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Problem Of The Week 7

with 7 comments


Two positive integers differ by  60.The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

A) 100                 B) 156                   C)  392                   D) 452                       E) none of these

P.S: If You are not able to solve this, Kindly refer to the concept 1 Perfect squares. Any student who has read that lesson and practised the problems should be able to solve this, if not reread the lesson!

Written by Implex

August 28, 2008 at 9:46 pm

7 Responses

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  1. Is the answer 156?

    Amit

    August 29, 2008 at 7:31 pm

  2. yeah, please post the full solution !!

    outtimed

    August 29, 2008 at 7:57 pm

  3. Two positive integers differ by 60.The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

    A) 100 B) 156 C) 392 D) 452 E) none of these

    a-b =60
    [sqrt(a)-sqrt(b)][sqrt(a)+sqrt(b)] = 60

    We can take sqrt(a) = A*sqrt(N) where N is a non perfect square factor of 60
    and sqrt(b) = B*sqrt(N)

    Now, N can be 3,5,15,20,12

    Since we have to find maximum sum we will consider N=3

    (A-B)(A+B)= 20 where A and B are positive integers.
    Only one case possible (A-B)(A+B) = 2.10
    A=6 B=4

    a= 36*3=108
    b= 16*3 = 48
    a+b= 156

    Is this the correct way to solve outtimed🙂

    Amit

    August 29, 2008 at 9:19 pm

  4. Can it not be 452?

    Deepshikha

    August 30, 2008 at 8:19 am

  5. Yes it has to be 156 only. 452 will make sum of square roots 30 which is the square root of an integer that is a perfect square.

    Deepshikha

    August 30, 2008 at 10:18 am

  6. If (a)^1/2+(b)^1/2=(k)^1/2 =>a+b=(1800/k)+(k/2)=>a=30+900/k+k/4 =K is of the form 4x =>a=30+(225/x)+x => for x=(75,3),a+b=156. For x=(45,5),a+b=100.For x=(25,9),a+b=68.

    Clearly, maximum holds for x=75=>k=300 and the sum is 156.

    Note: x=3 is an invalid value as it does not satisfy all the conditions

    Vineet Singh

    September 1, 2008 at 6:57 pm

  7. @amit:out of possibilities for N:ie 3 4 10etc
    3 , 5 ,12 , 20 are valid as they would only yield integral value for
    a and b.
    how you directly deduced that sum would be greater for N=3.

    amit gupta

    September 2, 2009 at 12:48 pm


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