ABCD+ABCD+ABCD+ABCD=DCBA ABCD=?

假設A不會零的話, A的數值是1或2, 因為大過2便不可能四個數加起來的答案還是四個位的.

A不會是1的, 因為四個相同的數字加起來, 一定是雙數, 而尾數是A, 那A唯一可能是2了.

如要得出答案尾數A是2的話, 那D可能是3或8, 但因為加數的千位數是2, 所以不可能是3了, 因加起來千位數必定是8或9, 所以唯一選擇, D=8

因為加數的千位數A=2, 所以為免百位數進位, 而影響千位數答案, 所以只有1或2的選擇.

那麼要是什麼值的C可以得出B的答案呢?

假設B=1, 4C+3 (3是8*4的進位)的個位數是1的話, C會是2,7, 但因為四個B加起來必定大過或等如4, 所以唯一選擇是7了

假設B=2, 4C+3的個位數是2的話, C不可能等如任何一個整整數, 因一個雙數加單數必定是雙數, 所以B=1

所以答案是 2178.


2006-12-22 11:20:38 補充：
樓上的, 你沒有答B和C的答案

Assume A, B, C, D are integers and only one of them can be zero.

From the sum of the unit column, we have
4D = A or 4D = 10 + A or 4D = 20 + A or 4D =30 + A (Eq 1)

From the sum of the thousandth column, we have
4A = D or 4A + 1 = D or 4A + 2 = D or 4A + 3 = D (Eq 2)

Noting that A and D are integers, from Eq 1
When 4D = A, A cannot be zero, A cannot be 1 (D would then be 1/4, not an integer), ...
A can only be equal to 4 or 8
so (A,D) = (4,1) or (8,2)
When 4D = 10 + A, (A,D) = (2,3) or (6,4)
When 4D = 20 + A, (A,D) = (4,6) or (8,7)
When 4D = 30 + A, (A,D) = (2,8) or (6,9)

Testing the possibility of each of the above by substituting into (Eq 2), the only possible values of (A,D) is (2,8), therefore A=2, D =8.

Similarly from the sum of the tenth column: 4C+3 = B or 4C+3=10+B or 4C+3=20+B or 4C+3=30+B. (Eq 3)
From the sum of the hundredth column: 4B=C or 4B+1=C or 4B+2=C or 4B+3=C (Eq 4)
From (Eq 3), (B,C) can only be one of the followings: (7,1), (1,2), (5,3), (9,4), (3,5), (7,6), (11,7), (1,7), (5,8), (9,9)
Testing the possibility of each of the above by substituting into (Eq 4), the only possible values of (B,C) is (1,7), therefore B=1, C =7.

Ans: ABCD = 2178