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Neopia's Fill in the Blank News Source | 16th day of Collecting, Yr 19
The Neopian Times Week 26 > Articles > Belated Solutions to the First 25 Lenny Conundrums

Belated Solutions to the First 25 Lenny Conundrums

by qg_jinn

Have you ever been stumped by a Lenny Conundrum? Well if you have, and are curious how to solve the problem (not just what the answer is, but how to arrive at the answer), I'm here to help!

This article will discuss the solutions to the first 25 Lenny Conundrums. Some questions do not require discussion (e.g. anagrams), but other require a bit of mathematical reasoning. It is clear that the mathematical sophistication of the questions has slowly increased over time. The math is not deep, and should certainly be accessible to anyone in high school. I have tried to keep the discussion as simple as possible, but in some places, I have had to make certain assumptions about background knowledge. Also, in a few solutions, I have included an "Extra" paragraph for those who want to know more.

In looking through the old conundrums, I have even found a couple of *cough*mistakes*cough* in NeoPets' official answers (see rounds 16 and 24)! Without further ado, let's get started...

Round 1. Answer: father

When the blue Nimmo says "my father's son", he must be referring to himself, because "brothers and sisters [he] has none". Thus, when he points to the green Nimmo, he is really saying "this Nimmo's father is myself"!

Round 2. Answer: 200 pounds

If a Skeith weighs 100 pounds plus half his weight, then the 100 pounds must be the other half of his weight. Since 100 pounds is half his weight, his full weight is 200 pounds.

Round 3. Answer: 34 years old

This question tests knowledge of Mulvinn the Wocky from the Neopedia, so it's not really a conundrum.

Round 4. Answer: N T

The sequence consists of all letters written using only straight lines.

Round 5. Answer: The digits are in alphabetical order of their spellings.

No discussion necessary.

Round 6. Answer: 120 times

The digit 3 appears as the hundreds digit of all numbers from 300 to 399 (100 occurrences). The digit 3 appears as the tens digit of all number from 330 to 339 (10 occurrences). Finally, the digit 3 appears as the ones digit of the numbers 303, 313, ..., 393 (10 occurrences). In total, the number of times you have to write the digit 3 when writing all the numbers from 300 to 400 is 100 + 10 + 10 = 120 times.

Round 7. Answer: "four" or "0"

A correct answer must represent a number, and also must take the same number of letters to write as the number that it denotes. Clearly the answers "four" and "0" satisfy these requirements. There are actually other good answers, such as "7 letters", but "four" and "0" seem to be the most pleasing because they are short. Can you find any other good answers?

Round 8. Answer: S

The sequence consists of the first letters of the words in the question: "What's the next letter in the following sequence?"

Round 9. Answer: 2616 legs

Note that all pets, petpets, and baby petpets have 4 legs each, so we'll count the total number of animals and multiply by 4 at the end. To make the equations below fit on one line, I will use the abbreviations M for Moehog, P for Puppyblew, and BP for baby Puppyblew.

First, there are 6 Moehogs. Then there are

(6 M) * (6 baskets per M) * (6 P per basket) = 216 Puppyblews.
Finally, there are
(216 P) * (2 BP per P) = 432 baby Puppyblews.
In total, there are
6 M + 216 P + 432 BP = 654 animals.
The number of legs is then
(654 animals) * (4 legs per animal) = 2616 legs.

Round 10. Answer: 19 years old

This one is best done by writing and solving equations. Let Siya's age be s, the age of her brother be b, and the age of her father be f. Since Siya is 9 years older than her brother and 25 years younger than her father, we get

b = s - 9 [equation 1],
f = s + 25 [equation 2].
The last statement is a bit tricky. In 7 years, Siya's father will be age f + 7, and Siya's brother will be age b + 7. Therefore, the statement translates to
(f + 7) = 3(b + 7) => 3b - f = -14 [equation 3].
Now we substitute equations 1 and 2 into equation 3.
3(s - 9) - (s + 25) = -14 => 2s = 38 => s = 19.
So we conclude that Siya must be 19 years old. (This means her brother is 10 years old and her father is 44 years old.)

Round 11. Answer: 2500 NP

The total fraction of Brucey's winnings that he put in the bank, invested in shares, and spent on chocolate neggs is

1/5 + 1/2 + 1/10 = 2/10 + 5/10 + 1/10 = 8/10 = 4/5.
Thus at the end of the day, he had 1/5 of his money left, which was 500 NP. So his total winnings must have been 2500 NP.

Round 12. Answer: 60 sandwiches

If you haven't taken a lot of math, here is one way you could reason this one out. The only numbers of sandwiches that can be split equally among 2 Skeiths with none left over are in the sequence 2, 4, 6, ... (multiples of 2). Similarly, the answer must also be in the sequence 3, 6, 9, ..., and the sequence 4, 8, 12, ..., and the sequence 5, 10, 15, ..., and the sequence 6, 12, 18, ... If you were to write out these sequences, you would see that the smallest positive number that appears in all the sequences is 60.

As a side note, some people might argue that 0 sandwiches is the correct answer, but the question says Mrs. Doyle made a "big, big pile of sandwiches", so the answer must be a positive number.

Extra. For those of you who know what the term least common multiple (LCM) means, the question is really just asking for the LCM of the numbers 2, 3, 4, 5, 6. The quick way to find it is to factor each number into primes: 2 = 2, 3 = 3, 4 = 2^2, 5 = 5, 6 = 2*3, and then multiply together the highest power of each prime that appears in any number in the list. In this case, the answer would be LCM = (2^2) * 3 * 5 = 60.

Round 13. Answer: PTERATTACK

No discussion necessary.

Round 14. Answer: TIGERSQUASH

No discussion necessary.

Round 15. Answer: 20 B.N.

This one is best done by writing and solving equations. Let a and b be the years in which Galat-Ra was born and in which he died. Let c and d be the years in which Hephaes-Ru was born and in which he died. We use negative numbers to denote the years before Neopia. Since Galat-Ra died 120 years after Hephaes-Ru was born,

b - c = 120 [equation 1].
Since their combined ages when they died was 100 years old,
(b - a) + (d - c) = 100 [equation 2].
Finally, Hephaes-Ru died in 40 B.N., which means
d = -40 [equation 3].
The question asks us to find a. To get the answer, we can rewrite equation 2 and then substitute the information from equations 1 and 3:
(b - a) + (d - c) = 100 => a = d + (b - c) - 100 = -40 + 120 - 100 = -20.
So we conclude that Galat-Ra was born in the year 20 B.N.

Round 16. NeoPets answer: 3 My answer: 2 or 3

You can reason this one out without equations, but equations make the thought process much less confusing (apparently NeoPets was confused), so I will explain the solution using equations. Let g be the number of green grundos, p be the number of purple grundos, r be the number of red grundos, and x be the number of grundos that are any other colors. The phrase "all but two of the grundos chasing Barca were green" can be restated as "the number of purple, red, and other-color grundos is two". Using this type of restatement, we can write down the following three equations:

p + r + x = 2,
r + g + x = 2,
g + p + x = 2.
Although we have three equations in four unknowns, there are not an infinite number of solutions because the unknowns can only have values 0, 1, or 2. The easiest way to solve the problem is to consider three cases separately.

The first case is when x = 0. Then solving the equations gives the solution g = p = r = 1. This means there is 1 green, 1 purple, and 1 red grundo, for a total of 3 grundos. This is the NeoPets answer.

The second case is when x = 1. Then solving the equations gives the solution g = p = r = 1/2. This solution is not valid, because half of a grundo would be dead, and not be able to chase Senator Barca.

The third case is when x = 2. Then solving the equations gives the solution g = p = r = 0. This means there are 2 grundos, neither of which are green, purple, or red. (For example, we could have two blue grundos, or we could have one blue and one yellow.) This is the answer that NeoPets missed.

Round 17. Answer: SPORKLE LEG

No discussion necessary.

Round 18. Answer: 21184 NP

This question just requires us to perform a series of calculations beginning with 5000 NP. (There is one confusing part, which is whether or not the Wheel of Excitement costs 100 NP to play before Heermeedjet gets the 20% increase. Only by working backwards from the official NeoPets answer was I able to deduce that the wheel in this question does not cost any money to play.) The first two calculations are:

5000 + (20% * 5000) = 6000,
6000 - (20% * 6000) = 4800.
Now we must add the interest for 10 days. 1% interest rounded down is just the bank balance without its last two digits. The balance changes as follows: 4800 -> 4848 -> 4896 -> 4944 -> 4993 -> 5042 -> 5092 -> 5142 -> 5193 -> 5244 -> 5296. Finally, the win in the Poogle Race pays the "odds" times the amount wagered:
5296 + (3:1) * 5296 = 21184.
So Heermeedjet ends up with 21184 NP.

Extra. Incidentally, if you subtracted the 100 NP cost of playing the Wheel of Excitement before performing the rest of the calculations, your answer would be 20760 NP. Personally, I feel this answer should have been accepted as well, but I'm not the one to make that call!

Round 19. Answer: SPACE USUKI

No discussion necessary.

Round 20. Answer: 28 red coins

This one is best done by writing and solving equations. Let r, g, p, and s denote the equivalent values in NP of the red, green, purple, and silver coins, respectively. Then the following equations can be written from the information in the question:

3r + 2 = g,
3g + 5 = p,
11r + 1 = p,
s = 4p.
If you understand how the equations come about, then you probably also know how to solve them, so I'm going to save space by omitting the process of solving the equations. The solution is r = 5, g = 17, p = 56, s = 224. If one pays for an 84 NP loaf of bread using a silver coin, then one can expect 224 - 84 = 140 NP worth of change. If the shopkeeper gives change all in red coins, then the change will consist of (140 NP) / (5 NP per red coin) = 28 red coins.

Round 21. Answer: 32 pyramids

This question requires knowledge of how to calculate the volume of a sphere and of a square pyramid. If you know the formulas, then you just have to find the total amount of clay available (volume of clay sphere) and the amount of clay it takes to make one little pyramid (volume of little pyramid), and then divide to see how many little pyramids can be made from the ball of clay. Here are the calculations:

volume of sphere = (4/3)R^3 = (4/3)(3.14...)(10^3) = 4188.79...
volume of pyramid = (1/3)(base area)(height) = (1/3)(8 * 8)(6) = 128
number of pyramids = 4188.79... / 128 = 32.72...
The number of whole pyramids that can be made is 32.

Round 22. Answer: 3,000,000 shops

This question requires an educated guess and a lot of luck!

Round 23. Answer: 6 flags

With 1260 NP, the sisters can buy (1260 NP) / (150 NP per ft) = 8.4 ft of cloth. This is a sheet measuring 8.4 x 5 ft. Just by drawing some pictures, one will realize that 6 flags can be made out of the cloth. First, divide it into two pieces, one 8.4 x 3 ft and the other 8.4 x 2 ft. Now 4 flags with their long sides touching will require an 8 x 3 ft sheet of cloth, so the larger piece can be made into 4 flags. Also, 2 flags with their short sides touching will require a 6 x 2 ft piece of cloth, so the smaller piece can be made into 2 flags. From drawing pictures, it seems impossible to make 7 flags, so the answer is 6 flags.

Extra. If we want to make the solution complete, we have to prove that 7 flags cannot be made. One way to prove this is as follows. The total amount of cloth that can be bought is 8.4 * 5 = 42 sq-ft. The amount of cloth that is required to make 7 flags, each 2 x 3 ft, is 7 * (2 * 3) = 42 sq-ft, the same number. Thus, if we are to make 7 flags, we must use up all the cloth. It is clear that the only way to use up all of a rectangular piece of cloth to make rectangular flags is by having the sides of the flags line up with the sides of the cloth. However one side of the cloth is 8.4 ft long, which cannot be completely used up by a combination of 2 ft and 3 ft flag sides. Since we can't use up the whole cloth, we can't make 7 flags.

Round 24. NeoPets answer: 20 seconds My answer: not enough information

NeoPets really botched this one! The question says "the initial launch velocity of the ball is 100 meters per second", but this only gives the speed at which the ball was launched. The velocity is given by both the speed and the launch angle, which NeoPets failed to provide. The ball stays in the air for different amounts of time depending on the launch angle, so the correct answer is there is not enough information.

Extra. So how did NeoPets come up with their bogus answer? Well, they simply made the assumption that the ball was launched straight up. This assumption doesn't match the question statement, which says the ball was fired "off into the distance". Nevertheless, if we follow this assumption, and also assume that the effects of air resistance are negligible, we have the following equation in the vertical direction (d = vertical distance from starting point, v = initial vertical speed, a = vertical acceleration, t = time):

d = vt + (1/2)at^2
In this case, the equation becomes
0 m = (100 m/s)t + (1/2)(-9.8 m/s^2)t^2 => t = 20.408... s
Rounding this number down, we get the NeoPets answer of 20 seconds.

Round 25. Answer: 11 bags

Here is one good way to reason this question out. There are numerous other ways. We start by asking what denominations (amount of NP in a bag) we need to fulfill requests of 1 NP, and then 2 NP, and then 3 NP, and so on, moving our way up.

To fulfill a 1 NP request, we must have a 1 NP bag.

Now to fulfill a 2 NP request, we need to add a second bag, either with 1 NP or with 2 NP in it. Adding a 2 NP bag is more efficient, since we can now fulfill a 3 NP request as well, by giving both bags. So we add a 2 NP bag, and the next request that we can't fulfill is 4 NP.

Again, we select the most efficient choice of a new bag, which is a 4 NP bag. By adding this bag, we can fulfill all requests up to 7 NP, because we were already able to fill all requests up to 3 NP, and for requests ranging from 4 NP to 7 NP, we can give the 4 NP bag plus the appropriate lower-denomination bags.

I hope you see a pattern developing. If we can fulfill all requests up to the sum of all the denominations we have currently made, and we add a new bag with one more NP than this sum, then we will once again be able to fulfill all requests up to the sum of all the denominations (including the new one).

Using this idea, we add an 8 NP bag to fulfill all requests up to 15 NP, and then add a 16 NP bag to fulfill all requests up to 31 NP, and then continue to add bags of 32, 64, 128, 256, and 512 NP. After adding the 512 NP bag, we are able to fulfill all requests up to 1023 NP, and have 60 NP left. We can put all the remaining 60 NP into a final bag, and be able to fulfill any request from 1024 NP to 1083 NP, because we can simply use the 60 NP bag plus the appropriate bags from the ones we've already made. In this process, we have made a total of 11 bags, which is the answer.

Extra. Our "answer" above is still a guess, because while we have shown that we can fulfill all requests using the 11 bags we made, we have not yet proven that it is impossible to accomplish the task with 10 or less bags! One way to prove this is as follows. I assume some knowledge of counting, in the mathematical sense. If we have 10 bags, the total number of ways to pick out some subset of those bags to fulfill a request is 10! = 10*9*...*2*1 = 1024. (Note the "!" is the mathematical symbol for the "factorial" operation.) This means that even if all the different combinations of the 10 denominations gave different sums, we can still only fulfill at most 1024 different requests. Since we must be able to fulfill 1083 different requests (from 1 NP to 1083 NP), 10 bags will not suffice. The same goes for any number of bags less than 10.

That's all folks! You now know how to solve the first 25 Lenny Conundrums. If I'm still actively playing NeoPets when the 50th Lenny Conundrum comes out, perhaps I will write another article with solutions to questions 26 through 50. However, if NeoPets continues to release Lenny Conundrums at the current the current pace, I doubt I'll ever lay eyes on the 50th Lenny Conundrum. Sigh...


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