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Problem Of The Week 4

with 4 comments


Let a,b be integers then find the no of pairs of (x,y) such that
[x]+2y=a and
[y]+2x=b  where  [z] means greatest integer <=z

a) 0          b) 1                 c) 2              d) 4                e) Cannot be determined

Written by Implex

August 27, 2008 at 5:35 pm

4 Responses

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  1. I could be wrong but I would say b)1.
    This is because here three cases can be considered
    1. Both x and y are integers
    2. Both x and y have o.5 as the decimal part.
    3. One of these is an integer and the other has a decimal part of 0.5
    In each of these cases the difference between 2x and 2y will have to be covered by the difference between [x] and [y], which is not possible even by taking minimum difference between x any y in each of the cases mentioned above.

    Deepshikha

    August 28, 2008 at 2:04 am

  2. I have not given the correct reasoning in my first response to this problem.This is what I mean.
    Firstly, we assign values to x any y to get a values of a and b.Now we have to find a pair of x and y for which the values of a and b are the same.for this four cases are possible
    1. If x,y are integers and the next values taken are integers with a min. difference of 1.
    In this case.
    [x] + 2y =a
    If both x and y are being integers are increased or decreased by 1, a cannot be obtained again.
    2.If x,y are both have 0.5 as the decimal part.
    Two cases are possible here
    a)If x and y are increased here lets say by 0.5 Then [x] and [y] will increase to the next integer and so will the value of 2x and 2y, so again obtaining a and b from the equations will not be possible.
    b) If decreased by 0.5 the [x],[y] remain same but again 2x and 2y change.
    3.if one of x and y has o.5 as the decimal part and the other is an integer, in this case as well the integer part will increase or decrease by 1 and will be doubled on multiplying by2.and the [x],[y]will be increased to the next integer or remain same depending on the next value chosen is greater or less.
    Hence there b)1

    Pardon me for my mistakes. I am fallible!!

    Deepshikha

    August 28, 2008 at 4:45 am

  3. Nice attempt, but you have misunderstood the problem !

    outtimed

    August 28, 2008 at 10:33 am

  4. I would proceed like this..
    say [x] = greatest integer <=x
    and {x} = fractional part of x.

    For positive integer
    So x = [x] + {x}.

    Please see its not very difficult to observe that for negative x.
    x = [x] + 1 + {x}.

    Consider the case where both x and y are positive.
    then 1st equation can be written as:-
    [x] + 2[y] + 2{y} = a …….. (Equ 1′)
    and second equation can be rewritten as :-
    [y] + 2[x] + 2{x} = b …………(Equ 2′)

    Now 1′ + 2′ gives:-
    3[x] + 3[y] + 2({x} + {y}) = a + b ……. Equ 3′

    and 1′ – 2′ gives
    [y] – [x] + 2({y} – {x}) = a – b ……….Equ 4′

    since {x} + {y} can take values 0, 0.5, 1.5 in order to satisfy 3′
    and {x} – {y} can take values 0, 0.5 in order to satisy 4′

    Now there are fe

    Duffer

    October 8, 2008 at 7:11 am


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