急!!!快d!thx!係趣味題來的~

101,111,121,131,141,151,161,171,181,191,
202,212,222,232,242,252,262,272,282,292,
303,313,323,333,343,353,363,373,383,393,
404,414,424,434,444,454,464,474,484,494,
505,515,525,535,545,555,565,575,585,595,
606,616,626,636,646,656,666,676,686,696,
707,717,727,737,747,757,767,777,787,797,
808,818,828,838,848,858,868,878,888,898,
909,919,929,939,949,959,969,979,989,999
共有90個這樣的數字

一些數字(如101),它的個位數字及百位數字互換位置後,數值仍不會改變。在100至999中,共有多少個這樣的數字?


101,111,121,131,141,151,161,171,181,191,
202,212,222,232,242,252,262,272,282,292,
303,313,323,333,343,353,363,373,383,393,
404,414,424,434,444,454,464,474,484,494,
505,515,525,535,545,555,565,575,585,595,
606,616,626,636,646,656,666,676,686,696,
707,717,727,737,747,757,767,777,787,797,
808,818,828,838,848,858,868,878,888,898,
909,919,929,939,949,959,969,979,989,999
共有90個這樣的數字

個位數字及百位數字互換位置後，數值仍不會改變，即個位數字及百位數字必須一樣。百位數字有9個選擇(不可能是0，可以是1至9)，十位數字有10個選擇(任何數字皆能)，而個位數字跟百位數字一樣，沒有任何選擇。所以在100至999中共有(9x10)=90個數。

應該有90個..

101,202,303,404,505,606,707,808,909,
111,212,313,414,515,616,717,818,919,
121,222,323,424,525,626,727,828,929,
131,232,333,434,535,636,737,838,939,
141,242,343,444,545,646,747,848,949,
151,252,353,454,555,656,757,858,959,
161,262,363,464,565,666,767,868,969,
171,272,373,474,575,676,777,878,979,
181,282,383,484,585,686,787,888,989,
191,292,393,494,595,696,797,898,999.

每個個位同百位都可以掉轉的..
Right ??

no
中間數字可以是 0~9 共10個
so從樓上可知
有 9*10=90個
（例：101 121 525 686 .....................)

9個"
例如:101,202,303,404,505,606,707,808,909
岩唔岩牙?