10+10+（100+10+10）×8－1×89 　　=20+960－89 　　=891（次） 　　答：数字1一共出现了891次。

　　100以内：1+10~19(共10个)+21+31+41+...+91(共8个)=19 　　100~200：共出现199-99+19=119 　　200.~300：19+10=29 　　300... 　　... 　　800...1共出现:29*7+119+19=341次

　　这样算吧，我们计算含有1的数，可知一位数中有1次，两位数中有11次，三位数中有120次，共132次不少加了20个，一共是152个

　　还是数一下吧：1.10.11.12.13.14.15.16.17.18.19.21.31.41.51.61.71.81.91.100.101.102.103.104.105.106.107.108.109.110.111.112.113.114.115.116.117.118.119.120.121.122.123.124.125.126.127.128.129.130.131.132.133.134.135.136.137.138.139.140.141.142.143.144.145.146.147.148.149.150.151.152.153.154.155.156.157.158.159.160.161.162.163.164.165.166.167.168.169.170.171.172.173.174.175.176.177.178.179.180.181.182.183.184.185.186.187.188.189.190.191.192.193.194.195.196.197.198.199.201.210.211.212.213.214.215.216.217.218.219.221.231.241.251.261.271.281.291.301.310.311.312.313.314.315.316.317.318.319.321.331.341.351.361.371.381.391.401.410.411.412.413.414.415.416.417.418.419.421.431.441.451.461.471.481.491.501.510.511.512.513.514.515.516.517.518.519.521.531.541.551.561.571.581.591.601.610.611.612.613.614.615.616.617.618.619.621.631.641.651.661.671.681.691.701.710.711.712.713.714.715.716.717.718.719.721.731.741.751.761.771.781.791.801.810.811.812.813.814.815.816.817.818.819.821.831.841.851.861.871.881.891。 　　然后边查数边找规律：一位数的1个；两位数的19个；100到199之间百位上的100个，个位上的10个，十位上的10个；200到299之间个位上的10个，十位上的10个；300到399之间个位上的10个，十位上的10个；……以此类推，1~899中，数字1共出现了1+19+100+（10+10）×8=280次。