Let * be a natural number $n$ a three-digit number, the sum of all three-digit numbers formed by the arbitrary permutation of its three non-zero digits minus $n$ equals 1990. Find $n$.


Let $n=\overline{a b c}$. In the sum, $a$ appears 2 times in the units, tens, and hundreds place, and the same is true for $b$ and $c$, so the sum is $222(a+b+c)$. 
1990 + 100 = 2090
222 \times 9 = 1998, 
222 \times 10 = 2220 = 1990 + 230, \\
222 \times 11 = 2442 = 1990 + 452, 
222 \times 12 = 2664 = 1990 + 674, \\
222 \times 13 = 2886 = 1990 + 896, 
222 \times 14 = 3108 > 2989 .
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Upon verification: when $a+b+c=11$, $n=452$ meets the condition.