Answer:
Let's call the three-digit number whose digits are all the same "n". We know that n must be between 111 and 999, inclusive.
Now let's consider a three-digit number x. We can write x as:
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x = 100a + 10b + c
where a, b, and c are the hundreds, tens, and one's digits of x, respectively. The sum of the digits of x is:
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a + b + c
If we subtract this sum from x, we get:
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100a + 10b + c - (a + b + c) = 99a + 9b
We want this to be equal to n, which means:
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99a + 9b = n
We know that n has three identical digits, so we can write:
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n = 111p
where p is a digit between 1 and 9. Substituting this into the equation above, we get:
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99a + 9b = 111p
11a + b = 37p
Since a and b are digits between 0 and 9, we can see that 37p must be between 0 and 99. The possible values of p are therefore 1, 2, and 3.
If p = 1, then 37p = 37, and we need to find two digits a and b such that 11a + b = 37. This gives us the solutions (a, b) = (3, 4) and (a, b) = (4, 3), which correspond to the numbers 343 and 434.
If p = 2, then 37p = 74, and we need to find two digits a and b such that 11a + b = 74. There are no such solutions since the largest possible value of 11a + b is 99.
If p = 3, then 37p = 111, and we need to find two digits a and b such that 11a + b = 111. This gives us the solution (a, b) = (10, 1), which corresponds to the number 1001. However, 1001 is not a three-digit number, so it does not count.
Therefore, the only two three-digit numbers that satisfy the conditions of the problem are 343 and 434. There are 2 such numbers.
Explanation: