Input the HCF and LCM of two numbers and calculate the numbers quickly and accurately.
Assume the two numbers are \( x \) and \( y \), with their highest common factor (\( \text{HCF} \)) and least common multiple (\( \text{LCM} \)) provided.
According to the basic properties of numbers, we know: \( x \times y = \text{HCF} \times \text{LCM} \) Denote the product of \( x \) and \( y \) as \( P \): \( P = \text{HCF} \times \text{LCM} \)
Let \( x = \text{HCF} \times a \) and \( y = \text{HCF} \times b \), where \( a \) and \( b \) are coprime integers (i.e., their greatest common divisor is 1). From this, we derive: \( a \times b = \frac{P}{\text{HCF}^2} = \frac{\text{HCF} \times \text{LCM}}{\text{HCF}^2} = \frac{\text{LCM}}{\text{HCF}} \)
Based on the calculated value \( a \times b = \frac{\text{LCM}}{\text{HCF}} \), determine all pairs of integers that satisfy the equation and are coprime.
Multiply the HCF by the determined values of \( a \) and \( b \) to compute \( x \) and \( y \): \( x = \text{HCF} \times a \) \( y = \text{HCF} \times b \)
Solution:
1. Compute the product \( P \):
\( P = \text{HCF} \times \text{LCM} = 179 \times 2685 = 480015 \)
2. Compute \( a \times b \):
\( a \times b = \frac{\text{LCM}}{\text{HCF}} = \frac{2685}{179} = 15 \)
3. Find coprime pairs for \( a \times b = 15 \):
Factorize 15 into pairs: (1, 15) and (3, 5). Both are coprime.
4. Calculate \( x \) and \( y \):
\( x_1 = 179 \times 1 = 179 \)
\( y_1 = 179 \times 15 = 2685 \)
\( x_2 = 179 \times 3 = 537 \)
\( y_2 = 179 \times 5 = 895 \)
Result: The numbers are (179, 2685) or (537, 895).
Solution:
1. Compute the product \( P \):
\( P = \text{HCF} \times \text{LCM} = 11 \times 7700 = 84700 \)
2. Compute \( a \times b \):
\( a \times b = \frac{\text{LCM}}{\text{HCF}} = \frac{7700}{11} = 700 \)
3. Find coprime pairs for \( a \times b = 700 \):
Factorize 700 into pairs:
Coprime pairs are (1, 700), (4, 175), (7, 100), (25, 28).
4. Calculate \( x \) and \( y \):
\( x_1 = 11 \times 1 = 11 \)
\( y_1 = 11 \times 700 = 7700 \)
\( x_2 = 11 \times 4 = 44 \)
\( y_2 = 11 \times 175 = 1925 \)
\( x_3 = 11 \times 7 = 77 \)
\( y_3 = 11 \times 100 = 1100 \)
\( x_4 = 11 \times 25 = 275 \)
\( y_4 = 11 \times 28 = 308 \)
Result: The numbers are (11, 7700), (44, 1925), (77, 1100) or (275, 308).