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riddles / easy / Re: Madman: Slasher Riddle |
Jan 11th, 2023, 11:08pm |
| Started by alien2 | Last post by alien2 |
Although Your fine reply made me giggle, regardless of the PG-13 content, I still wonder if it holds water, considering everything mentioned in the original riddle. But, you're right about excrement and are on the right track because of it, regarding the expected answer. Would it be easier if I said that you are alone in this room with Christina Aguilera (no excrement here, just pure explosive love)? Another clue: I believe slang plays a role; or something like it.
Just to be absolutely sure about what you're saying. So, say to the madman: "You're nuts". Take the "nuts" and feed the squirrel, out of the kindness of your heart and possible affinity for animals, thus defuse a life-threatening situation, at least considering the madman. But what then?
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| 4 |
riddles / what am i / Re: In Common To All Riddles |
Oct 28th, 2022, 6:04am |
| Started by alien2 | Last post by alien2 |
| Shouldn't it be "an" answer? Even so, the answers differ, so I wonder whether or not Your "answer" is satisfactory. I was thinking of something else, something autochtonous to most riddles, and cast in the same mold as Your answer. |
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| 5 |
riddles / microsoft / What time is the solar noon? |
Oct 13th, 2022, 10:54pm |
| Started by jollytall | Last post by jollytall |
It seems the forum is sort-of dormant. Still I try.
It is not a microsoft question, I came up with it. The logic of the answer is important.
At what time is the sun at its highest position (I guess properly it is called solar noon or culmination)? |
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| 7 |
riddles / hard / Re: power of 3 close to power of 2 |
Sep 29th, 2022, 7:09am |
| Started by jollytall | Last post by jollytall |
I think I have a proof for the ratio of 3^n/2^m (<1) can get larger than any 1-e.
Let's take 3^n, what is between 2^m and 2^(m+1). If 3^n/2^m > 3/2 then I use 3^n/2^(m+1) (>3/4) and call it less than the "closest" (by ratio) power of 2. If 3^n/2^m<3/2 than I call it more than the closest power of 2.
Starting with the special case 3^1 = 3, it is 3/2 and 3/4 for the two neighboring power of twos.
Now I call 3/4 the best lower number so-far and 3/2 the best upper number so-far (aL and aH).
In the next step I take aL*aH (with other words the power of 3 that is a sum of the lower and upper powers in the previous step and the same for the power of 2). In the initial case it is 3/4*3/2 = 9/8.
This aNext > aL and aNext < aH (simply because aH was multiplied with a number <1 and aL was multiplied with a number >1).
If aNext > 1 then it becomes aH, otherwise becomes aL. I can continue this iteration infinite times. In every step aH-aL decreases and converges to either a non zero number or to zero.
What we want to prove is that aL converges to 1. I use an indirect proof, assuming that aL converges to 1 - l where l > 0. aH either converges to 1 or to 1 + h. First I exclude the possibility that aH converges to 1 + h and then I exclude the possibility that aH converges to 1 (and aL to 1 - l).
Let in one step aL = 1 - l - e and aH = 1 + h + f (e<<l and f<<h).
aN = 1 - l - e + h - hl - he + f - fl + fe, but
aN < 1 + h, because
1 - l - e + h - hl - he + f - fl + fe < 1 + h
-l -e -hl - he + f -fl +fe < 0
f(1 - l + e) - e(1+h) < l(1+h), where all elements of the left side << than the elements on the right side.
So, aH cannot converge to 1 + h, it can only converge to 1.
Let in one step aL = 1 - l -e and aH = 1 + f.
aN = 1 - l - e + f - fl -fe. It is clearly < 1, i.e. aN becomes aL and aH does not change.
If in every step aH (= 1 + f > 1) does not change then after many (n) steps
aN = (1 - l - e) * (1 + f) ^ n, but (1 + f)^n can grow above any number including
(1 - l) / (1 - l - e), in which case aN > 1 -l what is a contradiction.
So, aL converges to 1, i.e. the power of 3 can get arbitrarily close (in terms of a ratio) to the next (larger) power of 2.
It raises to me one more question: Is this sequence of aL the best achievable or there can be other 3^n with lower n, where it is still a better ratio.
And still I have no idea about the absolute difference achievable.
P.S. What happened to the superscript. I can only write 3^n and not the proper superscript version. |
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| 9 |
riddles / hard / Identify your favourite childhood games |
Jun 29th, 2022, 10:57pm |
| Started by livinggod29 | Last post by livinggod29 |
Identify your favourite childhood games
1. Kevin runs to avoid being touched by Sahlan and Vhanz in hot pursuit
2. Declining number of seats ensure the inevitability of the last person standing.... or sitting
3. Count, then search
4. Throw and chart the course to retrieve it one foot
5. Demolish a pile and reconstruct before you are hit
6. Your ability to stand still will be a life saver
7. No need to invite a former American Idol judge but just do what he tells
8. Tiny round glass balls, you are crazy if you have lost them
9. This is the right opportunity to jump over the other riddlers
10. Get from Point A to Point B.... in a coarse bag
11. Keep off the ground at any cost
12. Back when you thought law enforcement would pursue the bad guys on foot
13. You snoop something....
14. Riddlers on their knees. Opponents running amok. Zsacksaw chases, tags Sanjay to continue the chase
15. Pull with all your might, bring victory to your side |
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| 10 |
riddles / easy / What are we? |
Jun 29th, 2022, 10:55pm |
| Started by livinggod29 | Last post by livinggod29 |
What are we?
1. I am always excited
2. I tell people where things are at
3. You can play a game on me
4. I'm very rich
5. I give percentages
6. I am always looking up
7. And.........
8. I look like a light in the sky
9. I keep things arranged that are to my right
10. I keep things arranged that are to my left
Share with your friends and family and see if they can figure out what are we? |
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