Question

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

Answer 1

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:

\[
\boxed{19}
\]