GATE2019 ME-1: GA-4

Question

The sum and product of two integers are $26$ and $165$ respectively. The difference between these two integers is ______

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Let the two integers be $x$ and $y.$

Given that:

• $x+y=26\qquad \to (1)$
• $xy=165\qquad \to(2)$

To be find$:x-y=?$

We know that $(x-y)^{2}=x^{2}+y^{2}-2xy$

$\implies(x-y)^{2}=x^{2}+y^{2}+2xy-4xy$

$\implies(x-y)^{2}=(x+y)^{2}-4xy$

Put the values from equation $(1)$ and $(2)$ and we get

$\implies(x-y)^{2}=(26)^{2}-4\times165$

$\implies(x-y)^{2}=676-660$

$\implies(x-y)^{2}=16$

$\implies(x-y)=\sqrt{16}$

$\implies x-y=\pm4$

$\implies x-y=4$  $(or)$  $x-y=-4$

$-4$ is ruled out as both the sum and product of the numbers are positive meaning both the numbers must be positive.

So, $(C)$ is the correct choice.

Comments

Your answer is right. I just want to point out something.

-4 is ruled out as both the sum and product of the numbers are positive meaning both the numbers must be positive.

This is not true.

If you solve the equations (1) and (2), you will get two solutions (x, y) = (15, 11) and (11, 15). In the second solution, $x - y = 11 - 15 = -4$ even though both (11,15) are positive integers.

The reason we rule out (11, 15) is that all the options are positive. i.e. the difference $x - y$ is positive, or $x > y$.

---

@Venky8 the question is just asking about the mod value of (x-y).

Answer 2

we know (x-y)^2=(x+y)^2-4xy

x+y=26,xy=165

so (x-y)^2=26^2-4*165=16

so x-y=4

Answer 3

factors of 165 = 5*3*11

X = 5*3 = 15

Y = 11

(X+Y = 15 + 11 = 26 ; XY = 15*11= 165)

Ans: 15-11 = 4