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This is a tough 'un. First of all I'm guessing the sum is abcd, not cbcd as you wrote.

(1) b<c, Since the sums that lead to b and c are identical (y+v in the tens place and v+y in the hundreds place), the only way that this can happen is if the sum in the units place (v+z) is at least 10, and y+v is less than 10. This means that v+z = v+ (v+3) = 2v+3 > 10. Also we know that v+y = v+(v-3)=2v-3 < 10. This doesn't tell us what v is, only that v is between 4 and 6. Not suff.

(2) z>7, since z=v+3 this means that v>4. Again not suff.

Taking them together all we know is that v is either 5 or 6. But there is no way to know exactly what is a=v+3.

I think answer is D. From the information given we have: d=2v+3 c= 2v-3 b= 2v-3 c=3+v = z 1) b<c => 2v-3<3+v => v<6 We know that c#b, unless z+v>10. From this b+1=c. From this v>3. So the only possible values of v are 4 and 5. For v=4 is impossible, so v=5 - suff 2) z>7 => c>7 => 3+v>7 => v>4. So v could have possible values 5,6,7,8,9. For every v>=6 the result will have 5 digits. So v=5.