The GREAT thread of riddleness

  • Thread starter JamesU
  • Start date
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Sorry it took me this long to reply.

Good work, Icebreaker. Your turn.
 

Icebreaker

yomamma said:
ICEBREAKER!! *slaps icebreaker* other people were supposed to solve the riddle, not you!
Well that was certainly odd.

Berislav said:
Good work, Icebreaker. Your turn.
Find a counterexample to the Riemann hypothesis.

Just kidding. Here's an easy one:

The length of the perimeter of a right triangle is 60 and the length of its altitude perpenticular to the hypotenuse is 12. Find the sides and show your work.

Good luck.
 
365
1
Icebreaker said:
The length of the perimeter of a right triangle is 60 and the length of its altitude perpenticular to the hypotenuse is 12. Find the sides and show your work.
Let a and b denote the lengths of the legs of the triangle, and let c denote the length of the hypotenuse.

60=a+b+c
a^2+b^2=c^2
ab/2=12c/2

Solving this system of equations, we get

a=15
b=20
c=25
 

Icebreaker

Yup. Your turn.
 

JamesU

Gold Member
732
3
Okay, moo of doom, your turn
 
365
1
Yikes, I don't really have anything...

Here goes...
Find the next row in this sequence:

1
2, 1
3, 3, 2, 1
4, 4, 4, 3, 2, 1
5, 9, 10, 11, 12, 12, 12, 12, 12, 12, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
???
 

JamesU

Gold Member
732
3
6, 5, 9, 10, 11, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1


I'm really not sure, I took a huge guess
 
365
1
Nope.

I'll give you a hint: it's more than twice as long as the last one. (sorry :P)
 

JamesU

Gold Member
732
3
Can you give a bigger hint?
 
365
1
There are 62 numbers in the sequence, and the highest number in the sequence is 32, which is repeated 17 times.

There's another big hint I'm willing to give if it's absolutely necessary.
 

JamesU

Gold Member
732
3
does it involve adding, subtraction, mult... you get it, so does it?
 
365
1
It involves subtraction, but not in the way you think. It also involves addition, but in an even weirder way than the subtraction.

(Basically, if you think the pattern is anything akin to add one, subtract two, etc. you're on the wrong track. There is no such rule, really.)
 
365
1
I have the feeling this puzzle is a bit too freakishly difficult, so here comes a big hint:

The sequence is very closely related to this one:

1
10, 1
11, 10, 2, 1
100, 11, 10, 3, 2, 1
...
???

If you're still stuck, I'll give the fifth row of the above sequence, although it kind of gives it away.
 

JamesU

Gold Member
732
3
30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
 
365
1
You have part of it, but you're missing the first 12 numbers, and one of the 32s.
 

JamesU

Gold Member
732
3
18, 19, 20, 21, 22, 23, 24, 25,, 26, 27, 28, 29, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
 
365
1
You have it right from 24 on, but the bit before that is wrong.
Think about how the first sequence might relate to the second sequence, and see if that helps. Tell me if you want the next row for the second sequence.
 

JamesU

Gold Member
732
3
20, 21, 22, 23, 24, 25, 24, 25,, 26, 27, 28, 29, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
 
365
1
Nope, colder.
 

JamesU

Gold Member
732
3
I give up, give another.
 
365
1
Here's the original sequence:

1
2, 1
3, 3, 2, 1
4, 4, 4, 3, 2, 1
5, 9, 10, 11, 12, 12, 12, 12, 12, 12, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Here's the second sequence:

1
10, 1
11, 10, 2, 1
100, 11, 10, 3, 2, 1
101, 100, 22, 21, 20, 15, 14, 13 , 12, 11, 10, B, A, 9, 8, 7, 6, 5, 4, 3, 2, 1

Two entries in the last row should be a big hint as to what kind of problem this is. How do these sequences relate? Dividing into columns may help.
 
854
16
To get the original sequence from the second sequence: find the representation in base 10 for each number in a row assuming that the number in the nth column is represented in base (n + 1). For instance, in the last row 101 base 2 is 5 base 10, 100 base 3 is 9 base 10, 22 base 4 is 10 base 10, 21 base 5 is 11 base 10, etc.
But I don't yet see how to get the second sequence.
 
854
16
Now I see how to get the second sequence.

101 -1 (base 2) = 100
100 -1 (base 3) = 22
22 - 1 (base 4) = 21
21 - 1 (base 5) = 20
20 - 1 (base 6) = 15
etc.
So the next line in the second sequence is:
110, 101, 100, 33, 32, 31, 30, 27, 26, 25, 24, 23, 22, 21, 20, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, V, U, T, S, R, Q, P, O, N, M, L, K, J, I, H, G, F, E, D, C, B, A, 9, 8, 7, 6, 5, 4, 3, 2, 1

Converting to the original sequence we get:

6, 10, 16, 18, 20, 22, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, ..., 32, 31, 30, ..., 1
 
Last edited:
365
1
Jimmy is correct. Good Job! You get the next puzzle.
 

JamesU

Gold Member
732
3
whew! :biggrin:
 

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