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Problem Of The Week 14

with 3 comments


Let X=[x] +{x} where [x] and {x} are respectively the integer and fractional parts of a real number x. Find the number of real number which exist such that they satisfy the following conditions:

a) 0<=x<=100

b) {x^2}={x}


A) 9000                  B) 9900                     C) 9990                    D) 9901                 E) 9991

Written by Implex

September 12, 2008 at 11:43 am

3 Responses

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  1. here is the official solution to this problem
    If x and x^2 have the same fractional part then x^2 – x is in N. Consider the quadratic function f(x) = x^2 – x. Since it is monotonous and strictly increasing on [0, 100], all the integer values between f(0) and f(100) will be assumed by f exactly once. For each positive integer m, the equation x^2 – x = m has precisely one positive solution. Now f(0) = 0 and f(100) = 9900. So the 9901 possible values of m are 0, 1, 2, …, 9900. The answer for 0<=x<=100 is 9901.

    outtimed

    September 13, 2008 at 4:45 am

  2. Didn’t understand the last 2 lines of the solution. Please elaborate.

    milind

    September 15, 2008 at 11:35 am

  3. f(x)=x^2-x
    now {x^}={x}
    so x^2-x is always a whole number in 0<=x<100
    now
    max f(x)=10000-100=9900 and min is 0
    and all these values will be achieved for one x
    so at max 9901 values

    outtimed

    September 15, 2008 at 12:16 pm


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