18* 18* 18=5832, 18* 18* 18* 18= 104976

Let's first study the possible "upper limit" of Wiener's age:

It is not difficult to find that the cube of 2 1 is four digits, while the cube of 22 is already five digits, so Wiener's age is at most 2 1 year;

Let's study the possible "lower limit" of Weiner's age: the fourth power of 18 is six digits, and the fourth power of 17 is five digits, so Weiner's age is at least 18 years old.

In this way, Wiener's age can only be one of the four numbers 18, 19, 20, 2 1. The remaining work is to screen one by one.

The cube of 20 is 8000, there are three repeated numbers 0, it doesn't matter.

Similarly, it is irrelevant that the fourth power of 19 equals 13032 1 and the fourth power of 2 1 equals 19448 1.

There is only one 18 in the end. Is this the correct answer? Check, the cube of 18 is equal to 5832, and the fourth power is equal to 104976, which is exactly ten Arabic numerals. What a perfect combination! This method of solving problems is called exclusion.

Hope to adopt. Thank you.