Find the values for the letters to make each of the folwing true. the letters A,B,C,and D represent even #'s?

Question

ABC+BBC=ECAB

AA+AA+BC=BDD

Answer 1

q2

AA + AA + BC = BDD

22A + 10B + C = 100B + 11D

11(2A - D) + C = (100 - 10)B = 90B

only B = 2 is possible, since B cannot be 0 (leading 0 is not allowed), and all are evenses.

11(2A - D) + C = 90*2 = 180

C = 4, from C < 10

2A - D = 16

only (A,D) = (8,0) is possible.

answer : A = 8, B = 2, C = 4, D = 0.

88 + 88 + 24 = 200

q1

ABC + BBC = ECAB

100(A+B) + 10(2B) + 2C = 1000E + 100C + 10A + B

E must be 1.

A(100 - 10) + B(100 + 2*10 - 1) = 1000 + C(100 - 2)

90A + 119B = 1000 + 98C

90(A + B - C) + 29B = 1000 + 8C

keeping the fact that all numbers (save E) are evenses in mind,

since 0 ≤ 8C ≤ 64

AND 0 ≤ 29B ≤ 232,

then 768 ≤ 90(A + B - C) ≤ 1064

we get 9 ≤ (A + B - C) ≤ 11,

but (even) + (even) - (even) = (even)

so A + B - C = 10

90*10 + 29B = 1000 + 8C

29B - 8C = 1000 - 900 = 100

only possibility is (B,C) = (4,2)

A + 4 - 2 = 10

A = 8

answer : A = 8, B = 4, C = 2, E = 1.

842 + 442 = 1284

Answer 2

I don't think this is possible.

The first obvious thing is that E is 1.  This is because two three-digit numbers are being added together and they result in  four digit number with E in the thousands column.  It can only be 1.  

Similarly, B must be 2.  The largest number you can get from adding two, two digit numbers together is 297, and as E is already 1, B can only be 2.

I'm certain my logic is rock-solid up to this point.

Now look at A.  If it's even then it must be 8, because in the top equation, A is added to B (which is 2) in the hundreds column and it gives an answer over 10.  We are told it's even, so the other possibility - 9 - is eliminated.

Logic still unimpeachable.

The units column of the top equation C + C >B  give C as 6 if it's even.

but ut;s here that things fall apart.  If C is 6 then the second equation falls to pieces.

Also D, using the units column of the second equation should  now fall into place and it doesn't.  Which is why I don't think this is possible.

Look at the Units column of the second equation a + a+ c = d

The second line is very similar a +a+b=d.  This must mean that b is one less than c, but that would make c 3.  And the rest all falls apart.

I don't think it's possible, unless someone can see the flaw in my logic.