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Topic: colcolpicle's puzzles (Read 87052 times)


Madball




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To be or not to be?
2B!2B=FF





pivotdude123

for predict the crate: rusubebEDIT: Found out it was practically the same as Werbad's. Crap.


« Last Edit: October 22, 2010, 04:28:03 PM by pivotdude123 »

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colcolpicle

Here's an interesting problem I heard recently. it's not rubicon related but it is math which is kind of related. How many numbers are there between 1000 and 10000 where each digit is at least one more than the previous digit. For clarification, the first number that has this property is 1234. Once you figure that out, can you figure out a formula for this problem between 10^n to 10^(n+1) for any n? Cross the streams kibomef: gigilys


« Last Edit: October 29, 2010, 11:54:24 PM by colcolpicle »

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Bucky

Cross the streams kibomef: feropym. The timing is rather tricky on this one. The [level]half furnace draw[/level] on the left side needed to replace the traditional version because that version was 2 ticks slow no matter what I tried. Here's an interesting problem I heard recently. it's not rubicon related but it is math which is kind of related.
How many numbers are there between 1000 and 10000 where each digit is at least one more than the previous digit. For clarification, the first number that has this property is 1234. Once you figure that out, can you figure out a formula for this problem between 10^n to 10^(n+1) for any n?
126. Formula is item n+2 in row 10 of Pascal's Triangle, or 0 if n>9 or n<1.Quick explanation for people with little math background: The number in each cell in Pascal's Triangle is the number of unique paths from the vertex to that point. I have a scheme where every number with this property corresponds to exactly one path and vice versa.
To generate the number for a path, start at the vertex with a counter at zero. At each step in the path, increment the counter, then write a digit with the value of the counter if the step was downright. You will always end up with the same number of digits because you perform a constant number of rightwards steps, and you will end up with digits in increasing order because the counter only increments.
To generate the path for a number, start at the vertex with a counter at 1 and a marker before the first digit. For each step in the path, if the next digit is equal to the counter, step downright and move the marker one digit; otherwise step downleft. Either way, increment the counter (and stop if the counter goes past 9). The digit 'after' the last one is larger than 9 for this. The path will end in the right spot because it will always have one downright step per digit and 9 steps total (because of the counter).
The reason why it's item n+2 is that the formula given asks about n+1 digit numbers, and the first item in each row corresponds to zerodigit numbers.


« Last Edit: October 30, 2010, 04:42:35 AM by Bucky »

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That is the most ingenious method of solving an impossible puzzle that I have ever seen.




Madball




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To be or not to be?
2B!2B=FF





colcolpicle

firing squad: gufycer : susyrarfiring squad easy: nideceh : gogigybfiring squad real: ricykerfiring squad real easy: debacidfor the math problem I used triangular numbers and figured out that it is actually the (10n)th n dimensional triangular number. this just happens to be equal to what bucky had done with pascal's triangle.


« Last Edit: November 02, 2010, 11:38:09 PM by colcolpicle »

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Rene



« Last Edit: November 03, 2010, 04:51:00 PM by Rene »

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