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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?

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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

其他解答:

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?63D0B758E2D502CC
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