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Neopia's Fill in the Blank News Source | 25th day of Relaxing, Yr 19
The Neopian Times Week 78 > Articles > Solutions to Lenny Conundrums 26 to 50

Solutions to Lenny Conundrums 26 to 50

by qg_jinn

Untitled Document GAMES ROOM - Have you ever been stumped by a Lenny Conundrum? If you have and are curious how to solve the problem (not just what the answer is, but how to arrive at the answer), we're here to help! Neopets doesn't seem to want to bother with putting up complete solutions, but we feel this is important for those who are interested and who find the questions a fun challenge.

This article will discuss the solutions to Lenny Conundrums 26 to 50. (For the solutions to the first 25 conundrums, please see qg_jinn's article Belated Solutions to the First 25 Lenny Conundrums.) Due to the incredible amount of work required to write this article, we needed to obtain special permission from Neopets to have two people work on this article together! Like the first 25 conundrums, some questions require a bit of mathematical reasoning. The math is not deep, and should certainly be accessible to anyone in high school (except Round 35 -- you'll need to have been exposed to basic calculus for that one). We've tried to keep the discussion as simple as possible, but in some places, we've had to make certain assumptions about background knowledge. Also, in a few solutions, we've included an "Extra" paragraph for those who want to know more.

In looking through the conundrums, we've even found a few mistakes in Neopets' official answers or their question statements (see rounds 35, 39, and 40)! Without further ado, let's get started...


Round 26. Answer: 310

With this one, you basically have to play around with different ideas until you find something that resembles words. Here's how we went about it [see Note 1]...

The question mark at the end and the "whbt" at the beginning provide somewhat of a hint. If we guess the clue is a sentence that asks a question, then perhaps the first word is "what", only one letter different from "whbt". So let's start by replacing b by a. (Note that if we follow this hint, we are forced to ignore spaces in the clue sentence, because "whbt" is part of the longer string "whbtdp".)

Now that we've started replacing letters, perhaps other letters should be replaced as well. To find out if this is indeed the case, we did a histogram (count) of letter frequencies. Here are the frequencies of the 26 letters in the clue sentence:

    abcdefghijklmnopqrstuvwxyz
    02010411040110050015042150
Here are the key observations that we made. First, the letters that appear the most in the clue sentence are f, j, p, t, v, y. With the exception of t, these letters are not the letters that appear the most in English [see Note 2]. Thus, some of these letters must be replaced. Second, the vowels a, e, i, o, u do not appear in the clue sentence at all. Thus, we must change some letters in the clue sentence into vowels if we are to make English words.

Now the question is what replacements should we make? This is where the initial hint (b is replaced by a) really helps. We note that b is the letter that comes right after a in the alphabet, and of the letters that appear the most in the clue sentence, we have f which follows e, j which follows i, p which follows o, and v which follows u. Let's make all these replacements (also take out spaces) and see what we get:

    whatdoyougetieyoumutiolyeiueaysixtytwo?
This isn't the final solution, but we can see some words, which gives us confidence that we're on the right track. We can make out "what do you get" at the beginning and "sixty-two" at the end, but what's that ugly chunk "ieyoumutiolyeiueay" in the middle with lots of vowels in it?

Well, remember how we replaced all the letters f, j, p, v in the clue sentence with vowels? This left the resulting sentence with none of these four letters. Perhaps these four letters are in the ugly chunk with lots of vowels, and we just made too many replacements. With this final key insight, we go back and see if we can undo some of the replacements we made in the ugly chunk to make some English words. It turns out that we almost can:

    whatdoyougetifyoumutiplyfivebysixtytwo?
We can now make out "what do you get if you mutiply five by sixty-two?". We said almost because "mutiply" is not actually a word! Neopets messed up the spelling, and it should really be "multiply". Anyway, so what do you get if you multiply five by sixty-two? Why, 310, of course!

Notes. (1) The reader should note this is not the idea that we tried first. We tried a couple of other things at first that did not pan out. This puzzle is not that easy. The main obstacle to our solving the puzzle quickly was that the clue sentence already had strings of connected letters in it, which led us to believe that those would become words in the solved puzzle. (2) A quick search on the Internet will tell you that the vowels a, e, i, o and the consonants t, n are the letters that appear the most in English.

Round 27. Neopets answer: 37 eyeballs   Our answer: not enough information

Not a conundrum. You just have to take a guess.

Round 28. Answer: 34 weeks

The first thing we should do is calculate the volume of the cave. This would be

    length * width * height
    = (100 m) * (75 m) * (50 m) = 375000 m3
One thing to watch out for is that the volume of the fungus spore is given in cubic inches, while the dimensions of the cave are given in cubic meters. We need to choose one set of units. Since the fungus spore's size is a simple one cubic inch, it would be easier for us to convert the size of the cave to cubic inches. This gives
    (375000 m3) * (100 cm/m)3 / (2.54 cm/in)3
    = 22883904035.5 in3

Since the fungus doubles in volume every week, the question now becomes how many times we need to double one cubic inch before it becomes 22883904035.5 cubic inches. One easy way to do this is to take a calculator, type in the number 1, and then keep multiplying the result by 2. Keep count of the number of times you multiply by 2. When you count to 34, you should see 17179869184 on your calculator, which is smaller than 22883904035.5. However, when you count to 35, you should see 34359738368 on your calculator, which is bigger than 22883904035.5. Thus, at some point between 34 weeks and 35 weeks, the fungus fills the cave. Rounding down means the answer is 34 weeks.

Extra. For readers that have studied logarithms, we can actually find out the exact number of weeks for the fungus to fill the cave. This number is simply the logarithm, in base 2, of 22883904035.5. This comes out to about 34.4136 weeks.

Round 29. Answer: 29th day

One way to reason about this problem is as follows. On each day at the omelette, there are some old faces of Neopets who have come to the omelette before, and there are also some new faces of Neopets who are just arriving for the first time. How many of each type are there? Well, the Neopets who have been here before are simply the Neopets who were here yesterday! And the Neopets who are just arriving took two days to get here, so they must have been informed of the omelette the day before yesterday. The number of such Neopets is the same as the number of Neopets that were here the day before yesterday, because each of those Neopets went on to inform a new friend!

To sum it up, on each day, the number of Neopets that come to the omelette is equal to the number of Neopets that were here yesterday, plus the number of Neopets that were here the day before yesterday.

Armed with a calculator, we can figure out how many Neopets came on each day. On the 1st day, there was only 1 visitor (Kawlet), and on the 2nd day, there was only 1 as well.

On the 3rd day, there were as many as the number on the 2nd day plus the number on the 1st day, or 1 + 1 = 2 visitors (Kawlet and Choosi).

On the 4th day, there were as many as the number on the 3rd day plus the number on the 2nd day, or 2 + 1 = 3 visitors (Kawlet, Choosi, and Zirpi).

Similarly, on the 5th day, there were 3 + 2 = 5 visitors. On the 6th day, there were 5 + 3 = 8 visitors. On the 7th day, there were 8 + 5 = 13 visitors. On the 8th day, there were 13 + 8 = 21 visitors. On the 9th day, there were 21 + 13 = 34 visitors. (You might recognize this sequence as the Fibonacci sequence, but even if you have never seen it before, you can still calculate the numbers.)

Here are some more... 10th day: 55 visitors, 11th day: 89 visitors, 12th day: 144 visitors, 13th day: 233 visitors, 14th day: 377 visitors, 15th day: 610 visitors, 16th day: 987 visitors, 17th day: 1597 visitors, 18th day: 2584 visitors, 19th day: 4181 visitors, 20th day: 6765 visitors, 21st day: 10946 visitors, 22nd day: 17711 visitors, 23rd day: 28657 visitors, 24th day: 46368 visitors, 25th day: 75025 visitors, 26th day: 121393 visitors, 27th day: 196418 visitors, 28th day: 317811 visitors, 29th day: 514229 visitors, 30th day: 832040 visitors, 31st day: 1346269 visitors, 32nd day: 2178309 visitors, and so on...

Okay, great, so we can find out how many visitors came to the omelette on each day, but we still haven't answered the question of when the omelette will run out. To answer this, we need to add up the visitors for each day, until we surpass one million, which is the total number of servings the omelette will provide. If we do this, we will find that after 28 days, there will have been 832039 visitors, which is less than one million, so the omelette has not run out yet. However, after 29 days, there will have been 1346268 visitors, which is greater than one million, so the omelette will run out on the 29th day.

Round 30. Answer: 45 Lenny Conundrums

This Lenny Conundrum was released on January 24th (24th day of Sleeping), and two previous Lenny Conundrums had already been released in 2002 (Year 4), on January 10th and 15th. Thus, a total of 3 Lenny Conundrums have been released at this point.

Now we follow the facts given in the question statement. Three weeks from January 24th brings us to February 14th, and the following Tuesday would be February 19th. Thus, Lenny Conundrums are consistently released every Tuesday starting from February 19th until the end of the year, except that no Lenny Conundrums are released in September (the month of Gathering).

These are the dates of the Lenny Conundrums: February 19th and 26th; March 5th, 12th, 1th, and 26th; April 2nd, 9th, 16th, 23rd, and 30th; May 7th, 14th, 21st, and 28th; June 4th, 11th, 18th, and 25th; July 2nd, 9th, 16th, 23rd, and 30th; August 6th, 13th, 20th, and 27th; October 1st, 8th, 15th, 22nd, and 29th; November 5th, 12th, 19th, and 26th; December 3rd, 10th, 17th, 24th, and 31st.

If we count up the dates, we find that 45 Lenny Conundrums are released throughout the whole of Year 4. (Of course, this problem is a far cry from reality!)

Round 31. Answer: 362880 combinations

There are a total of 9 books. One method to figuring out how many ways there are to arrange the books is by actually imagining the process of arranging them. First, imagine taking all the books off the shelf, and then putting them back on from left to right. At the start, all 9 books are off the shelf, so there are 9 books that we can choose from to place in the left most position.

Now after we do this, there are 8 books left to choose from for the second position on the shelf. Since every choice of the first book and every choice of the second book thereafter will produce a different of arrangement of the first 2 books, there are a total of 9 * 8 ways to arrange the first 2 books.

Next, there are 7 books left to choose from for the third position on the self. Once again, every arrangement of the first two books and every choice of the third book thereafter will produce a different arrangement of the first 3 books, so there are a total of 9 * 8 * 7 ways to arrange the first 3 books.

We can certainly see a pattern developing. If we continue on in this manner, there are 9 * 8 * 7 * 6 ways to arrange the first 4 books, and 9 * 8 * 7 * 6 * 5 ways to arrange the first 5 books, and so on until we get to

    9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880 ways
to arrange all 9 books on the shelf.

Extra. This number is usually referred to as "nine factorial".

Round 32. Answer: 175 NP per share

Let's say that Narila bought her ePlushie stock at x NP per share. Growing in value at 7.2% per month means that every month, the value of the stock becomes 1.072 times the value in the previous month. After 12 months of growing at 7.2% per month, the stock should now be worth x(1.072)12 NP per share. The problem statement gives us enough information to calculate this value:

    1128583 NP / 2800 shares = 403.065 NP/share
Thus, we get the equation
    x(1.072)12 = 403.065 NP/share
    ⇒   x = 403.065 / (1.072)12 = 175 NP/share.

Round 33. Answer: 250 minutes

First, let's figure out how many laps the Wheelie Bot will travel before running out of batteries. The total length traveled will be 10 km = 10000 m. The track has a radius of r = 1 m, so the track length (using the formula for the circumference of a circle) is 2πr = 6.2832 m. The total number of laps traveled will be

    (10000 m) / (6.2832 m/lap) = 1591.55 laps.

Now we should see how many laps are traveled in the first 3 hours, when the Wheelie Bot was going at 20 seconds per lap. We have

    (3 hr) * (3600 sec/hr) / (20 sec/lap) = 540 laps.
Therefore, after the first 3 hours, the Wheelie Bot can still go for 1591.55 - 540 = 1051.55 laps before it runs out of batteries.

Now we need to find out how much more time the Wheelie Bot will travel for at the new speed of 4 seconds per lap (5 times as fast as before). We have

    (1051.55 laps) * (4 sec/lap) / (60 sec/min) = 70.10 min.

Finally, we add up the times. The Wheelie Bot traveled for 3 hours at the slow speed, which is 180 minutes. Then it traveled for 70.10 minutes at the faster speed before running out of batteries. In total, it traveled for 250.10 minutes. Rounding down, we get 250 minutes.

Round 34. Answer: 7.543 cm

The height of the juice starts at 10 cm. After Maggie drinks an inch (2.54 cm), the height becomes 10 - 2.54 = 7.46 cm. Then we have to add on the height added by the bubbles, which we calculate below.

First, we add up the volumes of the three bubbles. The formula for the volume of a sphere is (4/3)πr3. Using this formula (remember the radius is half of the diameter), we find the volumes of the three bubbles to be 0.065450 cm3, 0.523599 cm3, and 1.767146 cm3, respectively. Adding them together gives a total volume of 2.356195 cm3. To find the height added by the bubbles, we have to divide their volume by the cross-sectional area of the cup, which is πr2 = 9π = 28.274334 cm2. Thus, the height contributed by the three bubbles is 2.356195 / 28.274334 = 0.083... cm.

Finally, the height of the juice after blowing the bubbles is 7.46 + 0.083 = 7.543 cm.

Round 35. Neopets answer: 31.274 ladles   Our answer: 31.206 ladles

We'll first give the outline of the solution, and then go into the details. From the information in the first paragraph and in the last sentence of the fifth paragraph, we can figure out how many ounces of syrup each color of Grundo needs to mutate. From the information in the second through fourth paragraphs, we can figure out how many Grundos of each color there are. Combining these, we can figure out how many total ounces of syrup are needed to mutate all the Grundos Then, using the information in the fifth paragraph and in the question, we can translate this into the number of cubic centimeters of syrup needed, and finally into the number of ladles needed.

Now on to the details.

First, let the number of ounces of syrup needed to mutate a green, red, blue, or yellow Grundo be denoted by g, r, b, or y, respectively. From the first paragraph, we get the equations

    r = 3g
    r = 2y
    b = (1/2)(3 + r)
This is only three equations for four variables, so we need one more piece of information. This comes from the last sentence of the fifth paragraph, which says that "a green Grundo, after drinking two ounces, mutates instantly". This gives us g = 2, and then from the equations above, we get r = 6, b = 4.5, and y = 3.

Next, we need to figure out the number of Grundos of each color, which we'll denote by G, R, B, and Y. If we follow the directions in the second through fourth paragraphs, we can find out the squares that each color of Grundo occupies. They are... Green: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Red: 9, 10, 12, 14, 18, 20, 24, 26, 30, 36, 38, 44, 48. Blue: 4, 6, 8, 16, 28, 32, 40, 42. Yellow: 1, 15, 21, 22, 25, 27, 33, 34, 35, 39, 45, 46, 49. Counting tells us G = 15, R = 13, B = 8, and Y = 13.

Now that we know the number of Grundos of each color and the number of ounces of syrup each color of Grundo needs to mutate, we can figure out the total number of ounces needed to mutate all the Grundos The number of ounces needed to mutate all the green Grundos is equal to the number of green Grundos times the number of ounces needed to mutate each one, and similarly for red, blue, and yellow Grundos Thus, the total number of ounces needed is

    Gg + Rr + Bb + Yy
    = 15 * 2 + 13 * 6 + 8 * 4.5 + 13 * 3 = 183 oz.

The final steps are to convert this number into cubic centimeters, and then into ladles of syrup. The question sentence tells us that "Mutox Syrup weighs 0.2 ounces per cubic centimeter". This means that the number of cubic centimeters needed is

    (183 oz) / (0.2 oz/cm3) = 915 cm3.
Now we just need to know the number of cubic centimeters of syrup in each ladle that is filled to a depth of 2 cm. As far as we can tell, we need calculus for this. (If somebody knows an easy way to find the answer without resorting to calculus, please let one of us know.) Now please refer to the following diagram, which is a side view of the ladle, with the dotted line being the top of the syrup:

 

We let the vertical axis (axis of symmetry) be the x-axis, with the center of the hemisphere being x = 0 and the bottom of the hemisphere being x = 3. Also, we let the radial axis be the y-axis, with the x-axis being y = 0. Because the ladle is a hemisphere, the cross-section is a circle, which means the curve in the diagram has equation
    x2 + y2 = 32   ⇒   y2 = 9 - x2

We integrate along the x-axis (using the formula for the area of a circle) to get the volume of syrup in a ladle filled to a depth of 2 cm. Here is the calculation:

Finally, we can calculate the number of ladles needed as

    (915 cm3) / (29.3215314335 cm3/ladle)
    = 31.2057370562 ladles.
Rounded off to three decimal places, the correct answer is 31.206 ladles. [Phew!]

Extra. Because Neopets couldn't figure out the exact correct answer on this one, they ended up accepting all answers of the form 31.2xx. Thus, many people who got the exact correct answer were denied their rightful gold trophies. [Sigh.]

Round 36. Answer: 10850 Calories

This one's just plain annoying. You have to either go into the Golden Dubloon and write down the names of all the foods and drinks on the menu, or else search the Internet for a list of all the foods and drinks on the menu. And then you have to count up the letters and do the calculation as instructed. The lists of foods and drinks on the menu are given in the next couple of paragraphs. There are 368 letters in all the foods and 110 letters in all the drinks, for a total of 368 * 25 + 110 * 15 = 10850 Calories.

Here is the list of foods on the menu: Double Stuffed Guppy, Crusty Clam Surprise, Tropical Breeze, Oyster Obsession, Tomato Cannon Ball, Shiver Me Shrimp, Caesar Salad, Baby Bloater, Headless Horsefish, Slithering Squid Surprise, Barnacle Bills Belt Busting Burger, Bilge Rat Madeira, Loretta Fontaines Perfect Pizza, Capn Threelegs Cutlass Crusade, Our Famous Krawk Pie, Pinanna Paradise, Blueberry And Oyster Ice Cream, Squid On A Stick, Joy Fun Pops, Kraku Berry Cove, Forbidden Plunder.

Here is the list of drinks on the menu: Grog Light, Grog in Four Fruity Flavours, Assorted Bottles of Pop, Man O War, Keel Haul, Cannon Fodder, Walk The Plank, Land Lubber, Hogshead.

Extra. The qualification of foods and drinks on the menu is important here, because there is one food menu item (Joy Fun Pops) and two drink menu items (Grog in Four Fruity Flavours and Assorted Bottles of Pop) that actually represent multiple foods or drinks. Specifically, Joy Fun Pops represents the foods Chocolate Fun Pop, Berry Fun Pop, and Mocha Fun Pop; Grog in Four Fruity Flavours represents the drinks Tchea Grog, ErgyFruit Grog, Bomberry Grog, and Kraku Berry Grog; Assorted Bottles of Pop represents the drinks Raspberry Pop, Lemon Pop, Grape Pop, and Cherry Pop.

Round 37. Answer: purple

Skip down to the question, which asks for the driver's hair color. Then type "Thyassa" in the search box and hit enter. Voila! You see a purple chia!

Round 38. Answer: 3 apples

This puzzle is a long string of 0's and 1's, which we have to figure out for ourselves how to decode. There are many things we could try, but hopefully, sooner or later, we'll hit upon the right idea. In this case, we have to break up the string of 0's and 1's into 8-bit groups (each 0 or 1 is one bit), and translate each group into the corresponding ASCII character. (ASCII is the American Standard Code for Information Interchange, and tables of mappings between characters and their values are easily found on the Internet.) When we do this, we get the string "Bobby the Chia has one apple. Suki the Usul gives Bobby two apples. How many apples does Bobby have in total?\0" (The weird '\0' character at the end just indicates the end of the string.) Now we just have to answer the question, which is easy: 1 apple + 2 apples = 3 apples.

Round 39. Neopets answer: 99   Our answer: No solution, because the question statement has an error

Based on the official answer, we believe that Neopets made an error in the question statement. (If there isn't any error, and we're just not seeing the pattern, please let one of us know.) We believe that the age of Doctor Sloth in the second row should be 89 instead of 88. Then the pattern would be: the answer in each row equals twice the number of letters in the phrase minus the number of letters in the last word of the phrase.

Specifically, the numbers of letters in each phrase are 34, 48, 39, and 52, respectively, and the numbers of letters in the last word of each phrase are 4, 7, 7, and 5, respectively. For the first row, 34 * 2 - 4 = 64; for the second row, 48 * 2 - 7 = 89; for the third row, 39 * 2 - 7 = 71; for the fourth row, 52 * 2 - 5 = 99.

Round 40. Neopets answer: 278116   Our answer: No solution, because the question statement has an error

In two consecutive conundrums, Neopets makes an error in the question statement. In this one, the solution is supposed to be that if all spaces are removed from the question statement, then the strings "two", "seven", "eight", "one", "one", and "six" can be found within the long string of concatenated words. The "two" comes from "bright woolen"; the "seven" comes from "pass even"; the "eight" supposedly comes from "the light" (but that actually gives "elight"); the first "one" comes from "shone"; the second "one" comes from "monotone"; the "six" comes from "six".

Anyway, as you can clearly see, Neopets never managed to spell "eight". They should have used a word like "weigh" or "sleigh" followed by a word that begins with the letter "t". Alternatively, they could have simply used one word, as in "height" or "freight".

Round 41. Answer: 126

This conundrum by itself is quite difficult, but is much easier for someone who has previously solved rounds 38 and 26. The solution below will assume that you have already read the solutions to rounds 38 and 26.

First of all, the Christmas tree shape doesn't have any special significance. The puzzle is just simply a long string of 0's and 1's. Immediately we should be reminded of round 38, where the puzzle was also of the same form. Now, if we immediately tried the same translation as in round 38, then we would get a bunch of nonsensical garbage. However, if we took a closer look at the actual strings of 0's and 1's, we might notice something very interesting. See, the final 8 bits (also known as a byte) in the string for round 38 were all 0's, and they translated into the character '\0' denoting the end of the string. However, in the string for this round, it is the first 8 bits that are all 0's. Thus, the bits in this string may have been reversed just to confuse us! If we reverse the string and then do the translation, we actually get something that might be a clue (although it makes no sense):

    FN JNSFM JTN.  FFT RVPFS JTBHW!
Here's where the experience with round 26 really comes in handy. If we stare at this string for a really long time, we might just notice that the last four letters are "TBHW", which is the reverse of "whbt", the first four letters from the clue string of round 26! Therefore, if we reverse this string (letter by letter this time, instead of bit by bit), we might be able to apply the translation from round 26! If we try this, we will arrive at:
    WHATISFOURTEENTIMESNINE
The question is, "What is fourteen times nine?" And the answer is obviously 126.

Round 42. Answer: Turmaculus

There are a few clues here that might lead us down the right track. One is the fact that Idylwylde is a knight; another is the fact that the grid is 8x8, the same size as a chess board; still another is the fact that, when all the numbers in the poem are added together, we get 63, which happens to be the number of lettered squares in the grid. All these clues add up to tell us that we probably have to make knight moves, starting from the right bottom square and landing on every square once.

Now, there are many ways to cover the board using knight moves, so Neopets has narrowed it down for us a little more. In the poem, they have told us the lengths of the words we should expect to find. If we try hard enough, we can eventually figure out the correct path through the board (see the "Extra" paragraph for the full path). The resulting clue will say, "I like a nice Kadoatie with a side of Turtums. I live in Meridell. What is my name?" The answer, of course, is the Turmaculus.

Extra. Knight moves are difficult to describe. We will describe them using directions. The eight directions a knight can jump in will be denoted NNE (north-northeast), ENE (east-northeast), ESE (east-southeast), SSE (south-southeast), SSW (south-southwest), WSW (west-southwest), WNW (west-northwest), and NNW (north-northwest). Now here are the 63 knight moves: WNW, ENE, NNW, NNE, WNW, SSW, ENE, SSE, WNW, NNE, SSE, WSW, ESE, NNW, NNE, WSW, WNW, SSW, SSE, ESE, SSE, WNW, NNE, SSE, WSW, NNW, NNW, ESE, SSW, WSW, NNE, WNW, SSE, ESE, NNE, WNW, WNW, NNE, ESE, NNE, WSW, WNW, SSE, ESE, NNE, WNW, WSW, SSE, SSW, ESE, WSW, NNE, SSE, WNW, NNE, NNW, NNE, ESE, SSE, ESE, SSE, WSW, NNE.

Round 43. Answer: Boris the Kacheek

In this puzzle, a clue sentence is hidden in the numbers held up by chias. In fact, each number represents one letter, and each row of chias represents one word. One way that we might have guessed this is the case, is by the fact that all the numbers range from 1 to 26. In order to decipher the code, we use the standard technique of counting letter frequencies (or in this case number frequencies). This technique has already been demonstrated in the solution to Round 26, so we need not go into the details here. In this puzzle, it turns out that the numbers 1, 5, 7, 12, and 21 appear the most, and these correspond to the common letters t, i, n, o, and e, respectively. The whole clue sentence turns out to say, "The winner of today's bingo competition is Boris the Kacheek."

Round 44. Answer: Fyora

This one is quite difficult. We somehow have to come up with the idea (more or less out of the blue) to arrange the letters in a grid with 8 rows and 7 columns. The letters are used to fill the grid one column at a time from left to right, and within each column from top to bottom. If we do this, the resuling grid is as shown:

    w h a t a w r
    q f g n s i s
    h g t h e j m
    n a m e r e a
    f g f o f m z
    p l l t h e d
    o f a e r i e
    c q u e e n i
In bold are the words hidden in the rows of the grid. Reading off the words gives "What is the name of the Faerie queen?" The answer, of course, is Fyora.

Round 45. Answer: 1051 damage

In the Battledome, a Neopet can only use 2 weapons per round. Therefore, 10 rounds of combat is equivalent to 20 total uses of a weapon (with the restriction that no weapon can be used more than 10 times). Now, we basically have to list what each weapon can do in the first 10 uses, and pick the best 20 damage amounts out of all the lists.

In the first 10 uses... the Wand of the Air Faerie does 47, 46, 45, 44, 43, 42, 41, 40, 39, and 38 damage; the Asparachucks do 50, 49, 48, 47, 46, 45, 44, 43, 42, and 41 damage; the Yo-Yo of Death, if Heran gets super lucky, does 106 (28+27+26+25), 90 (24+23+22+21), 74 (20+19+18+17), 58 (16+15+14+13), 42 (12+11+10+9), 26 (8+7+6+5), 10 (4+3+2+1), 0, 0, and 0 damage.

The best 20 damage amounts turn out to be all the numbers that are 42 damage or more. Heran uses the Wand of the Air Faerie 6 times for 47 + 46 + 45 + 44 + 43 + 42 = 267 damage. He uses the Asparachucks 9 times for 50 + 49 + 48 + 47 + 46 + 45 + 44 + 43 + 42 = 414 damage. Finally, he uses the Yo-yo of Death 5 times for 106 + 90 + 74 + 58 + 42 = 370 damage. The total damage that Heran can do to the Jelly Chia in 10 rounds is thus 267 + 414 + 370 = 1051 damage. As a final sanity check, we must make sure that the Jelly Chia can actually withstand 1051 damage, which it can, since it has 3800 hit points.

Round 46. Answer: 335.41 meters

If you've studied basic geometry, then you should have no trouble with this question. Using pencil and paper to draw out the path that the farmer takes is highly recommended. Also, always keep in mind that we want to try to keep the farmer moving away from his starting point as much as possible.

First, the farmer walks 100 + 50 + 50 = 200 meters away from his starting point. Then, he turns 90 degrees (for this first turn, it doesn't matter which way he turns) and walks 100 meters.

Then he turns 90 degrees again. For this second turn, he should turn in the opposite direction from his first turn, because this will keep him moving away from his starting point. After the second turn, he should be walking in the same direction as when he first started walking, and he walks an additional 100 meters in this direction. The total amount he has walked in this direction is thus 200 + 100 = 300 meters.

Then he turns 90 degrees for a third time. For this third turn, he should again turn in the opposite direction from his second turn, because this will keep him moving away from his starting point. After the third turn, he should be walking in the same direction as after the first turn, and he walks an additional 50 meters in this direction. The total amount he has walked in this direction is thus 100 + 50 = 150 meters.

Finally, the farmer turns toward his starting point and walks 30 meters closer. We shouldn't care about this part, because we want to know his farthest distance from his starting point.

Ignoring the last 30 meters, we find that the farmer has walked a total of 300 meters away from his starting point in one direction, and 150 meters away from his starting point in a perpendicular direction. Using the Pythagorean theorem, we can calculate the total distance between him and his starting point as the square root of (3002 + 1502), which comes out to about 335.41 meters.

Round 47. Answer: 123 points

In Spell-or-Starve, more letters in a word are worth disproportionately more points, so what we need to figure out are the longest word accepted by the game that has all vowels, and also the longest word accepted by the game that has only one consonant (because the wildcard could be a consonant). It turns out the longest word accepted by the game that has all vowels is 3 letters long (example: eau), and the longest word accepted by the game that has only one consonant is 5 letters long (example: queue). Sadly for Channy, words of 1 letter are worth 1 point each, while words of 3 letters are only worth 3 points each, so he gains no extra points for repeatedly using "eau" as opposed to repeatedly using "a". Oh, well. [Shrug.] However, a word of 5 letters is worth 15 points, and since Channy gets to double one word's score, it might as well be this word's score. Now, in hard mode, the board is 10x10 in size, which is 100 tiles. There are 2 blocks, which leaves Channy with 98 tiles to make words with. He uses up 5 tiles to make "queue", including the wildcard tile and the double-word-score tile. This leaves 93 tiles, each of which earn him only 1 point. Thus, the maximum number of points Channy can earn is 15 * 2 + 93 = 123 points.

Round 48. Answer: 150 words

This is by far the most insanely annoying Lenny Conundrum in the history of the contest!!! There are two difficult and time-consuming steps to solving this puzzle.

The first step is you must find all the English and Neopets words that can be spelled with the letters T, E, S, P, L, N, and A. Finding the list of English words is basically impossible by hand, but it's doable if you know how to write a computer program to do it. You can download a dictionary file of English words from the Internet. And then, you need to write a program to read through every word in the file and check whether it can be spelled using the seven given letters. Looking for Neopets words is even more annoying. You must scour the Neopets site for a couple of hours, looking for Neopets words that can be spelled with the seven letters. In the end, to your dismay, you would find none.

The second step is even more annoying than the first. If you've ever played Spell-or-Starve, you'll know that the dictionary is very incomplete. There are many common English words that are not accepted by the game. Therefore, after making the list of words that can be spelled with the seven letters, you now have to try every single word in Spell-or-Starve to see if the game actually accepts it! This involves many hours of repeatedly restarting the game on hard mode, looking for a good board (meaning that the seven letters are in abundance and appear close together). When you find a good board, you try to check off as many words as you can in the five minutes before the round ends. Sometimes, you have to play a few words before the board becomes good. Toward the end, when only a handful of unchecked long words are left, you have to play a board with the specific purpose of checking off one particular word.

If you labor through this whole process, you'll be rewarded with the final list of words, which you count up to get the answer, 150 words. Of course, since we've already done all the hard work, we can just give you this list! Here it goes... a, ale, ales, an, ant, antes, ants, ape, apes, apt, as, asp, aspen, at, ate, east, eat, eats, etnas, la, lane, lanes, lap, laps, lapse, last, late, lean, leans, leant, leap, leaps, leapt, least, lepta, lens, lent, lest, let, lets, nap, nape, napes, naps, neap, neaps, neat, nest, net, nets, pa, pal, pale, pales, palest, pals, pan, pane, panel, panels, panes, pans, pant, pants, pas, past, paste, pastel, pat, pate, paten, patens, pates, pats, pea, peal, peals, peas, peat, pelt, pelts, pen, penal, pens, pent, pest, pet, petal, petals, pets, plan, plane, planes, planet, planets, plans, plant, plants, plate, platen, platens, plates, plats, plea, pleas, pleat, pleats, sale, salt, sane, sap, sat, sea, seal, seat, sent, sepal, septa, set, slant, slap, slat, slate, slept, snap, spa, span, spat, spate, spelt, spent, splat, stale, staple, steal, step, tale, tales, tan, tans, tap, tape, tapes, taps, tea, teal, teals, teas, ten, tens. [Phew!]

Extra. Our opinion is that Neopets picked the wrong combination of seven letters for this problem, because the resulting answer was 150, a very round number. Due to this, we believe that most of the people who got gold trophies probably just had a lucky guess instead of actually doing the work to figure it out.

Round 49. Answer: Abracadabra

This one's a simple letter replacement puzzle. Take the three ancient "words" and smush them together into one big word, and then replace M with A, replace R with B, replace N with R, etc. It might be rather difficult if you aren't able to guess Abracadabra to begin with, though!

Round 50. Neopets answer: not yet released   Our answer: 156 positions

Before we discuss the solution, let's come up with some notation to talk about the game board. The size of the game board is 10x10. We're going to name the rows A through J from top to bottom, and the columns 1 through 10 from left to right. So the top left corner is A1 and the bottom right corner is J10. We recommend you draw a picture of the board and label the sides with A-J and 1-10 before continuing.

We're going to assume that Vex is playing the Darigan pieces in this game. (We make this assumption "without loss of generality", because even if Vex were playing the Meridell pieces, the problem would be entirely symmetrical and the answer would be the same.) Now, one thing we must not forget is that the Yellow Knight begins the game with Meridell pieces on squares E5 and F6, so that Vex can never place Darigan pieces on those squares.

There are four possible orientations for a winning position of five pieces: vertical, horizontal, diagonal from top left to bottom right, and diagonal from top right to bottom left. Below, we'll proceed through each one of these four possibilities and calculate the number of possible positions for the five winning pieces. At the end, we'll add up the four numbers we get to arrive at our answer.

First possibility: vertical positions. Let's just look at one column, say column 1. We can have a winning vertical position in column 1 in six ways, from A1-B1-C1-D1-E1 at the very top to F1-G1-H1-I1-J1 at the very bottom. The same goes for columns 2, 3, 4, 7, 8, 9, and 10 (columns 5 and 6 are special). So, for these 8 columns 1-4 and 7-10, there are 6 winning positions on each column, for a total of 8 * 6 = 48 winning positions. Now, in column 5, there is a Meridell piece that starts on square E5, which eliminates all the possible winning positions for the Darigan pieces except for the single position F5-G5-H5-I5-J5. Similarly, in column 6, the only winning position for the Darigan pieces is A6-B6-C6-D6-E6. Thus, columns 5 and 6 together contribute only two more possible winning vertical lines, for a grand total of 48 + 2 = 50 winning vertical positions.

Second possibility: horizontal position. This case is symmetrical to the first case, so there are 50 winning horizontal positions as well.

Third possibility: diagonal position going from top left to bottom right. Each of the squares in column 1 and each of the squares in row A are the top left square of some diagonal line of squares on the board. For each of these 19 diagonals, we have to see how many winning positions are in the diagonal. The diagonals that start with J1, I1, H1, and G1 don't contain any winning positions because they're too short. The diagonals that start with F1, E1, D1, C1, and B1 contain respectively 1, 2, 3, 4, and 5 winning positions, because each diagonal is one square longer than the previous. (If you're a bit confused, consider the diagonal starting with E1. It contains two winning positions, namely E1-F2-G3-H4-I5 and F2-G3-H4-I5-J6.) The diagonal that starts with A1, which is the longest diagonal, doesn't actually contain any winning positions for the Darigan pieces. This is because of the Meridell pieces that begin the game on squares E5 and F6. The diagonals that start with A2, A3, A4, A5, and A6 contain respectively 5, 4, 3, 2, and 1 winning positions. Finally, the diagonals that start with A7, A8, A9, and A10 are once again too short to contain any solutions. Adding up the numbers we've counted, there are a total of 1 + 2 + 3 + 4 + 5 + 5 + 4 + 3 + 2 + 1 = 30 winning diagonal positions going from top left to bottom right.

Fourth possibility: diagonal position going from top right to bottom left. There are 19 diagonals in this direction as well. The diagonals that start with A1, A2, A3, A4, G10, H10, I10, and J10 don't contain any winning positions because they're too short. The diagonals that start with A5, A6, A7 and A8 contain respectively 1, 2, 3, and 4 winning positions. The same goes for the diagonals that start with F10, E10, D10, and C10. The diagonals that start with A9 and B10 don't contain any solutions for the Darigan pieces because of the Meridell pieces on E5 and F6. Finally, the diagonal that starts with A10, which is the longest diagonal, has 6 winning positions. Adding up the numbers we've counted, there are a total of 1 + 2 + 3 + 4 + 6 + 4 + 3 + 2 + 1 = 26 winning diagonal positions going from top right to bottom left.

We get the final answer by adding up the answers from the four cases, which comes to 50 + 50 + 30 + 26 = 156 positions.

Extra. Actually, this problem is more easily solved by counting all the winning positions without regard to the starting Meridell pieces on E5 and F6, and then subtracting off the winning positions that require a piece on either E5 or F6. The final calculation for this solution method would be 60 + 60 + 36 + 36 - 36 = 156 positions. This solution method might be harder to explain, however, and that's why we gave the solution above.


Phew! Well, that's it! Kudos to you if you managed to read through our article in one sitting. If you have any questions about our solutions, feel free to Neomail either one of us (qg_jinn/melibunni). We probably won't be writing solutions to any future Lenny Conundrums; it simply takes too much time. In addition, Neopets has begun to put semi-solutions in the news to the previous conundrum upon releasing a new conundrum (although these are sometimes very incomplete). For anyone inclined to continue our work, we suggest writing up solutions for every 10 rounds instead of every 25 as we have. 25 rounds is just too long of a wait for the readers, and with 10 rounds, the work involved in writing the solutions should be much more manageable. Well, thank you for reading and good luck to everyone on future puzzles!


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