A five-digit number multiplied by four equals another five-digit number. It is easy to draw the conclusion that A must be less than 3, because if it is greater than 3, it must be multiplied by 4 to get 6 digits.

A is either 2 or 1.

However, the last e of the four abcde, the last digit becomes a, and any single digit of the four ABCDE can't be 1 in the end, only 2.

So a=2, this part is 2bcde×4=edcb2.

Then e is either 3 or 8, because 4 can only be multiplied by these two numbers to get that the last digit is 2.

Look at the number on the tens of thousands of digits, which is 4 times that of abcde, and becomes E, 4 * 2 = 8; In other words, the smallest e is 8.

So we get e=8, which is 2bcd8×4=8dcb2.

It can be seen that when 4×b is used, the number on the ten thousand digits is not rounded. If rounded, the number at10,000 will be greater than 8.

So b is either 2 or 1. Similarly, the last bit of 4×d is b,

Because 4×8=32, a 3 is added to the ten digits, that is to say, the last digit of 4×d+3 is either 1 or 2, and if D is not multiplied by 4 plus 3, the last digit becomes 2; So the last one is 1.

That is, b= 1, and this part is 2 1cd8×4=8dc 12①.

To satisfy b= 1, then d can only be 2 or 7 (the last bit of 2 * 4+3 is1; The last digit of 7*4+3 is 1.

Draw the conclusion that d is 2 or 7.

Look at the formula (1), 2 1cd8×4. The number on thousands must be greater than 4 (because thousands don't carry thousands).

So d=7 to this part is 2 1c78×4=87c 12②.

Hope is just ahead.

4×7+3 =3 1, that is to say, the last bit of 4×c+3 is C, and the last bit of 4× c+3 is C, which leads to a thousand times 4 to become 7.

This C is only 9, so this number is 2 1978*4=879 12 (calculated secretly with a calculator, it succeeded).

2 1978= We love mathematics.