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A, B, C, and D are positive digits such that $$\overline {ABCD} + \overline {BCD} +\overline {CD} +\overline {D} = 2012.$$ Find A+B+C+D.

Jan 3, 2021

#1
+808
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Hello ss333!

Background

The line on top of ABCD means to read it as a number instead of A*B*C*D.

Equation

line(ABCD) + line(BCD) + line(CD) + line(D) = 2012

(1000A + 100B + 10C + D) + (100B + 10C + D) + (10C + D) + (D) = 2012

1000A + 200B + 30C + 4D = 2012.

Letter A

First, let's try to find A.

A has to be 1 or 2, since if A was 3, then 1000*A = 3000, which is bigger than 2012.

However, if we try setting A as 2, we find out that B would be 0 since 200B + 30C + 4D = 12.

B can't be 0 since B has to be a positive digit, making A 1.

Equation

A = 1

1000(1) + 200B + 30C + 4D = 2012.

200B + 30C + 4D = 1012

100B + 15C + 2D = 506

Letter B

Following the same logic as letter A, B has to be in the range 1-5 since if B was 6, 100*B = 600, which is greater than 506.

However, if B were 5, then C would be 0, and if B were less than or equal to 3, then C would be greater than 10, which isn't a digit number, making B 4.

Equation

B = 4

100(4) + 15C + 2D = 506.

15C + 2D = 106.

Letter C

Following the same letter as A and B, C has to be in the range 1-7 since if C were greater than 7, then 15C > 106.

However, if C were less than 6, then D would be greater than 10 and not a digit.

Similarly, if C were greater than 6, then D would be a decimal number.

Thus C has to be 6.

Equation

15(6) + 2D = 106.

2D = 16

D = 8

A = 1

B = 4

C = 6

D = 8

1+4+6+8 = 19.