Question 1151097: A, B, C and D are distinct digits, and 4 x AAB = CDA. If C is less than D, find the sum A+B+C+D.
Answer by ikleyn(52781)   (Show Source):
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Since AAB is 3-digit number and CDA is 3-digit number, there are only 2 possibilities for A: A= 1 or A= 2.
But A is the last digit of the number CDA, which is multiple of 4; so it leaves only one possibility for A to be 2: A = 2.
Then B = 3 or B = 8.
Case 1). If B = 3, then 4 * 223 = 892; so, C= 8 and D= 9.
Case 2). If B = 8, then 4 * 228 = 912; so, C= 9 and D= 1.
The condition C < D is satisfied in case 1 ONLY; so, case 2) is denied.
Therefore, A+B+c+D = 2+3+8+9 = 22. ANSWER
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