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| Here's how we can solve this problem step-by-step:
Understanding the Problem
We have a four-digit number, A.
Moving the last digit of A to the front creates a new four-digit number, B.
The sum of A and B is a four-digit number, C.
The sum of the digits in A is 20.
The hundreds digit of C is 0, and the tens digit of C is 4.
Let's use variables:
Let the digits of A be represented as a, b, c, and d, where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the units digit.
So, A = 1000a + 100b + 10*c + d
And B = 1000d + 100a + 10*b + c
Formulating Equations
Sum of digits of A: a + b + c + d = 20
Structure of C: C = 1000*_ + 100*0 + 10*4 + _ (We don't know the thousands and units digits of C yet)
A + B = C:
(1000*a + 100*b + 10*c + d) + (1000*d + 100*a + 10*b + c) = C
Analyzing the Sum (A + B = C)
When we add A and B, let's consider the possible carry-overs:
Units column: d + c results in the units digit of C. There might be a carry-over to the tens column.
Tens column: c + b plus any carry-over from the units column results in 4 (the tens digit of C). There might be a carry-over to the hundreds column.
Hundreds column: b + a plus any carry-over from the tens column results in 0 (the hundreds digit of C). This means there must be a carry-over of 1 to the thousands column (since b + a cannot be negative).
Thousands column: a + d plus the carry-over of 1 from the hundreds column results in the thousands digit of C.
Solving the Equations
Hundreds column: Since b + a plus a carry-over results in 0, and we know there's a carry-over of 1 to the thousands column, we have:
b + a + 1 (carry-over) = 10
Therefore, a + b = 9
Tens column: We know c + b plus a possible carry-over from the units column results in 4. Let's consider two cases:
Case 1: No carry-over from the units column:
c + b = 4
Since a + b = 9, and a + b + c + d = 20, then d = 7.
But if d = 7, then in the units column, d + c (7 + c) would need to result in a number greater than or equal to 10 to have a carry-over. This would make c at least 3, and c + b would be greater than 4, contradicting our assumption of no carry-over. Therefore, this case is not possible.
Case 2: Carry-over of 1 from the units column:
c + b + 1 = 4
Therefore, c + b = 3
From equations a + b = 9 and c + b = 3, and knowing a + b + c + d = 20, we can deduce:
a = 9-b, c = 3-b and 9 + 3 - b + d = 20, then b + d = 8
Units column: Since there's a carry-over of 1 to the tens column, d + c must be greater than or equal to 10. Substituting c = 3-b, we get:
d + 3 - b >= 10
d - b >= 7
Combining d - b >= 7 and d + b = 8, we find that the only possible values are:
d = 8, b = 0
With d = 8 and b = 0, we can find the other values:
a = 9, c = 3
Calculating C:
A = 9038
B = 8903
C = A + B = 17941
However, the problem states that C is a four-digit number. This means our initial assumption that A is a four-digit number was incorrect. Let's adjust our approach.
Revised Approach: A as a Three-Digit Number
Let's assume A is a three-digit number (abc) and B becomes a four-digit number (bca) when the units digit is moved to the front.
A = 100a + 10b + c
B = 1000c + 100a + 10b
A + B = C
a + b + c = 20
C = 1000*_ + 100*0 + 10*4 + _
Analyzing the Sum (A + B = C) - Revised
Units column: c + b results in the units digit of C. There might be a carry-over.
Tens column: b + a plus any carry-over results in 4. There might be a carry-over.
Hundreds column: a + c plus any carry-over results in 0. There must be a carry-over of 1.
Thousands column: The carry-over of 1 is the thousands digit of C.
Solving the Equations - Revised
Hundreds column: a + c + 1 (carry-over) = 10, so a + c = 9.
Tens column:
Case 1: No carry-over from the units column: a + b = 4. But since a + b + c = 20, this would mean c = 16, which is not possible.
Case 2: Carry-over of 1 from the units column: a + b + 1 = 4, so a + b = 3.
Combining a + b = 3 and a + c = 9 with a + b + c = 20, we get:
3 + c = 20
Therefore c = 17, still not possible.
The issue is that if C is a 4 digit number, then A cannot be a 3 digit number, because the only way to have a carry of 1 into the thousands column is if A is a 4 digit number as well.
Let's try one more approach: a, b, c, d must not all be single digits
We know a+b = 9, c+b=3, and a+b+c+d=20. If b=0, then a=9, c=3, and d=8, which gives us C as 17941. This means that at least one of a, b, c, d must be greater than 9.
Since b is in both a+b=9 and c+b=3, we know that if b is a two-digit number, these equations cannot both be true. Therefore, b must be a single digit.
If c is a two digit number, then the addition of c+b in the tens column would need to be 4, but there is also a carry-over of 1 from the units column. This means that c+b=13, and if c=10, b=3, a=-1, d=8. This also doesn't work.
If d is a two-digit number, then when added to c, there must be a carry-over of 1. If d=10, then 10+c >= 10, which is true for any single-digit c. However, we know c=3-b, and if c is negative, then a must be greater than 9 to compensate, which we have already tested.
If c=0, then b=3, a=6, d=11. This gives us 6300+11, 360+11, and 6300+360+22 = 6682.
Answer
C represents the number 6682.
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