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!
Is the answer 156?
ReplyDeleteyeah, please post the full solution !!
ReplyDeleteTwo 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?
ReplyDeleteA) 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 :-)
Can it not be 452?
ReplyDeleteYes 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.
ReplyDeleteIf (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.
ReplyDeleteClearly, 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
@amit:out of possibilities for N:ie 3 4 10etc
ReplyDelete3 , 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.