從1至211中，數字「2」出現了多少次？
回答 (4)
2006-10-25 6:09 am
✔ 最佳答案
個位: 211 \ 10 = 21十位: 211 \ 10 - 1 = 2
百位: 12
Thus: 21 + 2 + 12 = 35
2006-10-24 22:10:10 補充：
Sorry打錯: 21 20 12= 53
2006-10-25 6:26 am
Only item 50 times
數字「2」出現 53 times,因22,122,202 各出現多一次
2
12
20
21
22
23
24
25
26
27
28
29
32
42
52
62
72
82
92
102
112
120
121
122
123
124
125
126
127
128
129
132
142
152
162
172
182
192
200
201
202
203
204
205
206
207
208
209
210
211
數字「2」出現 53 times,因22,122,202 各出現多一次
2
12
20
21
22
23
24
25
26
27
28
29
32
42
52
62
72
82
92
102
112
120
121
122
123
124
125
126
127
128
129
132
142
152
162
172
182
192
200
201
202
203
204
205
206
207
208
209
210
211
2006-10-25 6:07 am
altogether 53次
2006-10-24 22:13:04 補充：
2.12.20.21.22.23.24.25.26.27.28.29.32.42.52.62.72.82.92.102.112.120.121.122.123.124.125.126.127.128.129.132.142.152.162.172.182.192.200.201.202.203.204.205.206.207.208.209.210.211 The above has 50 numbers cosisting number 2However , 22 , 122 and 202 has double 2 , so altogether 53 次
2006-10-24 22:13:04 補充：
2.12.20.21.22.23.24.25.26.27.28.29.32.42.52.62.72.82.92.102.112.120.121.122.123.124.125.126.127.128.129.132.142.152.162.172.182.192.200.201.202.203.204.205.206.207.208.209.210.211 The above has 50 numbers cosisting number 2However , 22 , 122 and 202 has double 2 , so altogether 53 次
2006-10-25 6:07 am
51 times
2006-10-24 22:09:08 補充：
2.12.20.21.22.23.24.25.26.27.28.29.32.42.52.62.72.82.92.102.112.120.121.122.123.124.125.126.127.128.129.132.142.152.162.172.182.192.200.201.202.203.204.205.206.207.208.209.210.211
2006-10-24 22:09:08 補充：
2.12.20.21.22.23.24.25.26.27.28.29.32.42.52.62.72.82.92.102.112.120.121.122.123.124.125.126.127.128.129.132.142.152.162.172.182.192.200.201.202.203.204.205.206.207.208.209.210.211
收錄日期: 2021-04-12 21:13:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061024000051KK04490