The sum of three consecutive numbers is 41 less than 5 times the largest number. Find the largest number.

1 answer

Let the three consecutive numbers be \(x\), \(x+1\), and \(x+2\).

The sum of these three numbers can be expressed as:

\[ x + (x + 1) + (x + 2) = 3x + 3 \]

The largest of these three numbers is \(x + 2\). According to the problem, the sum of the three consecutive numbers is 41 less than 5 times the largest number. This can be set up as the equation:

\[ 3x + 3 = 5(x + 2) - 41 \]

Now, we will simplify the right side:

\[ 3x + 3 = 5x + 10 - 41 \] \[ 3x + 3 = 5x - 31 \]

Next, we will rearrange the equation to isolate \(x\):

\[ 3 + 31 = 5x - 3x \] \[ 34 = 2x \]

Now, we divide both sides by 2:

\[ x = 17 \]

Thus, the three consecutive numbers are:

\[ 17, 18, 19 \]

The largest number among them is:

\[ x + 2 = 17 + 2 = 19 \]

To confirm, we check the condition stated in the problem:

The sum of the numbers is:

\[ 17 + 18 + 19 = 54 \]

5 times the largest number is:

\[ 5 \times 19 = 95 \]

Now, we check if the sum is 41 less than this value:

\[ 95 - 41 = 54 \]

Since both sides match, the solution is verified. Therefore, the largest number is:

\[ \boxed{19} \]