A, B, C, and D are positive digits such that ABCD + BCD = 2012. Find A + B + C + D.

Guest Dec 22, 2021

#1**0 **

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?

tinfoilhat Dec 22, 2021