Guess 3 numbers


Source: Quantnet Forums

Problem:
A question which is asked on interview in some software development companies.
I guessed 3 natural numbers - x,y,z. You can ask me about two sums of these numbers with any coefficients - a,b,c. For example, you give me a, b and c and I tell you the result of the expression a*x+b*y+c*z. Give me the algorithm to find x,y and z.

Observation/Hint:
Irrespective of whether you get the solution, its interesting that you are solving a system of three variables using two equations. You are able to do that because the coefficient in the second equation depends on the answer of the first equation. :)

Update (November 6, 2011):
Solution: Posted by Harsh Pareek (Graduate Student at UT Austin, CSE IITB 2011 Alumnus), Rudradev Basak (IITD CSE Senior Undergraduate) and AnonymousD in comments!

Comments

  1. Harsh Pareek9:14 PM, November 05, 2011

    Crucial fact: they are natural numbers. If you knew the maximum number of digits any of them can have, say d, you could set a=1, b=10^-(d+1), c=10^-(2d+2), and you would be able to read the numbers off the decimal expansion for the sum.

    So, you use the first calculation to find the maximum number of digits, noting that it is okay to overestimate. Set a=1,b=1,c=1.
    Let the sum x+y+z=D. and d = ceil(log_10 D). Then 0<x,y,z<10^d.

    Then, set a=1,b=10^-d, c = 10^-2d.
    Let the sum be S.
    Then x = floor(S), y = floor((S-x)*10^d), z = floor( (S-x)*10^(2d) - y * 10^d)

    Finally, note that you can actually solve for n natural numbers x_1,x_2,...x_n with just 2 questions.

    ReplyDelete
  2. Rudradev Basak9:24 PM, November 05, 2011

    If we know T is greater than the values of x,y,z, Then it is sufficient to ask a question with a=1, b=T, c=T^2.
    To determine a feasible T there are lots of ways, simplest is to ask a question with a=b=c=1, and take the answer as T.

    ReplyDelete
  3. AnonymousD11:30 PM, November 05, 2011

    First get x+y+z=n, and then get
    x + n*y + n^2*y and write it down in base n to get x y z as its digits.

    ReplyDelete
  4. Pratik Poddar12:36 AM, November 06, 2011

    All solutions correct and same! Thanks!

    ReplyDelete

Post a Comment

Popular posts from this blog

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

Fraction Brainteaser

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions.  In total, Siddhant scores 25 out of 34.  Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34. Note that a) Vaibhav scores more than Siddhant b) Siddhant score better than Vaibhav in both individual topics -  5/6 > 20/25 and 20/28 > 6/9 How is it possible? 
Read more

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially) ...
Read more