The product of two consecutive natural numbers which are multliples of 3 is equal to 810. Find the two numbers.

Answer

To solve the problem of finding two consecutive natural numbers that are multiples of 3 and whose product equals 810, we can follow these steps:

### Step 1: Define the Variables
Let the first multiple of 3 be \( x \). Since we are looking for consecutive multiples of 3, the next consecutive multiple of 3 will be \( x + 3 \).

### Step 2: Set Up the Equation
According to the problem, the product of these two numbers is equal to 810. Therefore, we can write the equation:
\[ x \cdot (x + 3) = 810 \]

### Step 3: Expand the Equation
Expanding the left-hand side gives us:
\[ x^2 + 3x = 810 \]

### Step 4: Rearrange to Form a Quadratic Equation
Rearranging the equation to set it to zero:
\[ x^2 + 3x - 810 = 0 \]

### Step 5: Factor the Quadratic Equation
To factor the quadratic equation, we need two numbers that multiply to \(-810\) and add up to \(3\). After examining the factors of 810, we find:
\[ 30 \text{ and } -27 \]
Thus, we can write:
\[ (x + 30)(x - 27) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x + 30 = 0 \) → \( x = -30 \) (not a natural number)
2. \( x - 27 = 0 \) → \( x = 27 \) (a valid natural number)

### Step 7: Find the Consecutive Numbers
Now that we have \( x = 27 \), we can find the consecutive multiple of 3:
- The first number is \( 27 \)
- The second number is \( 27 + 3 = 30 \)

### Conclusion
The two consecutive natural numbers that are multiples of 3 and whose product is 810 are:
\[ \boxed{27 \text{ and } 30} \]

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