Originally posted by: yankeesfan
Originally posted by: silverpig
Originally posted by: DrPizza
I think that coming up with a math equation for which the answer is very simple to find for two of the values, and difficult to find for the 3rd value might be the way to go.
i.e. if he answers yes, the answer is 1.
if he answers no, the answer is 2
if he cannot answer the question because the problem is unsolvable in a realistic amount of time, then the answer is 3
see:
http://en.wikipedia.org/wiki/Complexity_classes_P_and_NP
Give him some instructions like:
Assign the value of your number to the variable x. Compute (x-1)^(51200329123/443). Does the number which results from this computation have a 0 in the "ones" (first to the left of the decimal) place?
If x = 1, then the number is 0, and he'll answer yes.
If x = 2, then the number is 1, and he'll answer no.
If x = 3, then he'll have to do a lot of crunching and you'll know it's 3.
I think that this is the best way to go. Good job Dr. Pizza and Silver Pig.
That answer works for me
🙂 My solution started with (x-2) , with this question:
is the product of (x-1) and the sum of numbers of any subset constructed with at least two elements from the larger set and of which your number must be one of the elements of the subset with the larger set being: {your number, a lot of really large positive or negative even numbers)} ever equal to zero?
If his number is 1, then it's trivially simple: the product is equal to zero.
If his number is 3, then it's relatively simple to deduce that the product cannot be equal to zero. Because (x-1)=2, and the second factor in the product *must* be an odd number, because it is the sum of one odd number and any number of even numbers.
If his number is 2, then determining if there is a subset of numbers such that the sum of that subset = 0 can be a problem for which no solution can be guaranteed in any reasonable amount of time.
I realized at this point, that I might have to modify the set part a little bit. Here's a similar problem (hopefully I don't screw up the numbers; I have this on a worksheet in school to use as a day killer for my advanced math students on those days when more than half the class is absent):
Given a set of 100 15 digit numbers, are there any two subsets such that the sum of the numbers in each of those subsets is the same?
It's very simple to prove that the answer is "yes" by the pigeonhole concept... Every sum must be between 15* 100000000000000 and 15*999999999999999. That's a lot of possible answers when you sum up subsets. (roughly 1.5*10^16) However, there are more subsets than there are possible sums. (100C1 + 100C2 + 100C3 +...) A whole heck of a lot more possible sums. So, imagine a giant post office with a box for every possible sum. Every time you get the sum of a subset, write it on a piece of paper and put it in the appropriate box. Not every box will necessarily have a sum in it when you're finished, but since there are more pieces of paper than boxes, it's necessary that at least 1 of the boxes has more than one slip of paper in it. Compute the number of subsets there are with 50 elements in it out of 100 elements from your original set... then figure that you can find 10^9 (a billion) sums every second. Calculate the number of years it's going to take, simply to calculate the sums of 50 number subsets.
You may need to modify my original subset restriction slightly in order for it to work out to this level of difficulty.
edit: note, the original question said "what is one yes or no question that you could ask to discern which number it is I'm thinking of" - since my solution provides a problem for which there is a yes or no answer possible for 2 of the number selections, and a complete lack of silence for the next few billion years (or an answer of "uhhhhh, I'll have to get back to you on that), then you *can* discern the original number.
