.6*40 = 64 pounds of food on Saturday .15*40 = 6 extra pounds of food on Species days If one looked at a calendar during this round they'd see that the Month of Eating (Y14) contained five Sundays; five Mondays; four Tuesdays, Wednesdays, Thursdays, Fridays, and Saturdays; and five Species days. So the amount of food Plumpy ate that month was: 64*5 + 40*5 + 64*4 + 40*4 + 64*4 + 24*4 + 64*4 + 6*5 = 1574 Note that although it says Plumpy eats "no more, no less" on Mondays/Wednesdays, the extra cake on Species days applied even to Shoyru Day and Krawk Day, which were both Mondays. Also note that although the answer currently given onsite is 1102, this was obtained by incorrectly using 28 as Plumpy's weight. This was the answer that was originally deemed correct, but a second wave of neomails was later sent out for people who submitted 1574 and 1574 is the answer that appeared in the news. Round 456:
Category: Cryptology Difficulty: Disturbingly tough Solution: Let me mention that this round went on for three weeks because no one answered it correctly.** The first step in solving this is to lay down a system. This system is given by the first two lines of the conundrum, which by the way were both HINTS. Given that Jelly  Linae = JY it follows that Peach  Peadackle = H, Rainbow Negg  Orange Negg = IBW, and so on. Continuing this we get: 1 = H + IBW 2 = N + GSL 3 = CT + FLV 4 = GT – EMN 5 = Y – URDNK 6 = WDC – LCKR 7 = MQU – TGL "Great," you say, "but what do we do next," you ask? Okay kid, this is where it gets complicated. The second hint tells us C+E = H. If you're keen, you might notice that C is the 3rd letter of the alphabet, E is the 5th, H is the 8th, and 3+5=8. We can use this system to add/subtract letters as long as the resulting number ranges from 1 to 26. But what exactly do we do when we have groups of letters to add/subtract? We have to think back to algebra and use the distributive property. For example, H+IBW becomes H+I, H+B, and H+W. So let's start! H+W is out of the question for now unless we were to say 27=A, 28=B, etc. We will not do this as it is just unnecessary work. So let's only consider H+I and H+B H+I = 8+9 = 17 = Q H+B = 8+2 = 10 = J Now we'll look at the second equation, which when distributed gives: N+G, N+S, and N+L. We can rule out N+S because it goes past Z. N+G = 14+7 = 21 = U N+L = 14+12 = 26 = Z Now we move on to the third equation where the amount of work increases a bit. Distributing gives six options: C+F, C+L, C+V, T+F, T+L, T+V We do not have to consider the last two additions. C+F = 3+6 = 9 = I C+L = 3+12 = 15 = O C+V = 3+22 = 25 = Y T+F = 20+6 = 26 = Z The fourth equation also gives us six options: GE, GM, GN, TE, TM, TN Pairs in which the first letter comes before the second can be ruled out as subtracting would then yield a negative number. The remaining pairs to consider are: GE, TE, TM, and TN. GE = 75 = 2 = B TE = 205 = 15 = O TM = 2013 = 7 = G TN = 2014 = 6 = F The fifth equation gives five viable options: YU, YR, YD, YN, and YK. YU = 2521 = 4 = D YR = 2518 = 7 = G YD = 254 = 21 = U YN = 2514 = 11 = K YK = 2511 = 14 = N The sixth equation is even worse than the previous five. It has twelve possible pairs, however, it can easily be noted that none of the four pairs beginning with C are valid subtractions and the only valid pair beginning with D is DC. The five pairs which remain after our pruning are: WL, WC, WK, WR, and DC WL = 2312 = 11 = K WC = 233 = 20 = T WK = 2311 = 12 = L WR = 2318 = 5 = E DC = 43 = 1 = A I'm sure you're exhausted by now. I know I am! Unfortunately we've still got one entire equation to solve until we're able to begin finding the answer and this one is even worse than the rest. By now I can trust you're able to find all nine pairs in the seventh equation and spot the bad ones. We've got seven pairs to subtract: MG = 137 = 6 = F ML = 1312 = 1 = A QG = 177 = 10 = J QL = 1712 = 5 = E UT = 2120 = 1 = A UG = 217 = 14 = N UL = 2112 = 9 = I Now that we've finally solved all seven equations we can move on to the final stage. Our work above has revealed this information: 1 = QJ 2 = UZ 3 = IOYZ 4 = BOGF 5 = DGUKN 6 = KTLEA 7 = FAJENI Now we can find a word in the same way we did above in Round 448. Unfortunately there are two answers this time: QUIGGLE and QUIGUKI. But which one is correct? We don't know, so we just submit one and hope it's right. ** Although the official answer lists both words, some people solved the puzzle in the first two weeks the conundrum was live and submitted Quiggle, but it was not accepted as correct until the third week. Round 457:
Category: Math Difficulty: Easy Solution: This problem is really just basic algebra. We can set up a system of two equations in two variables that can be solved very easily using elimination, substitution, or other methods. The equations we need are: 25g + 5s = 1250 25g  5s = 1000 We can subtract the second equation from the first and get 10s = 250. Of course s = 250/10 = 25. Round 458:
Category: Facepalm. Difficulty: Facepalm. Solution: Facepalm. Round 459:
Category: Cryptology Difficulty: Easy
Solution: This is a shift cipher, but we only have to shift the letters by one. Replace each letter in the puzzle with the letter that comes after it in the alphabet. The letter Z becomes A. The decoded message is:
JEFF THE KYRII REALLY HATES APPLES
HE WAS OFFERED ONE THIS MORNING AND ANOTHER TONIGHT
HE WAS EVEN OFFERED THREE LAST WEEK AS WELL AS TWO YESTERDAY
IF TODAY IS WEDNESDAY HOW MANY APPLES WERE OFFERED TO JEFF THIS WEEK
Let's count. He was offered one this morning and another tonight. Our current total is two. He was offered two yesterday, putting us at four total so far. The three from last week do not count because the question asks specifically about this week, so our answer is four. Some people reencoded the answer by replacing each letter in the word FOUR with the letter before, giving ENTQ as a possible answer. Both answers were marked as correct, though not everyone who submitted ENTQ got credit for it. Round 460:
Category: Math Difficulty: Very Easy Solution: Every minute she goes up 10 meters and down 5 meters. This is a net gain of 5 meters per minute. Thus, to climb 240 meters she has to climb for 240/5 = 48 minutes. Round 461:
Category: Patterns Difficulty: Easy Solution: The relationship between all the things listed isn't hard to see, especially if you've ever done some restocking. All the phrases are contained in items classified as Candy on Neopets. Although this commonality was pretty easy to spot, guessing the answer the question wanted was really a matter of getting lucky. The official answer turned out to be "Chocolate Factory." Note that Grundo Thief referred to the item Grundo Thief Candies, which is r101 and has never been sold in the Chocolate Factory. Round 462:
Category: Riddle Difficulty: Easy Solution: The trick to this one is that "peek" and "pirate" both refer to the letters in the answer(s). Also, we can get from the context that we're probably looking for birdlike petpets. A little bit of searching would lead to the observation that Pirakeet contains the letters needed to spell PEEK, but Piraket requires us to add the letter E. Both also share the letters needed to spell the word PIRATE. Thus, our answer is Pirakeet and Piraket. Round 463:
Category: Patterns/Guessing Difficulty: Easy Solution: This one is similar to Round 461, but instead of just finding the common factor, we also have to guess at what the fourth item is. The first step to solving this conundrum is to identify that all three things listed are the names of Usuki Dolls. Now of course we have to guess which Usuki is considered "the best treasure of all." Apparently Hannah thinks the best treasure of all is herself, so she sent a Usuki Pirate Wench. Round 464:
Category: Cryptology Difficulty: Extremely Difficult Solution: The solution is given on the answer page; no explanation will be given here. Round 465:
Category: Guessing/Luck Difficulty: Hard Solution: Guess. Round 466:
Category: Physics Difficulty: Easy Solution: The key here is to notice boats are buoyant. The water will never be any higher relative to the boat, so the water will never ascend the ladder. Zero hours doesn't make much sense given this information though; a better answer would have been never. Round 467:
Category: Problemsolving Difficulty: Easy Solution: Let's find the oddonesout: (1) Aroota is the odd one out, as its name does not begin with AL as the others do. (2) Similar to above, the petpet that doesn't fit is Blugar because it does not begin with BA. (3) Cubett is the different one, as it is only petpet of these four that cannot be painted. In fact, the other three even have a color in common: Ghost. (4) Drackobunny is the odd one out because it comes from the Cooking Pot on Mystery Island whereas the others come from the main shops. (5) Flippy is the only petpet of the four without wings. When we put together the listed letters, we get the word Lutra. Round 468:
Category: Math Difficulty: Easy Solution: The P3's in Experimental Habitarium Three divide every 60 seconds. The Habitarium was half full at 12:13:00, so in 60 more seconds it divides and reaches full capacity. Thus it becomes full at 12:14:00 NST Round 469:
Category: Classic Riddle Difficulty: Easy Solution: This is a very old riddle that has taken many forms. We assume the ball floats and say that he pours water in the hole to make it rise to the top. Round 470:
Category: Guessing/Luck Difficulty: Hard Solution: Guess. Round 471:
Category: Patterns Difficulty: Easy Solution: When this puzzle went live, the morphing potions listed were some of the newest ones. The morphing potions listed were from the weeks immediately preceding the conundrum and they're listed in order of release. In addition, of the two morphing potions in the news, all those listed in this conundrum were positioned on the right in the news. The release dates were: Wraith Gelert Morphing Potion  16th of June
Speckled Kau Morphing Potion  19th of June
Snow Peophin Morphing Potion  26th of July
Camouflage JubJub Morphing Potion  22nd of August
Eventide Kougra Morphing Potion  7th of September
Pirate Techo Morphing Potion  13th of September
Water Kau Morphing Potion  18th of September
Now we need to find the next batch of morphing potions that was released and the answer will be the one that was pictured on the right in the news. The next morphing potions were on the 29th of September and they were Baby Blumaroo Morphing Potion and Baby Shoyru Morphing Potion. Baby Shoyru Morphing Potion is on the right, so it's our answer.
Round 472:
Category: Math Difficulty: Medium Solution: Believe it or not, no trialanderror or guessing is needed to solve the puzzle. I will show a solution involving only logic. Before we proceed let's lay down some vocab: Augend  The number in the first row that follows the pattern EEO Addend  The number in the second row that follows the pattern EOO Sum  The number in the third and final row that follows the pattern OEOE Now that we've got that vocab out of the way, we'll proceed to solve the conundrum. We're told that the digits of the augend have a sum of 9. Possible numbers for the augend are: 207, 243, 261, 405, 423, 603, 621, and 801. In order for the sum to be 4 digits long, the hundreds digits of both the augend and the addend must total to 10 or more. The first digit of the sum must be a 1, as it's impossible to produce a 2 to carry over. This rules out 261, 621, and 801 for the augend. We know there's no carry over in the previous columns because both O+O=E and E+O=O are already true. This means the second digit in the augend cannot be 0 as that would cause the sum to contain a repeat of the second digit of the addend. This rules out 207, 405, and 603 for the augend. So far we've narrowed down the augend to 243 or 423. Let's try 243: The final digit of the addend must be less than 7 and odd. We've already used 1 and 3 so it has to be 5. As previously mentioned, the first digit of the augend plus the first digit of the addend must cause a 1 to carry over. The only possible number for the first digit of the addend is therefore 8, but this has already been used as the last digit of the sum. We have shown that the augend cannot be 243, so we move on to 423. Again, we start out by recognizing the ones digit of the addend must be 5. This time though, 6 is sufficiently large for the first digit of the addend. We now have two digits left to place: 9 and 7. Two plus seven is nine, so the 7 goes in the addend and the 9 in the sum. We have now created the following (correct) addition: 423+675=1098. Therefore the sum (and the final answer) is 1098. Round 473:
Category: Classic Riddle Difficulty: Easy Solution: This is another very old riddle; this riddle comes from a nursery rhyme. There are many many answers to the riddle, though one of the answers is far more common than the others. The things we encountered on our way to Meridell were headed the other direction, thus we are the only one going to Meridell. Therefore the answer is one. Round 474:
Category: Math Difficulty: Medium Solution: Below I will use the the symbol ^ to mean bitwise XOR (exclusive or). This is not to be confused with a bitwise AND. Two important things to note about XOR are that: 1) Z ^ Z = 0 (Z is any number... any number XOR itself is 0, because 0^0=0 and 1^1=0) 2) Y ^ 0 = Y (Y is another number... any number XOR 0 is itself, as 0^0=0 and 1^0=1) Therefore, 3) Y ^ Z ^ Z = Y ^ 0 = Y Using rule 3, we can isolate B by saying B = (A ^ C ^ D) ^ (A ^ B ^ C ^ D) = 1000011010 ^ 0011101011 = 1011110001 We're looking for B ^ C ^ E and we have B, so we need C ^ E You may see that (A ^ C ^ E) ^ A = C ^ E So, B ^ (C ^ E) = B ^ (A ^ C ^ E) ^ A = 1011110001 ^ 0000101010 ^ 0100010101 = 1111001110 Thus our final answer is B ^ C ^ E = 1111001110 You may have noticed that this solution was equivalent to A ^ (A^C^D) ^ (A^C^E) ^ (A^B^C^D), which is simply XOR'ing all four of the given bitstrings. Round 475:
Category: Math Difficulty: Easy Solution: Although the conundrum is incorrect in its assessment of the height of a binary tree (using the standard definition of height, a complete binary tree of height 5 actually has 63 nodes), it gives enough information to solve it the way we're intended to. Using the conundrum's definition of height, we see that a complete binary tree of height H has 2^H  1 nodes. Therefore a complete binary tree of height 97 has 2^97  1 nodes by the definition used in the puzzle. Computing this gives us the answer: 158456325028528675187087900671. That sure is a lot of nodes! Round 476:
Category: Guessing/Luck Difficulty: Hard Solution: Guess... Round 477:
Category: Problemsolving Difficulty: Easy Solution: I could use formal logic (and a bunch of crazy symbols) here, but that'd just scare you. Oh, and it'd overcomplicate the problem too! Instead, we'll just consider the three very obvious cases and prove that two of them create contradictions to show that the third is the correct answer. First case: Farmer Larry is telling the truth  If Larry is telling the truth, then Russell is lying. If Russell is lying, then either Larry or John is telling the truth. If John is lying (he must be for we have only one truthful farmer), then Larry is telling the truth. We're back where we started, so this is a viable solution. Second case: Farmer John is telling the truth  If John is telling the truth, then Larry is lying. If Larry is lying, then Russell is telling the truth. We cannot have two truthtellers, so this is not a viable solution. Third case: Farmer Russell is telling the truth  If Russell is telling the truth, then both Larry and John are lying. If Larry is lying, then Russell is telling the truth. This is okay, but we still have to see what implication John's statement has. If John is lying, then Larry is telling the truth. This is also a contradiction, so it is also not a viable solution. Our only viable solution was for Farmer Larry to be telling the truth. Therefore the answer must be Farmer Larry. Round 478:
Category: Lateral Thinking Difficulty: Easy Solution: "Wallace caught something he couldn't throw away" hints at the fact Wallace caught a disease. If we go to the hospital and read the descriptions we will find that Wallace's symptoms are associated with Jitters.
Round 479:
Category: Cryptology Difficulty: Medium Solution: The puzzle tells us to take the line apart, so that's what we do. If we split it into pieces we get: RBI OE NAMI UL RHA UMA RON IKATIA OV IZE. Now we have ten anagrams which unscramble to: BRI MAIN LU HAR MAU ORN TAIKAI VO ZEI. These are the names of the ten basic (brown) codestones. Round 480:
Category: Guessing/luck Difficulty: Easy Solution: We just have to make a logical guess as to what the answer is here. The image show is from The Lair of the Beast in Tyrannia, so our best guess is that we will encounter the lair beast.
Round 481:
Category: Math Difficulty: Easy Solution: Here is another puzzle that doesn't appear on its answer page. It was: A plushie factory in Neopia Central produces 150 blue Kacheek plushies per hour. The workers produce plushies 8 hours a day, 21 days a month at a cost of 250,000 NP per work month. Overtime production costs 1,750 NP per hour. If the plushies are priced at 15 NP each, the factory can manage to sell all the plushies produced during the 8 hour day. The factory marketer insisted that they can sell an additional 3,000 plushies per month if they lower the price of each plushie to 14 NP. A clever Blumaroo noted that this would actually decrease profits.
How many NP would this actually reduce profit by?
First, let's find the gain in sales caused by this change: Originally, they were selling 25,200 plushies (150 plushies/hour * 8 hours/day * 21 days) at 15 NP each, earning 378,000 NP total from sales. Selling 3000 more plushies at 14 NP each is 394,800 NP from sales (28200 plushies * 14 NP/plushie) which means there was a net gain of 16,800 NP from sales. Now let's consider the losses caused by increased expenses: The 250000 NP work month is constant because we're only considering a onemonth span of time. The extra expenses will come from the overtime. They will need 20 overtime hours (3000 plushies / 150 plushies per hour) to make the 3000 extra plushies, which means 35,000 NP invested in overtime hours. Subtracting our losses from our gains, we get that this change would reduce profits by 18,200 NP. Round 482:
Category: Problemsolving
Difficulty: Easy
Solution: There are several hints to the answer throughout the poem. The first clue is in the first line. What twinkles in the night? Stars, of course! The line about the compass rose is ambiguous (as they can have 4, 8, 16, or even 32 points) so we'll ignore it. The last two lines hint at paper, and the third to last line makes us think of origami. If we do a quick search we will find three items that are origami stars: Eight Point Paper Star, Four Point Paper Star, and Paper Star. Reading their descriptions and comparing them to the riddle, we can see the best answer is Four Point Paper Star which has the description, "folded to make them stronger." Round 483:
Category: Problemsolving Difficulty: Hard Solution: These are anagrams! Working them out, we get the names of four items. In order they are: Hot Dog with Mustard & Relish, Starting Spells, Apple and Custard Drops, Halloween Moehog Dastardly Cloak. At first these items may seem entirely unrelated, but if you look closer you will notice that all of their names contain the word STAR. Round 484:
Category: Patterns Difficulty: Tricky Solution: The phrases are the answers to every other Mystery Picture round released on a Tuesday and judged on a Thursday. The rounds were: Lime Lozenge was the answer to round 1185 judged on Thursday, February 21st.
Tiki Tack was the answer to round 1181 judged on Thursday, February 7th.
Tug O War was the answer to round 1177 judged on Thursday, January 24th.
Merry Go Round was the answer to round 1173 judged on Thursday, January 10th.
Now it is certainly tempting to assume the next phrase in the sequence is the answer to round 1169 because it was four rounds before 1173, but looking through the news archive we see some funny stuff happened due to the holidays. As a result, the round we actually need is 1170 which was judged on Thursday, December 20th. The answer was Winter Starlight Celebration so this is the answer to the conundrum. Round 485:
Category: Math Difficulty: Easy Solution: In order to win three pieces of candy rocks, Timmy had to win three more times than Capara. Capara won thrice, so Timmy won six times. Therefore they played 3+6 = 9 games. Round 486:
Category: Word problem Difficulty: Easy Solution: Only the first paragraph and last sentence are significant. The rest was just silly timewasting banter. We just have to look at the opponents from the third wave of the war for the obelisk and submit the name of the challenger from The Brutes. This enemy is of course the Brutal Mercenary. Round 487:
Category: Patterns Difficulty: Easy Solution: The bottles shown from left to right are: Bottle of Green Sand, Bottle of Grey Sand, Bottle of Orange Sand, and Bottle of Pink Sand. Looking at the names, it's easy to see they're sorted alphabetically by color. So the next bottle of sand has to be the color of sand that comes directly after pink alphabetically. At first you may think this color is either purple or rainbow, but the Tiki Tack Man does not sell sand in those colors. The answer is therefore Bottle of Red Sand. Round 488:
Category: Math Difficulty: Easy
Solution: The middle number in each triplet is the sum of the digits in both the first and third number in the triplet. For example, in the first triplet: 2+8 = 10 = 7+3.
Thus the missing number in the fourth triplet satisfies the equation 3+6 = ? = 5+4. Of course, the question mark is a 9. Round 489:
Category: Math Difficulty: Easy Solution: Because the figure is a "perfectly shaped step pyramid" the number of blocks in each level is the square of an odd integer. Thus, because there are 7 tiers in the pyramid, the number of blocks in the pyramid is the sum of the squares of the first 7 odd numbers. There are two ways to find this sum. The first way is to actually add all seven squares: 1+9+25+49+81+121+169 = 455. The second way is to remember the really very super extremely useful series from our math classes in school: the sum of (2k1)^2 from k=1 to n is (1/3)(4n^3  n). Plugging in n=7 here also gives 455, but who would want to do it that way? Round 490:
Category: Logic Difficulty: Easy Solution: This is yet another conundrum that doesn't show up in the archive. The puzzle was: Zaira was tidying up at the Neggery when she made an interesting discovery. After measuring some of the neggs, she realized the following:
The Basic Power Negg is heavier than the Sweet Heart Negg.
The Kaleideonegg is lighter than the Blue Furry Negg.
The Sweet Heart Negg weighs the same as the Blue Furry Negg.
If the above statements are all true, which negg would be the heaviest of the four?
We can combine these three statements into one: The Kaleidonegg is lighter than the Blue Furry Negg and the Sweet Heart Negg which are both lighter than the Basic Power Negg. The above sentence lists the neggs by weight from lightest to heaviest. Thus Basic Power Negg is the heaviest. Round 491:
Category: Word problem Difficulty: Easy Solution: The words in order are: Jinjah, Turtmid, Itchi, Scado, and Urchull. Thus the missing letters are J, T, I, D, and U and the answer is an anagram of JTIDU. Notice all the words are petpets and petpetpets. A fiveletter petpet name using the aforementioned letters is Djuti. Round 492:
Category: Patterns Difficulty: Hard Solution: The image of a Beekadoodle tells us the answer is probably related to petpets. The letters (and numbers) are the first characters of the names of petpets that won the PPL. (rounds 498 and 497) The T,K,P,4,S clearly comes from round 498. Going to the previous round we see the first two petpet names begin with B and 0, so the next three petpet names give us the answer: OPM.
Round 493:
Category: Patterns Difficulty: Easy Solution: The numbers are the game ID's of the games resulting from searching "Kreludor" in the Games Room. The fifth game that shows up when one does so is TROTRODS which has the ID 480. Round 494:
Category: Math Difficulty: Easy Solution: Let's say each Aisha eats one unit of food per day. They start with 28*135 = 3780 units of food. After nine days, they have 19*135 = 2565 units of food remaining. At this point, they gain 36 more Aishas and reduce their ration to .75 units per day each. This means the 2565 remaining units will last 2565/(171*.75) = 20 more days. Thus the food will last 9+20 = 29 days. Round 495:
Category: Cryptology Difficulty: Easy Solution: The message given is merely missing its vowels. It's clear that the last word begins with the K and ends with the S. When we insert vowels (and spaces) we get: You will defend yourself against a flock of enraged Kaus. The last word is Kaus. Note that this message comes from the Island Mystic. To make up for not including the final 5 solutions leading up to round 500, we have created 5 puzzles of our own! Use everything you've read about in the past 95 LC solutions in order to solve these tough conundrums. Puzzle #1:
What is the product of all the numbers in the next row of the sequence? 1 3 2 4 6 6 6 4 2 5 7 1 1 2 2 3 3 3 2 2 1 1 4 4 4 4 3 3 3 4 4 4 4 # # # # # # # # # # # Puzzle #2:
Plumbeard recently pillaged several ships and he now has quite the spoils, but he doesn't know where to bury them. When he landed on the beach of Krawk Island, he noticed the trees were arranged on a grid in threedimensional space (the beach was hilly!). He decided to bury the treasure at a point precisely in the middle of two trees, but he also wanted to bury it at a point on the gridlines. What is the minimum number of arbitrarily arranged trees Plumbeard must find to ensure that this is possible? Puzzle #3:
At the beginning of this year Lisha had no Neopoints, and so she made the following resolution: she was going to maximize her income for the year. In the first week of Month of Sleeping Lisha was successful and made 2856 NP. The second week of her Neopoint collecting was also a success and she made 1005174 Neopoints. The next week was also a success for her and she happened to receive a total of 7017 Neopoints that week. By the end of the Month of Sleeping Lisha had amassed a grand total of 3205258 NP, WOW! Assuming the second month was also a success for Lisha, how many Neopoints did she have by the end of the Month of Awakening? Puzzle #4:
Hildegard  a very clever Hissi  decided to use her cleverness to count cards in Scarab 21. As she was approaching the stall to play, she noticed a cloaked figure muttering inaudible incantations. Hildegard got through half of her game before noticing the deck had been rigged. The cloaked figure had left behind a message for her! The top half of the deck was in the following order: Ace, Ace, Ten, Jack, Queen, King, Nine, Jack, Jack, Queen, Six, Nine, Two, Three, Eight, Ten, Four, Five, Nine, Ten, Two, Seven, Ace, Queen, Six, King Hildegard did as the message said: she walked to a nearby stall and made a purchase. What did Hildegard buy? Puzzle #5:
What do all of the following have in common? 80 E in GK 6 F at the B 1 S of OPD ***ANSWERS AND SOLUTIONS*** Puzzle #1:
Answer: 452984832 Solution: The first row is arbitrary. In each successive row, each number is the number of times the number directly above it is in the previous row. So the first four numbers in the missing row will be 8's because there are eight 4's in the previous row. The next three numbers will be 3's because there are three 3's in the previous row. The last four numbers will also be 8's because there are eight 4's in the previous row. So the missing row has eight 8's and three 3's, and the product is 8^8 * 3^3 = 452984832 Puzzle #2:
Answer: Nine Solution: If we were to number the grid, there would be eight possible ways to place each tree in 3D space: (Even, Even, Even), (Even, Even, Odd), (E, O, E), (E, O, O), (O, E, E), (O, E, O), (O, O, E), and (O, O, O). In order for the midpoint of two integer points to lie at another integer point, each coordinate in both of the endpoints must have the same parity. Thus, since there are eight ways to arrange the parities, in order to ensure we have two points with the same parities we need nine points. Puzzle #3:
Answer: 3311585 Solution: Lisha's two sources of income were the Lenny Conundrum and the Mystery Picture. She won both Mystery Pic competitions and the Lenny Conundrum each week. So we go through the news archive and add up the NP she earned in the Month of Awakening and add it to the 3205258 NP she already had. Puzzle #4:
Answer: Two Erisims Solution: Did you notice that every other card was higher than the one before it? Good, that wasn't a coincidence! Take the list and pair up the cards. (Ace, Ace), (Ten, Jack), (Queen, King), etc. Then convert each card to a number (unless it already is one, of course!) like so: (1,1), (10,11), (12,13), etc. Now add together the numbers in each pair of cards and convert that number to a letter: 1+1 = 2 = B, 10+11 = 21 = U, 12+13 = 25 = Y, and so on. Doing so for all thirteen pairs reveals the message: BUY TWO ERISIMS. Puzzle #5:
Answer: Tyrannia Solution: Each line described something in Tyrannia 80 E in GK = 80 eggs in Grarrl Keno 6 F at the B = 6 factions at the Battleground 1 S of OPD = 1 slice of omelette per day
Assuming the LC has not been held up any time since round 495, coming up next week is the 500th round of the Lenny Conundrum. The past '100th' rounds of the LC have been particularly tough, so we can expect something challenging next week! Even though it will be quite difficult, there will likely be an exclusive and worthwhile prize if history repeats itself. We hope you've enjoyed reading about these indepth solutions. The Lenny Conundrum is a fun weekly game, and it gets easier with practice and looking at previous puzzles. Hopefully you can use the problemsolving skills you've read here to better understand how the LC works, and how to answer quickly enough to get your very own trophy and avatar.
