In the addition problem find the digit represented by C. Each time a letter appears it represent the same digit. A does not equal 0 (zero).
AB
AB
AB
+ AB
______
CA
please explain your results
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To find the digit represented by C in the addition problem, let's look at the problem step-by-step.
We start with adding two digits (A and A) in the units place. Since each letter represents the same digit, this can be written as 2A + A + A + A = CA, where A is a single digit.
Simplifying the equation, we have 4A = CA. We can now analyze the possibilities:
If C = 1, then 4A = 1A. However, we know that A cannot be zero, so C cannot be equal to 1.
If C = 2, then 4A = 2A. In this case, A must be zero (A = 0), and the result is 40 + 40 + 40 + 40 = 160. However, the given condition states that A cannot be zero, so this is not a valid solution.
If C = 3, then 4A = 3A. In this case, A must be zero (A = 0), and the result is 30 + 30 + 30 + 30 = 120. However, the given condition states that A cannot be zero, so this is not a valid solution either.
If C = 4, then 4A = 4A. In this case, A can be any digit from 1 to 9 (excluding zero), and the result will satisfy the equation. For example, if we choose A = 2, then the result is 42 + 42 + 42 + 42 = 168.
Therefore, the digit represented by C is 4, and any digit from 1 to 9 (excluding zero) can be represented by A.
To find the value of the digit represented by C in the addition problem, we need to solve the sum. Let's break it down step by step:
Given:
AB
AB
AB
+ AB
_____
CA
1. Start by adding the rightmost column, which consists of the digit B added four times. Since A does not equal 0, the maximum value of B can be 9 (to yield a two-digit result for the sum). Therefore, the maximum sum for the rightmost column is 9 + 9 + 9 + 9 = 36.
2. The sum of the rightmost column gives us a value in the units place (6). Write this down and carry over any tens place value (if applicable).
AB
AB
AB
+ AB
_____
C6
3. Move to the next column (where A is located). Since A represents the same digit, adding the column means adding A four times, which gives us a sum ranging from 4A to 9A (depending on the value of A).
4. We need to consider the carryover from the previous column, which may represent a value in the tens place. If this carryover exists, it should be added to the sum from step 3. Let's call the carryover X.
AB
AB
AB
+ AB
_____
XC6
5. In the tens place, the maximum sum will be 9A + X. Since the maximum value for AB is 99, the maximum sum cannot exceed 9(9) + 9 = 90 + 9 = 99.
6. Now, let's look at the sum in the tens place: (9A + X). If this sum is less than 100, it can only be 10X or 20X, as it needs to be a two-digit number. If it is greater than or equal to 100, then X must be 1 and the sum is 100 + (9A mod 10).
7. Finally, let's analyze the leftmost column (where C is). To obtain C, we add 1 + (9A mod 10) to the sum in step 6. If 9A mod 10 is 0, then C is 1. Otherwise, C is 2.
Therefore, the value of C in the addition problem could be either 1 or 2, depending on the value of A.