1
00:00:00,050 --> 00:00:04,010
判断380是否被2 3

2
00:00:04,010 --> 00:00:07,260
4 5 6 9 或者10整除

3
00:00:07,260 --> 00:00:10,150
题目跳过了7和8 因此我们不用考虑它们

4
00:00:10,150 --> 00:00:11,860
先看一下2

5
00:00:11,860 --> 00:00:16,600
380能被2整除吗

6
00:00:16,600 --> 00:00:18,550
我把2写在这

7
00:00:18,550 --> 00:00:22,670
某数要想被2整除

8
00:00:22,670 --> 00:00:27,200
必须是偶数 也就是说

9
00:00:27,200 --> 00:00:30,790
它的个位-- 我再写一下380

10
00:00:30,790 --> 00:00:34,570
偶数的个位必须是偶数

11
00:00:34,570 --> 00:00:36,800
所以这一位必须是偶数

12
00:00:36,800 --> 00:00:42,760
也就是说它是0 2 4 6 8中的一个

13
00:00:42,760 --> 00:00:47,270
380的个位是0 所以它是偶数

14
00:00:47,270 --> 00:00:53,900
也就是说它可以被2整除 所以2符合要求

15
00:00:53,900 --> 00:00:55,730
2是可以的

16
00:00:55,730 --> 00:00:58,140
再判断3

17
00:00:58,140 --> 00:01:00,510
一种快速的方法是--

18
00:01:00,510 --> 00:01:02,300
我写下3和问号--

19
00:01:02,300 --> 00:01:06,020
一种快速方法是把各位的数加起来

20
00:01:06,020 --> 00:01:09,010
如果得到的和可以被3整除

21
00:01:09,010 --> 00:01:11,070
那么这个数就可以被3整除

22
00:01:11,070 --> 00:01:12,080
我们试着做一下

23
00:01:12,080 --> 00:01:14,950
380 把各位加起来

24
00:01:14,950 --> 00:01:22,210
3+8+0是-- 3+8是11 再加0

25
00:01:22,210 --> 00:01:24,180
还是11

26
00:01:24,180 --> 00:01:25,580
如果你们不会计算11

27
00:01:25,580 --> 00:01:26,900
是否被3整除

28
00:01:26,900 --> 00:01:28,780
可以再把这两位加起来

29
00:01:28,780 --> 00:01:30,860
也就是1+1

30
00:01:30,860 --> 00:01:32,380
得到2

31
00:01:32,380 --> 00:01:35,530
无论是11还是2

32
00:01:35,530 --> 00:01:37,670
都不被3整除

33
00:01:37,670 --> 00:01:45,670
所以不被3整除 或许在未来的视频中

34
00:01:45,670 --> 00:01:47,520
我会解释这种方法的根据

35
00:01:47,520 --> 00:01:49,860
你们应该会想了解其原因

36
00:01:49,860 --> 00:01:56,740
这两个不被3整除 所以380不被3整除

37
00:01:56,740 --> 00:02:05,630
380 不被3整除 3不符合

38
00:02:05,630 --> 00:02:07,220
380不被3整除

39
00:02:07,220 --> 00:02:11,890
现在考虑一下4

40
00:02:11,890 --> 00:02:14,910
判断4的整除性

41
00:02:14,910 --> 00:02:16,880
我用橙色来写

42
00:02:16,880 --> 00:02:20,170
我们想知道380能否被4整除

43
00:02:20,170 --> 00:02:24,040
现在不论你们知道或者不知道

44
00:02:24,040 --> 00:02:26,540
100可以被4整除

45
00:02:26,540 --> 00:02:27,700
这是没问题的

46
00:02:27,700 --> 00:02:30,060
原题是380

47
00:02:30,060 --> 00:02:34,100
300可以被4整除 我们只需要

48
00:02:34,100 --> 00:02:37,070
计算剩下的80能否被4整除

49
00:02:37,070 --> 00:02:38,120
另一种思路是

50
00:02:38,120 --> 00:02:49,300
看最后两位能被4整除吗？

51
00:02:49,300 --> 00:02:52,050
这是从100被4整除这一事实得来的

52
00:02:52,050 --> 00:02:54,510
所以百位或者以上的部分

53
00:02:54,510 --> 00:02:55,460
肯定可以被4整除

54
00:02:55,460 --> 00:02:57,660
只需考虑剩下的数

55
00:02:57,660 --> 00:03:03,680
在本例中 也就是只需考虑80能否被4整除

56
00:03:03,680 --> 00:03:05,550
可以打眼一看

57
00:03:05,550 --> 00:03:07,790
8一定可以被4整除

58
00:03:07,790 --> 00:03:11,640
8除以4是2

59
00:03:11,640 --> 00:03:15,110
80除以4是20 所以这个可以

60
00:03:15,110 --> 00:03:17,430
答案是肯定的

61
00:03:17,430 --> 00:03:19,310
因为80可以被4整除

62
00:03:19,310 --> 00:03:24,400
所以380也可以被4整除 4符合

63
00:03:24,400 --> 00:03:25,420
再来算一下5

64
00:03:25,420 --> 00:03:27,960
我向下拉一下屏

65
00:03:27,960 --> 00:03:29,110
看看5怎么样

66
00:03:29,110 --> 00:03:32,300
被5整除的数有什么规律呢？

67
00:03:32,300 --> 00:03:33,640
我们来写一下5的倍数

68
00:03:33,640 --> 00:03:38,690
5 10 15 20 15

69
00:03:38,690 --> 00:03:41,390
所以如果某数被5整除-- 我可以继续列--

70
00:03:41,390 --> 00:03:48,750
也就意味着它最后一位是5或者0 对吧？

71
00:03:48,750 --> 00:03:58,910
每个5的倍数其个位都是5或者0

72
00:03:58,910 --> 00:04:05,470
380个位是0 所以它可以被5整除

73
00:04:05,470 --> 00:04:08,180
再来讨论6

74
00:04:08,180 --> 00:04:10,630
看一下6的情况如何

75
00:04:10,630 --> 00:04:13,410
我们想知道380是否被6整除？

76
00:04:13,410 --> 00:04:15,710
要想被6整除

77
00:04:15,710 --> 00:04:19,080
必须要被组成6的数字整除

78
00:04:19,080 --> 00:04:23,710
6=2×3

79
00:04:23,710 --> 00:04:26,990
所以如果一个数被6整除 那么

80
00:04:26,990 --> 00:04:30,260
它也必须被2和3整除

81
00:04:30,260 --> 00:04:34,180
如果一个数同时被2和3整除 那么它也可以被6整除

82
00:04:34,180 --> 00:04:37,370
380是可以被2整除的

83
00:04:37,370 --> 00:04:38,600
但我们已经确定

84
00:04:38,600 --> 00:04:40,810
它不被3整除

85
00:04:40,810 --> 00:04:45,050
如果它不被3整除 它就不能被6整除

86
00:04:45,050 --> 00:04:47,880
所以6被排除了

87
00:04:47,880 --> 00:04:50,330
380不能被6整除

88
00:04:50,330 --> 00:04:54,290
再看一下9

89
00:04:54,290 --> 00:04:56,570
9的整除性

90
00:04:56,570 --> 00:04:59,230
你们可以得出一个类似的结论

91
00:04:59,230 --> 00:05:02,020
如果某数不被3整除

92
00:05:02,020 --> 00:05:04,200
它也不可能被9整除

93
00:05:04,200 --> 00:05:08,080
因为9=3×3

94
00:05:08,080 --> 00:05:09,720
所以要被9整除

95
00:05:09,720 --> 00:05:13,050
必须至少可以被3整除两次

96
00:05:13,050 --> 00:05:15,540
这个数必须要被3整除两次

97
00:05:15,540 --> 00:05:19,130
这在本题并不成立 所以9也可以排除

98
00:05:19,130 --> 00:05:21,900
但如果我们不知道380不能被3整除

99
00:05:21,900 --> 00:05:24,570
还有另一种方法 它和计算

100
00:05:24,570 --> 00:05:27,640
3的整除性很像

101
00:05:27,640 --> 00:05:29,320
可以把各位加起来

102
00:05:29,320 --> 00:05:34,140
3+8+0 得到11

103
00:05:34,140 --> 00:05:36,440
11被9整除吗？

104
00:05:36,440 --> 00:05:41,910
不可以

105
00:05:41,910 --> 00:05:45,330
所以380也不被9整除

106
00:05:45,330 --> 00:05:46,980
在判断3的时候 用的是一种方法

107
00:05:46,980 --> 00:05:49,390
只不过是判断各位之和能否被3整除

108
00:05:49,390 --> 00:05:51,800
对于9则要判断是否被9整除

109
00:05:51,800 --> 00:05:56,560
最后要看的是10

110
00:05:56,560 --> 00:05:58,310
判断一下10的整除性

111
00:05:58,310 --> 00:06:00,090
这在某种意义上说是最简单的一个

112
00:06:00,090 --> 00:06:02,170
10的倍数有什么特点呢？

113
00:06:02,170 --> 00:06:06,950
10 20 30 40 可以继续写下去--

114
00:06:06,950 --> 00:06:08,820
它们都以0结尾

115
00:06:08,820 --> 00:06:11,680
或者说如果某数以0结尾 它就可以被10整除

116
00:06:11,680 --> 00:06:15,590
380以0结尾

117
00:06:15,590 --> 00:06:17,850
也就是说其个位是0

118
00:06:17,850 --> 00:06:19,890
所以它可以被10整除

119
00:06:19,890 --> 00:06:22,030
所以除了3 6和9以外

120
00:06:22,030 --> 00:06:24,500
380可以被剩下的所有数整除