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Mathematical sorting problem
Increment: the minimum number is three digits at the beginning of 1, so the decimal number can be 2-8, and if the decimal number is 2, it can be 3-9, that is, seven situations; If the decimal number is 3, then 4-9 can be used, that is, there are 6 situations; And so on: ten digits is four, and there are five situations; Ten is five, and there are four situations; Ten is six, and there are three situations; Ten is seven, there are two situations; Ten digits are eight, and there is a case of 1. There are ***28 cases, that is, 28 hundred digits are 1. Similarly, when the percentile is 2-7, there are 2 1, 15, 10, 6, 3 and 1, and the * * count is1+3+6+1.

Decreasing: the maximum number is three digits starting with 9, so the decimal digit can be 1-8, and if the decimal digit is 8, it can be 0-7, that is, there are eight situations; If the decimal number is 7, you can use 0-6, that is, there are 7 situations; And so on: ten digits is six, and there are six situations; Ten is five, and there are five situations; Ten is four, and there are four situations; Ten is three, and there are three situations; Ten is two, there are two situations; The decimal number is 1, and there are 1 cases. There are 36 cases, that is, 36 digits are 9. Similarly, when the percentile is 2-8, there are 1, 3,6, 10,15,2128, and * * is1+3+/kloc-0.

So, there are 204 such figures.

There are 28+2 1, 2 * *+0+ 1 = 50, all greater than 32 1, 320, 3 10 less than 32 1, so 326538.