A, B, C, and D are positive digits such that ABCD + BCD = 2012. Find A + B + C + D.
+133 Expand everything in base 10 form. 1000A + 100B + 10C + D + 100B + 10C + D = 2012.
1000A + 200B + 20C + 2D = 2012
Divide by 2.
500A + 100B + 10C + D = 1006.
D has to be 6.
500A + 100B + 10C = 1000
Divide by 10.
50A + 10B + C = 100
C has to be 0.
50A + 10B = 100
5A + B = 10
(A,B) = (1,5) or (2,0)
Solutions: ABCD = 1506 or 2006.
We cannot have a solution without a 0, so there is no solution.
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In case you mean A*B*C*D + B*C*D = 2012:
B*C*D(A+1) = 2012
One of these must include 503 (prime), which is not a digit. So this is not it either.
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I found another question that has a realistic solution at https://brainly.com/question/17335668, is this perhaps what you meant?
