2 1978*4=879 12

Let cluster =a small =b love =c number =d learning = e.

A must be an even number, which can be found in the relationship between "learning" and "learning"

A cannot be greater than 2, because 5 digits remain unchanged.

So a=2, e=3 or 8.

B is less than 5 because 5*4 will enter 2, but a*4 is already 8.

At this time, on the premise that A, b.c.d.e are not equal.

There are only eight situations. Try one at a time.

40000 a+4000 b+400 c+40d+4e = 10000 e+ 1000d+ 100 c+ 10 b+e

(2)

1089*9=980 1

Spring: 1

Xia: 0

Autumn: 8

Winter: 9

A 4-digit× 9 or 4-digit, and the thousand digits can only be1; And because 1×9=9, spring and winter can be deduced.

∴ 1 summer and autumn 9×9=9 autumn and summer 1

∫11* 9 = 9999, so this number does not exceed11.

Then, if summer is 1, it is only 1 109*9, but the calculation does not meet the meaning of the question, so summer is 0.

10 Autumn 9*9=9 Autumn 0 1

9 Autumn 0 1 is a multiple of 9, so autumn is 8.

Taking these four values into account, we find that the equation holds, so spring = 1, summer =0, autumn =8 and winter =9.