RE: 20 Questions Type Puzzel
January 20, 2015 at 9:55 pm
(This post was last modified: January 20, 2015 at 9:56 pm by TheRealJoeFish.)
That is *correct*!!! YAY WE HAVE A WINNER!!!
I think there's an easier way to conceptualize it: if x is the nth number in the sequence, then the (n+1)th number differs from x by one of the factors of x. This is equivalent to what you said: you find two factors (y and z) of the nth number, x, and the (n+1)th number has one more or one less of either y or z.
So, for example, the factors of 10 are 1, 2, 5 and 10. Any of the following numbers could follow 10:
0 (10-10), 5 (10-5), 8 (10-2), 9 (10-1), 11 (10+1), 12 (10+2), 15 (10+5), 20 (10+10).
Note that weird things happen when 0 comes up, which I realized after I made the first sequences; I would've had to do some post hoc explanation in that case, and ultimately would have allowed any number to come after 0, because, *technically*, all integers are factors of 0.
I think there's an easier way to conceptualize it: if x is the nth number in the sequence, then the (n+1)th number differs from x by one of the factors of x. This is equivalent to what you said: you find two factors (y and z) of the nth number, x, and the (n+1)th number has one more or one less of either y or z.
So, for example, the factors of 10 are 1, 2, 5 and 10. Any of the following numbers could follow 10:
0 (10-10), 5 (10-5), 8 (10-2), 9 (10-1), 11 (10+1), 12 (10+2), 15 (10+5), 20 (10+10).
Note that weird things happen when 0 comes up, which I realized after I made the first sequences; I would've had to do some post hoc explanation in that case, and ultimately would have allowed any number to come after 0, because, *technically*, all integers are factors of 0.
How will we know, when the morning comes, we are still human? - 2D
Don't worry, my friend. If this be the end, then so shall it be.
Don't worry, my friend. If this be the end, then so shall it be.
