Question

The HCF of three numbers is 8 and the sum of these numbers is 80. List the possible set of such three numbers.

Answer 1

Let's denote the three numbers A, B, and C.

Given that the highest common factor (HCF) of these three numbers is 8 and their sum is 80, we can consider possible combinations of numbers that satisfy these conditions.

Since the HCF is 8, all three numbers must be divisible by 8. Additionally, the sum of the numbers is 80, so we need to find combinations of three numbers that satisfy both conditions.

Let's list the possible combinations:

1. (8, 16, 56): In this case, A = 8, B = 16, and C = 56. All three numbers are divisible by 8, and their sum is 8 + 16 + 56 = 80.
2. (16, 8, 56): Here, A = 16, B = 8, and C = 56. Again, all three numbers are divisible by 8, and their sum is 16 + 8 + 56 = 80.
3. (24, 8, 48): In this combination, A = 24, B = 8, and C = 48. All three numbers are divisible by 8, and their sum is 24 + 8 + 48 = 80.
4. (8, 24, 48): Similarly, A = 8, B = 24, and C = 48. All three numbers are divisible by 8, and their sum is 8 + 24 + 48 = 80.

These are the four possible sets of three numbers that satisfy the given conditions: (8, 16, 56), (16, 8, 56), (24, 8, 48), and (8, 24, 48).