习题五解答 1.解:分类计算,并将有数字“1”的数枚举出来. “1”出现在个位上的数有: 1,11,21,31,41,51,61,71,81,91, 101,111,121,131,141,151,161,171,181,191 共20个; “1”出现在十位上的数有: 10,11,12,13,14,15,16,17,18,19 110,111,112,113,114,115,116,117,118,119 共20个; “1”出现在百位上的数有: 100,101,102,103,104,105,106,107,108,109, 110,111,112,113,114,115,116,117,118,119, 120,121,122,123,124,125,126,127,128,129, 130,131,132,133,134,135,136,137,138,139, 140,141,142,143,144,145,146,147,148,149, 150,151,152,153,154,155,156,157,158,159, 160,161,162,163,164,165,166,167,168,169, 170,171,172,173,174,175,176,177,178,179, 180,181,182,183,184,185,186,187,188,189, 190,191,192,193,194,195,196,197,198,199 共100个; 数字“1”在1至200中出现的总次数是: 20+20+100=140(次). 2.解:采用枚举法,并分类计算: “3”在个位上:3,13,23,33,43,53,63,73,83,93共10个; “3”在十位上:31,33,35,37,39共5个; 数字“3”在1至100的奇数中出现的总次数: 10+5=15(次). 3.解:枚举法:12,17,22,27,32,37,42,47,52,57,62,67,72,77,82,87,92,97共18个. 4.解:分段统计,再总计. 页数 铅字个数 1~9共9页 1×9=9(个)(每个页码用1个铅字) 10~90共90页 2×90=180(个)(每个页码用2个铅字) 100~199共100页 3×100=300(个)(每个页码用3个铅字) 第200页共1页 3×1=3(个)(这页用3个铅字) 总数:9+180+300+3=492(个). 5.解:列表枚举,分类统计: 10 1个 20 21 2个 30 31 32 3个 40 41 42 43 4个 50 51 52 53 54 5个 60 61 62 63 64 65 6个 70 71 72 73 74 75 76 7个 80 81 82 83 84 85 86 87 8个 90 91 92 93 94 95 96 97 98 9个 总数1+2+3+4+5+6+7+8+9=45(个). 6.解:枚举法,再总计: 101,111,121,131,141,151,161,171,181,191共10个.