SOLUTION: A, B, C, and D, are distinct digits, and 4(AAB) = CDA. If C is less than d, find the sum of a +b+c+d

Question 1105587: A, B, C, and D, are distinct digits, and 4(AAB) = CDA. If C is less than d, find the sum of a +b+c+d
Answer by ikleyn(52800) (Show Source): You can put this solution on YOUR website!
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A, B, C, and D, are distinct digits, and 4(AAB) = CDA. If C is less than d, find the sum of a +b+c+d
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1.  From 4*(AAB) = CDA,  it follows that  A is an even number/digit different from zero and not more than 2.

    It leaves only one possibility for A:  A = 2.



2.  Since A = 2, it leaves only two possibilities for C:  C= 8  or  C= 9.

    Since, in addition,  C is less than D,  it implies that C= 8  and  D= 9.



3.  Then CDA = 892,  and it implies that  AAB =  = 223.



4.  Thus we just found that  AAB = 223,  CDA = 892,

    A= 2,  B= 3,  C= 8,  D= 9.

    Then A + B + C + D = 2 + 3 + 8 + 9 = 22.


Answer.  A + B + C + D = 22.

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