Many of you submitted solutions, but the only ones finding all three solutions were Lou Cairoli, Fransesc Suñol y Esquirol.

We begin by observing that A, B, or N = 0 trivializes the problem.  Also if D = 0, then C = 0, which violates the "distinct" condition.

Re-writing the first equation, we have A*N = 10*B + C  (1) .  Similarly, the second equation becomes 10A*N + D*N = 110B + C.  Substituting this from (1), we get 100B + 10C + D*N = 110B + C and then D*N = 10B - 9C  (2).  Combining (1) and (2), we get

from which we determine that 5 | (A - D) or 5 | N.

Case 1:  If 5 | N, then A*N = BC implies C = 0 or C = 5 , but if C = 0 we see from (3) that either N = 0 or A = D, neither of which is allowed.  Hence C = 5. So (A - D)*N = 50. So N is either 10, 25, or 50.  But N = 10 or 50 implies that C = 0, which we've already ruled out. So N = 25 and A - D = 2. Since A*N < 100, we get A < 4 from which it quickly follows that A = 3 and D = 1.  So the only solution in this case is A = 3, D = 1, N = 25, B = 7, C = 5.

Case 2:  If 5 | (A - D), then since A and D are distinct, single-digit integers, A - D = 5. So A ³ 6 and D £ 4.  From (3) we have 5N = 10C, or N = 2C.  So N is even.  Also (1) then tells us that C is even, too. Interestingly, whether C is 2, 4, 6, or 8 (not 0 since C = 0 implies N = 0), then A = 8.  See table below.
 
 

Since A = 8, then D = 3, and C must be 2, 4, or 6.  But if C = 2, then B = 3 = D, which is out, so C = 4 or 6 which yields the final two solutions:  A = 8, D = 3, N = 8, B = 6, C = 4  and  A = 8, D = 3, N = 12, B = 9, C = 6.

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