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The problem in the image asks to find the values of \( A \), \( B \), and \( C \) given that:

1. \( A \) and \( B \) are consecutive numbers.
2. The sum of the two numbers \( AB \) and \( AC \) gives \( 1BB \).

Let's break this down:

### Step 1: Translate the digits to numbers
- \( AB \) is a two-digit number where \( A \) is the tens digit and \( B \) is the units digit. Thus, \( AB = 10A + B \).
- \( AC \) is a two-digit number where \( A \) is the tens digit and \( C \) is the units digit. Thus, \( AC = 10A + C \).
- \( 1BB \) is a three-digit number where \( 1 \) is the hundreds digit, and both \( B \)'s are the tens and ones digits. Thus, \( 1BB = 100 + 10B + B = 100 + 11B \).

### Step 2: Set up the equation
We now know the sum of \( AB \) and \( AC \) equals \( 1BB \). So the equation is:
\[
(10A + B) + (10A + C) = 100 + 11B
\]
Simplifying this equation:
\[
20A + B + C = 100 + 11B
\]

### Step 3: Use the information that \( A \) and \( B \) are consecutive
We know \( A \) and \( B \) are consecutive numbers, so we can write \( A = B - 1 \).

### Step 4: Substitute \( A = B - 1 \) into the equation
Substituting \( A = B - 1 \) into the equation \( 20A + B + C = 100 + 11B \), we get:
\[
20(B - 1) + B + C = 100 + 11B
\]
Expanding this equation:
\[
20B - 20 + B + C = 100 + 11B
\]
\[
21B - 20 + C = 100 + 11B
\]
Simplifying further:
\[
21B + C = 120 + 11B
\]
Now subtract \( 11B \) from both sides:
\[
10B + C = 120
\]

### Step 5: Solve for \( B \) and \( C \)
From the equation \( 10B + C = 120 \), we can see that \( C = 120 - 10B \).

Now, since \( B \) is a digit (between 0 and 9), let’s test possible values for \( B \).

- For \( B = 7 \):
  \[
  C = 120 - 10(7) = 120 - 70 = 50
  \]
  This is not possible because \( C \) must be a single digit.

- For \( B = 6 \):
  \[
  C = 120 - 10(6) = 120 - 60 = 60
  \]
  Again, not possible.

- For \( B = 5 \):
  \[
  C = 120 - 10(5) = 120 - 50 = 70
  \]
  This also does not work.

- For \( B = 4 \):
  \[
  C = 120 - 10(4) = 120 - 40 = 80
  \]
  No.

- For \( B = 3 \):
  \[
  C = 120 - 10(3) = 120 - 30 = 90
  \]
  



Si A y B son consecutivos halla A, B y C.

Solve the problem where A and B are consecutive digits and the sum of AB and AC gives 1BB. This algebraic puzzle focuses on forming and solving equations involving consecutive numbers. Suitable for high school students.

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