The consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectively, they add up to 74. find these numbers.

Answer 1

Let the consecutive integers be \( n, n+1, n+2 \).

According to the problem, when these integers are multiplied by 2, 3, and 4 respectively and added together, the sum is 74. This leads to the equation:

\[ 2n + 3(n + 1) + 4(n + 2) = 74 \]

Now, let's simplify this equation step by step:

1. Expand the terms: \[ 2n + 3n + 3 + 4n + 8 = 74 \]

2. Combine like terms: \[ (2n + 3n + 4n) + (3 + 8) = 74 \] \[ 9n + 11 = 74 \]

3. Subtract 11 from both sides: \[ 9n = 74 - 11 \] \[ 9n = 63 \]

4. Divide by 9: \[ n = \frac{63}{9} = 7 \]

So the consecutive integers are: \[ n = 7, \quad n+1 = 8, \quad n+2 = 9 \]

Thus, the three consecutive integers are 7, 8, and 9.

To verify:

• Multiply 7 by 2: \( 2 \times 7 = 14 \)
• Multiply 8 by 3: \( 3 \times 8 = 24 \)
• Multiply 9 by 4: \( 4 \times 9 = 36 \)

Now, add them up: \[ 14 + 24 + 36 = 74 \]

Since the sum matches the requirement of the problem, the consecutive integers are indeed 7, 8, and 9.