Guess my Number
May 26th, 2007Warning! Spoiler!
The Solution
Let X be the sum of the numbers of the men modulo 100. Then X is in the set { 0, 1, …, 99 }. The first man gives the unique number x1 such that the sum of the numbers modulo 100 will be 0. The second man gives the unique number x2 such that the sum of the numbers modulo 100 will be 1. etc. They are obviously bound to succeed.
The Afterthoughts
Can the men employ a different strategy? Well, instead of considering x1+x2+…+x100 (where xi is the number of the ith man) the men can consider a different function, such as 7*x1+3*x2+…+11*x100. Obviously, if the coefficients are relatively prime to 100, then this strategy will work as well. How many of these strategies are there?
Well, to guarantee that at least one of the men gets it right, one of the men has to say a number such that the sum of all the numbers will equal the correct sum modulo 100. I.e. no matter what strategy the men select, it must go through each of the sums modulo 100 { 0, 1, …, 99 }. It is then evident that there are exactly 100! (1*2*…*100) different strategies for the men.
As of guaranteeing that more than one man will get it right – it is obviously impossible, since if two people get it right that means that not all the numbers in the set { 0, 1, …, 99 } were generated as the sum modulo 100, and obviously this might have been the correct sum.
Feel free to post solutions (or solution related comments) to this page!
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May 27th, 2007 at 6:02 pm
Gee, I feel stupid now… I’ve tried thinking about this for over SEVEN minutes and still no solution
I’ll allocate another five minutes for this later and maybe have better luck (maybe just need more coffee)
May 27th, 2007 at 9:00 pm
Hi Matan,
I put the “5 Minute Riddle” icon not in order to make you feel stupid in case you fail to solve it in that amount of time, but to indicate that this riddle is indeed much easier to solve than most others in my site.
The time it takes to solve a riddle varies greatly from one person to the next, and the 5-minute icon is my estimation of the expectancy of the duration of solving the riddle. My estimation of the variance is about 30 minutes. If you fail to solve the riddle in more than one hour, only then should you start feeling stupid (
).
May 28th, 2007 at 10:44 am
IMHO, this riddle is harder much harder than the ants and regexp riddles. Even harder than the rectangles one (though solved it without all that evil vertices and edges shit)