let the numbers be x and y

let the LCM of two numbers is L

and HCF of two number is H

Be Question

H+L=680

L=84H

85H=680

H=8

also HL=xy(scine HCF times LCM is the product of two numbers is(Basic algebraic law,valid for only two numbers)

Therefore L= 680-8=672

L=672×8=5376

5376÷56=96

other number is 96

Step-by-step explanation:

Sure, I can help you with that.

Let the other number be x. We know that the HCF of two numbers is always less than both the numbers, and the LCM is always a multiple of both the numbers.

Since 56 is a factor of both the numbers, the HCF of the two numbers must be a factor of 56. So the HCF is 4.

We also know that the LCM is 84 times the HCF, so the LCM is 84 * 4 = 336.

The sum of the HCF and the LCM is 680, so 4 + 336 = 340.

The other number is x, and we know that x + 56 = 340.

Solving for x, we get x = 284.

Therefore, the other number is 284.

Here is the step-by-step explanation:

1. Let the other number be x.

2. The HCF of the two numbers must be a factor of 56. So the HCF is 4.

3. The LCM is 84 times the HCF, so the LCM is 84 * 4 = 336.

4. The sum of the HCF and the LCM is 680, so 4 + 336 = 340.

5. x + 56 = 340.

6. Solving for x, we get x = 284.

Therefore, the other number is 284.

I hope this helps! Let me know if you have any other questions.