If A B B × 999

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 $A = 1$, $B = 0$, $C = 9$.

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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.