The end of E×4 is a, and the end of x 4 can only be even, so A=2.

2× 4 = 8. If the carry problem is considered, E is 8 or 9, 4× 8 = 32, 4× 9 = 36, and the last digit is 2, so E=8.

We can know that B×4 has no carry, so b is either 1 or 2, but A=2, ABCDE is different, so b = 1.

E×4 goes into 3, and the number of digits plus D×4 is B = 1, so the number of digits of D×4 is 8, D is 2 or 7, and A is 2, so D = 7.

D × 4+3 = 3 1 in 3, C × 4+3 mantissa is C, D-B × 4 = 3, and we can deduce that C = 9.

So ABCDE is 2 1978.