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.
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