**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 **

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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.

outtimedSeptember 13, 2008 at 4:45 am

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

milindSeptember 15, 2008 at 11:35 am

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

outtimedSeptember 15, 2008 at 12:16 pm