Input the product and HCF of two numbers, and compute the numbers instantly.
Assume the two numbers are \( x \) and \( y \), with their product denoted as \( P \) and their highest common factor as \( \text{HCF} \).
Let \( x = k \times a \) and \( y = k \times b \), where \( k \) is the highest common factor (\( \text{HCF} \)), and \( a \) and \( b \) are coprime integers (their greatest common divisor is 1).
Using the product \( P \), we get: \( x \times y = (k \times a) \times (k \times b) = k^2 \times (a \times b) = P \) From this, we derive: \( a \times b = \frac{P}{k^2} \)
Factorize \( a \times b \) and identify all pairs of factors that are coprime integers.
Use the proportionality constant \( k \) to compute the two numbers: \( x = k \times a \) \( y = k \times b \)
Solution:
1. Identify the proportionality constant:
\( k = HCF = 15 \)
2. Compute \( a \times b \):
\( a \times b = \frac{2025}{15^2} = \frac{2025}{225} = 9 \)
3. Find coprime pairs for \( a \times b = 9 \):
Factorize 9 into pairs:
(1, 9)
(3, 3)
Among these, (1, 9) is coprime.
4. Calculate \( x \) and \( y \):
\( x = 15 \times 1 = 15 \)
\( y = 15 \times 9 = 135 \)
Result: The two numbers are 15 and 135.
Solution:
1. Identify the proportionality constant:
\( k = HCF = 40 \)
2. Compute \( a \times b \):
\( a \times b = \frac{19200}{40^2} = \frac{19200}{1600} = 12 \)
3. Find coprime pairs for \( a \times b = 12 \):
Factorize 12 into pairs:
(1, 12)
(2, 6)
(3, 4)
Among these, (1, 12) and (3, 4) are coprime.
4. Calculate \( x \) and \( y \):
\( x_1 = 40 \times 1 = 40 \)
\( y_1 = 40 \times 12 = 480 \)
\( x_2 = 40 \times 3 = 120 \)
\( y_2 = 40 \times 4 = 160 \)
Result: The two numbers are (40, 480) or (120, 160).