A number is three times another number and their HCF is 8. What is the sum of the squares of the numbers?
- A640
- B512
- C1024
- D1000
Solution & Step-by-step Explanation
Let the two numbers be x and y.
Given that one number is three times the other:
y=3x
Since their HCF is 8, both numbers must be multiples of 8, and the coprime factors are 1 and 3.
Thus, the numbers are:
x=1×8=8
y=3×8=24
We need to find the sum of the squares of the numbers:
Sum of squares=x
2
+y
2
=8
2
+24
2
Sum of squares=64+576=640
Given that one number is three times the other:
y=3x
Since their HCF is 8, both numbers must be multiples of 8, and the coprime factors are 1 and 3.
Thus, the numbers are:
x=1×8=8
y=3×8=24
We need to find the sum of the squares of the numbers:
Sum of squares=x
2
+y
2
=8
2
+24
2
Sum of squares=64+576=640