Solutions to Lenny Conundrums 51 to 60 by cmcoury |  |
GAMES ROOM - As an avid follower of the Lenny Conundrum, I greatly enjoyed
qg_jinn's article Belated
Solutions to the First 25 Lenny Conundrums and qg_jinn and melibunni's Solutions
to Lenny Conundrums 26 to 50. I therefore found it only fitting that someone
should carry on their work, as seeing solutions to the problems is valuable,
interesting, and just plain fun. Here, I present, without further ado, solutions
to Lenny Conundrums 51 to 60, along with a bit of extra information with questions
57 and 59.
Round 51. Answer: 5832
On the rolls which are enchanted, a 6 will "come up" (appear to come up) on
a 1, 2, or 6, which is 3 out of the 6 rolls, for a probability of 1/2. On the
rolls that are not enchanted, a 6 will come up 1/6 of the time.
There are 2 possible ways to get an 11 on a roll of 2 dice: a 5 and a 6, and
a 6 and a 5. Since the dice have to total 11 for all four rounds (the calculation
for only one of the four rounds is more complex, but can be done) the odds of
an 11 on each roll can be multiplied together to get the total odds.
Roll 1 is the same as roll 2, for this purpose. There are 6 possible 11's
((1,5); (2,5); (6,5); (5,1); (5,2); (5,6)), and the odds of rolling an 11 are
6/36 = 1/6. On roll 3, the enchantment has worn off the first die but not the
second. Therefore, there are only 4 possible 11's, ((6,5); (5,1); (5,2); (5,6))
and the odds of rolling an 11 are 4/36, or 1/9. For roll 4, there is no enchantment,
and the odds of rolling an 11 are 1/18.
The odds of 11 on all four rolls, therefore, are 1/6 * 1/6 * 1/9 * 1/18, or
1/5832, giving 5832 to 1 odds.
Round 52. Answer: 11
In order to solve this, you have to know about the two shortcuts in the castle.
Detailed below is one possible solution, consisting of 11 moves. Each move will
result in your solving one anagram.
1. Move one space up and one to the left. You will be just under the left
corner of the bedroom.
2. Move three spaces up to the bedroom. This is one of the two shortcuts.
3. Move three spaces left, to the cracked space.
4. Move one space left and two down, totaling three, to the dining room.
5. Move three spaces up toward the kitchen.
6. Move two spaces up into the kitchen.
7. Move three spaces right, towards the library. Again, this is a cracked space.
8. Move one space down and two right, totaling three, towards the library.
9. Move one space up and two right, totaling three, into the library.
10. Move two spaces down to the cracked space.
11. From here, there is a secret passage to the crypt. As you have been through
all four rooms, you should be able to get into the crypt, solve the last anagram,
and escape from the castle.
Round 53. Answer: "Escape from Meridell Castle"
This one simply(?) was a long and vicious anagram; there was no particular
trick to it. Looking for words such as 'Meridell' in the anagram text was one
possible way to make the anagram simpler.
Round 54. Answer: 170 minutes
One Blumaroo can chop down 1/3 of a tree in 30 minutes, so the Blumaroo can
chop down 1/9 of a tree in 10 minutes.
At time t=0 minutes, there are 4 Blumaroos chopping down trees. Therefore,
4/9 of a tree is chopped down.
At time t=10 minutes, another Blumaroo is added, and between 10 and 20 minutes,
5/9 of a tree is chopped down.
So, we want to know the time at which 180/9 of a tree (20 trees) is chopped
down.
4/9 + 5/9 + 6/9 + ... + x/9 = 180/9, or 4 + 5 + 6 + ... + x = 180.
We discover that x = 19, and that therefore 19/9 of a tree was chopped down
in the last interval, which is the 16th interval, and that therefore the last
interval was the interval between 160 and 170 minutes. Therefore 170 minutes
were required to do the job.
Round 55. Answer: none.
As it says in the May 5 news, "The answer to the last one was ... hehe there
was no answer it was just lots of random words and numbers. Sorry about that,
we couldn't resist it :) It won't happen again!" Even the most persistent reader
could not solve this one.
Round 56. Answer: 18300.
If you view the source of this page (currently found at http://www.neopets.com/games/conundrum_feature.phtml?round=56
), you find
<!--
turn to page 2
//-->
in the comments.
Going to the URL http://www.neopets.com/games/conundrum_feature.phtml? round=56&page=2
begins a sequence of following comments in similar fashion.
For each question, go to the page of the answer to get the next question.
Page 2 - Turn to page 1+1+1+1+1 = 5.
Page 5 - The total number of leaves in all three Meri Acres Farm clovers =
3 + 4 + 12 = 19.
Page 19 - How many Chombies live in Neopia? 280,000 (at the time.)
Page 280,000 - What is the estimated cost of a wand of the Air Faerie? 35,200
NP.
Page 35,200 - How much does Jhudora's bewitched ring cost? 11,000,000 NP.
Page 11,000,000 - How many items in Neopia contain the word Usuki? multiply
that number by 100 There were 183 items at the time.
Page 18,300 - 18300 is the answer to this weeks conundrum, congratulations
:)
Round 57. Answer: 8.
There are two numerical sequences here: one for the even days, and one for
the odd days.
On odd days 1, 3, 5, 7, and 9, Mrs. Bunnikins has 1, 1, 2, 3, and 5 Mootix.
This is the Fibonacci sequence, in which a number is found by adding the two
before it. Therefore, on the eleventh day, Mrs. Bunnikins has 3 + 5 = 8 Mootix.
Interestingly enough, on the even days Mrs Bunnikins has 1, 4, 9, 16, and
25 Mootix. Should you wish to, you can find the number of Mootix in Mrs. Bunnikins'
fur on an even day by taking the day number, dividing it by two, and squaring
it. Therefore, on day 12, she would have (12/2)^2, or 36, Mootix.
Round 58. Answer: 12558.
Each of the 7 Kyrii has 2 legs, for 14 Kyrii legs.
There are 7*7, or 49 Snorkles, for 196 Snorkle legs.
There are 49*7, or 343 baby Snorkles, for 1372 baby Snorkle legs.
There are 343*8 Cootys (as _every_ Snorkle, baby or not, has seven Cootys),
or 2744 Cootys, for 10976 Cooty legs.
Adding together the four leg numbers gives the answer total of 12558.
Round 59. Answer: Chokato
This is a seven letter word in trinary. Taking * as 2,
10*00 = 81*1 + 9*2 = 99
10*1* = 81 + 9*2 + 3 + 2 = 104
11010 = 81 + 27 + 3 = 111
10*** = 81 + 9*2 + 3*2 + 1*2 = 107
101*1 = 81 + 9 + 3*2 + 1 = 97
110** = 81 + 27 + 3*2 + 1*2 = 116
11010 = 81 + 27 + 3 = 111
These are the ASCII codes for the letters c, h, o, k, a, t, o.
Alternatively, the puzzle can be solved without this knowledge with the clue
"Oops it seems that Neopet V2 has his binary in a twist." Take the resulting
values and "twist" them - try and make them look like letters. By subtracting
96 from each value, the results are (3, 8, 15, 11, 1, 20, 15.) These correspond
to the letters c, h, o, k, a, t, o.
Round 60. Answer: 0 NSU
Zorlon and the Fuzzles are in space. Sound waves do not carry in space, and
therefore the NSU detector cannot register the roar of the red fuzzle.
There you have it. Special thanks to qg_jinn and melibunni for their work,
and especially to qg_jinn, without whom this article would not have been possible.
Feel free to ask me (cmcoury)
any questions you might have, and good luck!
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