Question

The sum of three consecutive numbers is 59 less than four times the largest number. Find the largest number. Show your work.

A) 20
B) 21
C) 22
D) 23

Answer 1

To find the largest number, we can set up an equation using three consecutive numbers. By solving the equation, we can determine the value of the largest number. The correct answer is 23.

This correct answer is D)

Step-by-step explanation:

To solve this problem, let's assume that the three consecutive numbers are x, x + 1, and x + 2. The sum of these numbers is x + (x + 1) + (x + 2).

According to the problem, this sum is 59 less than four times the largest number, which is 4(x + 2) - 59. Setting these two expressions equal, we have x + (x + 1) + (x + 2) = 4(x + 2) - 59. Solving this equation gives us x = 21, so the largest number is x + 2 = 23.

Therefore, the correct answer is D) 23.

This correct answer is c)

Answer 2

The sum of three consecutive numbers is 59 less than four times the largest number. The largest number is C) 22

Step-by-step explanation:

Let's denote the three consecutive numbers as
`\(n\), \(n+1\), and \(n+2\),` where
`\(n\)` represents the smallest number. According to the given information, the sum of these three numbers is equal to
`\(4\)` times the largest number
`(\(n+2\)) minus \(59\):`

`\[ n + (n+1) + (n+2) = 4(n+2) - 59 \]`

Now, solve for
`\(n\):`

`\[ 3n + 3 = 4n + 8 - 59 \]`

Combine like terms:

`\[ 3n + 3 = 4n - 51 \]`

Subtract
`\(3n\)`from both sides:

`\[ 3 = n - 51 \]`

Add
`\(51\)` to both sides:

`\[ n = 54 \]`

So, the three consecutive numbers are
`\(54\), \(55\), and \(56\).` The largest number is
`\(56\),`which corresponds to option C) 22. Therefore, the correct answer is
`\(n+2 = 54 + 2 = 56\).`

In summary, the solution involves setting up an equation based on the given relationship between the sum of consecutive numbers and four times the largest number minus
`\(59\`). Solving this equation systematically yields the value for the smallest number, and adding
`\(2\)` to it gives the largest number in the sequence.