Round 264

Date Released: May 29th, 2008.
Date Rewarded: June 5th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

What is the missing word in this sequence?
allowed
here
entry
________
you
here
you

Please enter only a single word, with no other information. Otherwise, your entry will be disqualified.

Prize


Top Secret Skeith Stuff

Click to show/hide solution

Check out the Neoboards and the description of each board:

Avatars/Neosignatures: ... Item pools are a scam and are not allowed!
Battledome: ... Anything and everything Battledome is in here.
Beauty Contest: ... regarding YOUR OWN Beauty Contest entry!)
Customisation: ... that would be embarassing.
Evil Things and Monster Sightings: ... This board is for you.
Fan Clubs: ... Discuss it here!
Games: ... Yup, the Games board is for you.

Each word in the sequence is the last word of each board's description. Therefore, answer should be embarassing. Note that the answer is a common misspelling word.

Results: 3983 people guessed the correct answer earning themselves 503 NP each.

Round 263

Date Released: May 22nd, 2008.
Date Rewarded: May 29th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

What is the next number in this sequence?
1
721
727
5767
5791
5792
___

Please enter only a single number, with no other information. Otherwise, your entry will be disqualified.

Prize


Megachilli Hot Dog

Click to show/hide solution

First, you subtract the number to the previous number (assume that the number before the first one is 0). The sequence becomes:

1 721 727 5767 5791 5792 -> 1 720 6 5040 24 1.

We realize that 1 = 0! = 1!, 6 = 3!, 24 = 4!, 720 = 6!, 5040 = 7!, where ! is factorial [1].

Therefore, the original sequence can be decomposed at:
1 = 1! or (0!)
721 = 1! + 6!
727 = 1! + 6! + 3!
5767 = 1! + 6! + 3! + 7!
5791 = 1! + 6! + 3! + 7! + 4!
5792 = 1! + 6! + 3! + 7! + 4! + 1!
next number = 1! + 6! + 3! + 7! + 4! + 1! + n!

So, the sequence becomes:

1! (or 0!) 6! 3! 7! 4! 1! (or 0!) n!

Temporarily ignore the factorial sign and 'or 0!' part, we can rewrite the sequence as:

1 6 3 7 4 1 n

The sequence is the number of letters in each word of the first sentence of the puzzle.

"A Chomby was walking down a path one day..."

Therefore, n is 4. The next number should be: 1! + 6! + 3! + 7! + 4! + 1! + 4! = 5816.

Results: 1167 people guessed the correct answer earning themselves 1714 NP each.

References:
[1] Factorial

Round 262

Date Released: May 15th, 2008.
Date Rewarded: May 22nd, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

How many people did not properly follow the instructions on last week's Lenny Conundrum, and submitted more than a single word when the Conundrum specifically required a one-word answer?

Please round to the nearest 50, and please submit only a number, otherwise you will be disqualified like all those people from last week.

Prize


Rules to Rule By

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It's just a guessing game basically. You can guess number in 50 increment: 0, 50, 100, 150, 200, 250... The answer is 750.

Results: 180 people guessed the correct answer earning themselves 11112 NP each.

Round 261

Date Released: May 8th, 2008.
Date Rewarded: May 15th, 2008.

What's the next word in this sequence?

Cream
Radish
And
______

Submit only a one-word answer.

Prize


Radish Meringue

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Check out


The 1^st item in Appertisers: Maraquan Cream Broth.
The 2^nd item in Main Courses: Spicy Radish Salad.
The 3^rd item in Desserts: Peachpa and Stramberry Shell.

We observe that the words in the sequence are the second word in the 1^st, 2^nd and 3^rd items on the 1^st, 2^nd and 3^rd menu. Therefore, the next word should be the second word in the 4^th items on the 4^th menu (Cocktails): Phear Juice Tonic.

Results: 3477 people guessed the correct answer earning themselves 576 NP each.

Round 260

Date Released: May 1st, 2008.
Date Rewarded: May 8th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm³, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s², what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.

Prize


Steel Negg

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Using the stress equation [1]:

rho = F/A

where rho = 400 MPa = 400,000,000 kg/m·s² is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.

The crossed area is a regular pentagon can be calculated by the following equation [2]:

A = 5/4 t² tan(54°) = 0.000688190960236 m²

where t = 2 cm = 0.02 m is the side of the pentagon.

Now, we can calculate the force that the steel bar can withstand:

F = rho * A = rho * 5/4 t² tan(54°)

Using Newton's gravity law [3], we have:

F = m * g

Therefore, the mass that the steel bar can widthstand is:

m = F /g = rho * 5/4 t² tan(54°) / g = 28060.793485650833 kg

where g = 9.81 m/s² is the gravitational acceleration.

Don't round anything yet. The mass is contributed by 3 components: the big sphere m₁, small spheres m₂, and the steel bar m₃. Since all of them are steel, they have the same density. We have:

m₁ + int(h/3) * m₂ + m₃ = m
<=>[V₁ + int(h/3) * V₂ + A * h] * d = m

where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm³ = 7,800 kg/m³ is the density of steel, V₁, V₂ are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.

We can calculate V₁, V₂ by the sphere volume formula [4]:

V₁ = 4/3 * pi * (r₁)³ = 0.065449846949787 m³
V₂ = 4/3 * pi * (r₂)³ = 0.014137166941154 m³

where the radius r₁ = d₁/2 = 25 cm = 0.25 m, r₂ = d₂/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:

V₁ + int(h/3) * V₂ + A * h = m/d
<=>int(h/3) * V₂ + A * h = m/d - V₁

We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:

h/3 * V₂ + A * h = m/d - V₁

<=>(V₂/3 + A) * h = m/d - V₁
<=>h = (m/d - V₁)/(V₂/3 + A) = 654.0200901109209 m

Now round h down to an integer. We then have h = 654 m. Next, test h into the equation int(h/3)* V₂ + A * h = m/d - V₁. If the left hand side is smaller than the right hand side, the steel bar can withstand the gravitational forces. The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer.

Using h = 654 m, the left hand side: int(h/3)* V₂ + A * h = 3.5320. The right hand side: m/d - V₁ = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V₂ + A * h = 3.5327. The right hand side: m/d - V₁ = 3.5321. (not good). Therefore, the answer is h = 654 m.

Results: 986 people guessed the correct answer earning themselves 2029 NP each.

References:
[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume