## Problem Of The Week 7

**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?

AmitAugust 29, 2008 at 7:31 pm

yeah, please post the full solution !!

outtimedAugust 29, 2008 at 7:57 pm

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🙂

AmitAugust 29, 2008 at 9:19 pm

Can it not be 452?

DeepshikhaAugust 30, 2008 at 8:19 am

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.

DeepshikhaAugust 30, 2008 at 10:18 am

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 SinghSeptember 1, 2008 at 6:57 pm

@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 guptaSeptember 2, 2009 at 12:48 pm