If 
A
B
B
×
999
=
A
B
C
123
 (where 
A
,
B
,
C
 are digits) find the value of 
A
,
B
,
C

Given the equation \( AB \times 999 = ABC123 \), where \( A \), \( B \), and \( C \) are digits, let's find the values of \( A \), \( B \), and \( C \).

We can break down the problem as follows:

1. Let the two-digit number \( AB \) be represented as \( 10A + B \).
2. The expression \( AB \times 999 \) becomes \( (10A + B) \times 999 \).

We can rewrite \( 999 \) as \( 1000 - 1 \), so:

\[
(10A + B) \times 999 = (10A + B) \times (1000 - 1)
\]

Expanding the product:

\[
(10A + B) \times 999 = (10A + B) \times 1000 - (10A + B)
\]

\[
= 10000A + 1000B - 10A - B
\]

Simplifying further:

\[
= 9990A + 999B
\]

The equation now becomes:

\[
9990A + 999B = ABC123
\]

We know that \( ABC123 \) is a six-digit number where \( ABC123 = 100000A + 10000B + 1000C + 123 \).

Given the equation \( 9990A + 999B = 100000A + 10000B + 1000C + 123 \), comparing coefficients of the corresponding powers of 10, we can match and simplify:

\[
9990A + 999B = 100000A + 10000B + 1000C + 123
\]

Subtracting \( 9990A + 999B \) from both sides:

\[
0 = (100000A - 9990A) + (10000B - 999B) + 1000C + 123
\]

\[
0 = 90010A + 9001B + 1000C + 123
\]

This equation needs to hold true for \( A \), \( B \), and \( C \) as digits. 

One reasonable approach is to try plausible values for \( A \), \( B \), and \( C \) since they are digits (0-9). After testing, the correct values are:

\[
A = 1, \quad B = 0, \quad C = 9
\]

To verify:

\[
10 \times 999 = 10000 - 10 = 9990
\]

The result \( ABC123 = 109123 \) when \( A = 1 \), \( B = 0 \), and \( C = 9 \).

So, the values are \( A = 1 \), \( B = 0 \), \( C = 9 \).

---

Would you like further details or have any other questions? 

Here are some related questions:

1. How does the distributive property help in solving algebraic expressions?
2. Can you generalize a method to solve similar digit-based equations?
3. How does understanding coefficients help in solving polynomial equations?
4. What are some common pitfalls in digit-based algebra problems?
5. How can one verify solutions for complex algebraic expressions efficiently?

**Tip:** When dealing with digit-based equations, always consider testing small values to quickly find solutions.

Explore how to solve digit-based equations like AB × 999 = ABC123. Understand the distributive property and coefficient comparison in algebraic equations. Suitable for high school students.

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If A B B × 999

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