The HCF and LCM of two number are 4 and80. One of the number is 4 more than the other.Find the smaller numbe

To solve this problem, we need to use the relationship between the highest common factor (HCF), the least common multiple (LCM), and the two numbers. The product of two numbers is equal to the product of their HCF and LCM.

Let's denote the two numbers as x and y. According to the problem, we know that:

1. HCF * LCM = x * y
2. x = y + 4

We can substitute these values into the equations:

1. 4 * 80 = x * y
2. x = y + 4

From the first equation, we get x * y = 320.

Substitute the second equation into the first to get (y + 4) * y = 320.

This simplifies to y^2 + 4y - 320 = 0.

This is a quadratic equation, and we can solve it using the quadratic formula y = [-b ± sqrt(b^2 - 4ac)] / (2a).

In this case, a = 1, b = 4, and c = -320.

So, y = [-4 ± sqrt((4)^2 - 41(-320))] / (2*1) = [-4 ± sqrt(16 + 1280)] / 2 = [-4 ± sqrt(1296)] / 2 = [-4 ± 36] / 2 = 16 or -20

Since y cannot be negative, the smaller number y = 16.