Try to be an eagle and then put forward one hypothesis after another.
× 4 We assume that the contention is 1, so any number multiplied by 4 cannot be equal to an odd number, so the contention can only be 2.
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How many times is 4=2 (or 12, 22, 32, 42)? There are only three and eight numbers.
Suppose the eagle is 2 (see ten thousand places), then 2×4=8≠2, then the eagle is 8.
Look at thousands more. When ×4= young, we can synthesize 10,000 digits. When we find that ×4 can't be carried, it can only be 0, 1 (the digits can't be the same, so there is no 2). Look at it, 8 (Eagle) ×4=32, and enter three. If it is 0, young× 4+3 equals several tens (multiples of 10), 10-3=7, young× 4 equals several seventeen (odd numbers), but it is not true, so it is 65438.
After the unit enters three, chicks× 4 = dozens 1. If you don't score three, the chicken × 4 equals dozens 18, then the chicken is 2 or 7. Suppose the chicken is 2, and the chicken × 4 will not carry. X×4= a number with X, there is none, so young is 7.
Look at thousands, 4×2 (contention) =8 (eagle), and thousands do not carry. 1 (when) ×4=4, and the young one is 7, so one percent should enter 3. 4×X= 30, X=8 or 9, and the numbers cannot be repeated, so the decimal number is 9.
Answer: for =2, when = 1, small =9, young =7, eagle =8.