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Topic: Number sequence (Read 929 times) |
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Thomson
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As Indiana Jones explored yet another dangerous cave, he stumbled himself on a door with numbers engraved on it.
The first one was simply "9". The next, "19". Then, "1,119", "3,119", and "132,119".
He deduced that to advance he must figure out the next number in the sequence
Can you help Indy figure out the next number?
I'll post some follow-up riddles if you guys manage to crack this one. 
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ThudnBlunder
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Re: Number sequence
« Reply #1 on: Jan 14th, 2005, 11:14pm » |
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: 1113122119
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| « Last Edit: Jan 14th, 2005, 11:32pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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JocK
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Re: Number sequence
« Reply #2 on: Jan 15th, 2005, 3:51am » |
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on Jan 14th, 2005, 8:28pm, Thomson wrote:I'll post some follow-up riddles if you guys manage to crack this one.
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You're not gonne ask us about the asymptotic rate of growth of the number of digits in the n-th term, and finding a link to the polynomial
x^71 - x^69 - 2*x^68 - x^67 + 2*x^66 + 2*x^65 + x^64 - x^63 - x^62 - x^61 - x^60 - x^59 + 2*x^58 + 5*x^57 + 3*x^56 - 2*x^55 - 10*x^54 - 3*x^53 - 2*x^52 + 6*x^51 + 6*x^50 + x^49 + 9*x^48 - 3*x^47 - 7*x^46 - 8*x^45 - 8*x^44 + 10*x^43 + 6*x^42 + 8*x^41 - 5*x^40 - 12*x^39 + 7*x^38 - 7*x^37 + 7*x^36 + x^35 - 3*x^34 + 10*x^33 + x^32 - 6*x^31 - 2*x^30 - 10*x^29 - 3*x^28 + 2*x^27 + 9*x^26 - 3*x^25 + 14*x^24 - 8*x^23 - 7*x^21 + 9*x^20 + 3*x^19 - 4*x^18 - 10*x^17 - 7*x^16 + 12*x^15 + 7*x^14 + 2*x^13 - 12*x^12 - 4* x^11 - 2*x^10 + 5*x^9 + x^7 - 7*x^6 + 7*x^5 - 4*x^4 + 12*x^3 - 6*x^2 + 3*x - 6
... are you? 
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| « Last Edit: Jan 15th, 2005, 3:52am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Thomson
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Has this riddle been put up before? o.o
Jock that was one of the questions....Heres some more. 
For those who've answered the riddle
Follow-up 1
After hearing the story of Indy Dr.Karl asked himself "Instead of taking a single digit at a time, What if you look at the whole group of numbers, take an exhaustive view and say the total of each?"
132,119 -- would normally become --> 1,113,122,119
132,119 -- would now be ---->31,131,219 --->23,411,219
23,411,219 ->2,213,142,119 ->3,241,131,419 --->
2,312,244,119---> 3213312419 ->3322311419 ->3322311419
-->3322311419
After numerous testings he concluded that all numbers, end in a cycle(meaning they repeat the same numbers(or group of numbers over and over again).
ex. 9 ends in the cycle 3322311419 ->3322311419
He assumed that all numbers will eventually give a number in which
Number x ->Number a -> number a
Number x ->Number a -> number b -> number a
Number x -> number a -> number b -> number c -> number a
etc. etc.
Can you prove or disprove this?
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towr
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Re: Number sequence
« Reply #4 on: Jan 16th, 2005, 7:29am » |
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on Jan 15th, 2005, 3:34pm, Thomson wrote:| Can you prove or disprove this? |
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Prove what? That it cycles or ends? ::Each string of numbers is at most 20 digits long (less actually). Each string maps to only one other. So at most we can go through 10^20 different ones before we have to repeat. Thus either we encounter a cycle, or we end up repeating the same string (which is a cycle of length one)::
Quote:| Has this riddle been put up before? o.o |
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Well, yes, kind of, a very similar one at least.
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Wikipedia, Google, Mathworld, Integer sequence DB
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Nigel_Parsons
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Re: Number sequence
« Reply #5 on: Jan 16th, 2005, 8:41am » |
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Alternative answer to Thud&Blunder's
:: 31131219
This assumes we are stating how many of each digit appears in the previous, starting with the first digit without differentiating the order of duplicated digits.::
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ThudnBlunder
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Re: Number sequence
« Reply #6 on: Jan 16th, 2005, 8:44am » |
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Quote:| Has this riddle been put up before? |
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Yes, and my previous post links to it.
on Jan 16th, 2005, 8:41am, Nigel_Parsons wrote:| Alternative answer to Thud&Blunder's...This assumes we are stating how many of each digit appears in the previous, starting with the first digit without differentiating the order of duplicated digits |
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Sorry, I don't understand this.
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| « Last Edit: Jan 16th, 2005, 11:56am by ThudnBlunder » |
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