Explanation:
Let the two numbers be x and y.
We know that:
HCF(x,y) × LCM(x,y) = x × y
Substituting the given values:
4 × 288 = x × y
Simplifying:
x × y = 1152
Now we need to find two numbers whose product is 1152 and HCF is 4. One way to do this is to list all the factors of 1152 and find a pair of factors whose HCF is 4. However, we can also solve this problem by prime factorization.
Prime factorization of 1152:
1152 = 2^7 × 3^2
To find the two numbers, we need to divide these factors into two groups, one group for x and the other group for y. We can choose any combination of factors, as long as their product is 1152. However, we also need to ensure that the HCF of x and y is 4.
One possible way to do this is to choose one factor of 2 from the prime factorization of 1152 for x and the remaining factors for y:
x = 2^1 × 3^a
y = 2^6 × 3^b
where a and b are non-negative integers.
Multiplying x and y and equating to 1152, we get:
2^1 × 3^a × 2^6 × 3^b = 1152
Simplifying:
2^7 × 3^(a+b) = 1152
Since 1152 = 2^7 × 3^2, we have:
2^7 × 3^(a+b) = 2^7 × 3^2
Equating the exponents of 2, we get:
7 + 0 = 7
a + b = 2
Since the HCF of x and y is 4, we need to ensure that both x and y have a factor of 2^2 = 4. Thus, we choose a = 2 and b = 0:
x = 2^1 × 3^2 = 12
y = 2^6 × 3^0 = 64
Therefore, the two numbers are 12 and 64.