All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Getting Floored With 28 (Posted on 2008-09-06)
The greatest integer ≤ Y is denoted by [Y] , and {Y} = Y - [Y].

How many distinct real Y satisfy this equation, whenever 1 ≤ Y ≤ 28 ?

{Y2} = {Y}2

Note: While a solution may be trivial with the aid of a computer program, show how to derive it without one.

 See The Solution Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 4
See later comment for correction to the below:

From any integer to the next integer of the value of y, {y}^2 goes from equalling 0 to approaching 1.

The values of {y^2} during the same intervals depend on the interval in question. From y=1 up to but not including y=2, y^2 goes from 1 to approaching 4, thus incorporating 3 run-ups in its {} function, thus entailing 3 crossings (equalities) of the two functions.

Between 2 and 3, y^2 goes from 4 to 9, for 5 crossings or equalities.

Between 3 and 4, it goes from 9 to 16, for 7 crossings.

We see as we continue, that this will entail:

Sigma [y=1 to 27] (2y+1)

and also that this equals 28^2 - 1 crossing points, or equalities.

The first crossing point in each interval is in fact at a zero for each function, when y is an integer. The above computation therefore includes the beginning points of each of the 27 intervals.  However, the domain asked for includes the rightmost endpoint also, so we need to add 1. The answer becomes 28^2 = 784.

To check that we're doing this right, consider the range 1 to 2 inclusive, where we'd say the answer would be 2^2 = 4: crossovers would exist at the two endpoints as well as somewhere between sqrt(2) and sqrt(3) and somewhere between sqrt(3) and 2. Similarly up the line.

Edited on September 7, 2008, 2:19 am
 Posted by Charlie on 2008-09-06 15:22:22

 Search: Search body:
Forums (0)