积极的整数据span class="katex">
N据/span>是可分开的据/p>
-
2据/span>如果最后一位据span class="katex">
N据/span>是2,4,6,8或0;据/li>
-
3.据/span>如果是数字的总和据span class="katex">
N据/span>是3的倍数;据/li>
-
4.据/span>如果最后2位数字据span class="katex">
N据/span>是4的倍数;据/li>
-
5.据/span>如果最后一位据span class="katex">
N据/span>是0或5;据/li>
-
6.据/span>如果据span class="katex">
N据/span>由2和3分开;据/li>
-
7.据/span>如果减去两位数的两位数据span class="katex">
N据/span>从剩余的数字提供7个(例如,658可被7分开7,因为65-2×8 = 49,这是7的倍数);据/li>
-
8.据/span>如果是最后3位数字据span class="katex">
N据/span>是8的倍数;据/li>
-
9.据/span>如果是数字的总和据span class="katex">
N据/span>是9的倍数;据/li>
-
1据/span>0.据/span>如果最后一位据span class="katex">
N据/span>是0;据/li>
-
1据/span>1据/span>如果是交替的数字的差异据span class="katex">
N据/span>是11的倍数(例如,2343以11分开11,因为2 - 3 + 4-3 = 0,这是11的倍数);据/li>
-
1据/span>2据/span>如果据span class="katex">
N据/span>由3和4分开。据/li>
以下是一些示例问题,可以使用上述一些可分配规则来解决。据/p>
在不执行实际划分的情况下,显示下面的数字是整数:据/p>
1据/span>2据/span>1据/span>那据/span>4.据/span>8.据/span>1据/span>那据/span>4.据/span>8.据/span>1据/span>那据/span>4.据/span>6.据/span>8.据/span>。据/span>
从可分性规则中,我们知道,如果它是由3和4所以的话,我们将被12即12即将到来。因此,我们只需要检查1,481,481,468以3和4可分离。据/p>
应用可分割测试3,我们得到了据span class="katex">
1据/span>+据/span>4.据/span>+据/span>8.据/span>+据/span>1据/span>+据/span>4.据/span>+据/span>8.据/span>+据/span>1据/span>+据/span>4.据/span>+据/span>6.据/span>+据/span>8.据/span>=据/span>4.据/span>5.据/span>那据/span>可被3分开。因此,1,481,481,468由3分开。据/p>
应用可分性测试4,我们得到了最后两位数,68可被4分开。因此,1,481,481,468也可被4即。据/p>
现在,由于我们知道1,481,481,468由3和4所以,它可被12即12。因此,据span class="katex">
1据/span>2据/span>1据/span>那据/span>4.据/span>8.据/span>1据/span>那据/span>4.据/span>8.据/span>1据/span>那据/span>4.据/span>6.据/span>8.据/span>将是一个整数。据span class="katex">
□据/span>
找到所有可能的值据span class="katex">
一种据/span>这样的数字据span class="katex">
9.据/span>8.据/span>一种据/span>6.据/span>是一个倍数据span class="katex">
3.据/span>。据/span>
从可分性规则,数字据span class="katex">
9.据/span>8.据/span>一种据/span>6.据/span>是一个倍数据span class="katex">
3.据/span>如果且仅当其数字的总和据span class="katex">
9.据/span>+据/span>8.据/span>+据/span>一种据/span>+据/span>6.据/span>=据/span>2据/span>3.据/span>+据/span>一种据/span>是一个倍数据span class="katex">
3.据/span>。据/span>自从据span class="katex">
0.据/span>≤.据/span>一种据/span>≤.据/span>9.据/span>,这意味着据span class="katex">
一种据/span>=据/span>1据/span>那据/span>4.据/span>那据/span>7.据/span>是所有可能的值。据span class="katex">
□据/span>
没有表演分裂,解释为什么数字据span class="katex">
9.据/span>8.据/span>7.据/span>6.据/span>5.据/span>4.据/span>3.据/span>2据/span>1据/span>是一个倍数据span class="katex">
9.据/span>。据/p>
通过可分性的规则据span class="katex">
9.据/span>那据/span>自从据span class="katex">
9.据/span>+据/span>8.据/span>+据/span>7.据/span>+据/span>6.据/span>+据/span>5.据/span>+据/span>4.据/span>+据/span>3.据/span>+据/span>2据/span>+据/span>1据/span>=据/span>4.据/span>5.据/span>是一个倍数据span class="katex">
9.据/span>那据/span>
9.据/span>8.据/span>7.据/span>6.据/span>5.据/span>4.据/span>3.据/span>2据/span>1据/span>是一个倍数据span class="katex">
9.据/span>。据/span>□据/span>
没有执行实际划分,表明据span class="katex">
8.据/span>7.据/span>4.据/span>5.据/span>6.据/span>3.据/span>9.据/span>9.据/span>是据strong>不是据/strong>可分离据span class="katex">
1据/span>1据/span>。据/p>
应用可分配规则据span class="katex">
1据/span>1据/span>那据/span>奇数位置的数字总和之间的差异据span class="katex">
(据/span>8.据/span>+据/span>4.据/span>+据/span>6.据/span>+据/span>9.据/span>=据/span>2据/span>7.据/span>)据/span>和甚至地方的数字总和据span class="katex">
(据/span>7.据/span>+据/span>5.据/span>+据/span>3.据/span>+据/span>9.据/span>=据/span>2据/span>4.据/span>)据/span>是据span class="katex">
2据/span>7.据/span>-据/span>2据/span>4.据/span>=据/span>3.据/span>那据/span>哪个不可分割据span class="katex">
1据/span>1据/span>。因此据span class="katex">
8.据/span>7.据/span>4.据/span>5.据/span>6.据/span>3.据/span>9.据/span>9.据/span>不可分割据span class="katex">
1据/span>1据/span>。据span class="katex">
□据/span>
对于什么值据span class="katex">
一种据/span>和据span class="katex">
B.据/span>是据span class="katex">
1据/span>2据/span>一种据/span>B.据/span>多倍据span class="katex">
9.据/span>9.据/span>还是据/span>
自从据span class="katex">
9.据/span>9.据/span>=据/span>9.据/span>×据/span>1据/span>1据/span>,数字必须是多个据span class="katex">
9.据/span>和据span class="katex">
1据/span>1据/span>。据/span>
可分配规则据span class="katex">
9.据/span>告诉我们据span class="katex">
1据/span>+据/span>2据/span>+据/span>一种据/span>+据/span>B.据/span>是一个倍数据span class="katex">
9.据/span>。据/span>因为它是一个数字据span class="katex">
3.据/span>到据span class="katex">
2据/span>1据/span>那据/span>它必须是据span class="katex">
9.据/span>要么据span class="katex">
1据/span>8.据/span>。据/span>
现在,可分性规则据span class="katex">
1据/span>1据/span>告诉我们据span class="katex">
1据/span>-据/span>2据/span>+据/span>一种据/span>-据/span>B.据/span>是一个倍数据span class="katex">
1据/span>1据/span>。据/span>因为它是一个数字据span class="katex">
-据/span>1据/span>0.据/span>到据span class="katex">
8.据/span>那据/span>一定是据span class="katex">
0.据/span>。据/span>
解决据span class="katex">
{据/span>1据/span>+据/span>2据/span>+据/span>一种据/span>+据/span>B.据/span>=据/span>9.据/span>1据/span>-据/span>2据/span>+据/span>一种据/span>-据/span>B.据/span>=据/span>0.据/span>那据/span>没有整数解决方案。据/p>
解决据span class="katex">
{据/span>1据/span>+据/span>2据/span>+据/span>一种据/span>+据/span>B.据/span>=据/span>1据/span>8.据/span>1据/span>-据/span>2据/span>+据/span>一种据/span>-据/span>B.据/span>=据/span>0.据/span>那据/span>我们得到了据span class="katex">
一种据/span>=据/span>8.据/span>那据/span>B.据/span>=据/span>7.据/span>。据/p>
因此,唯一的解决方案是据span class="katex">
1据/span>2据/span>8.据/span>7.据/span>和据span class="katex">
一种据/span>=据/span>8.据/span>和据span class="katex">
B.据/span>=据/span>7.据/span>。据/span>
□据/span>
是据span class="katex">
6.据/span>5.据/span>9.据/span>7.据/span>3.据/span>3.据/span>9.据/span>0.据/span>可分离据span class="katex">
2据/span>1据/span>0.据/span>还是据/span>
我们不知道分配规则的210.但是,我们很容易看到据span class="katex">
2据/span>1据/span>0.据/span>=据/span>2据/span>×据/span>3.据/span>×据/span>5.据/span>×据/span>7.据/span>,所以,如果65973390可被2,3,5,7分开,那么它可被210分开。据/p>
- 由于65973390的最后一位数字为0,因此它可被2个。据/li>
- 自从据span class="katex">
6.据/span>+据/span>5.据/span>+据/span>9.据/span>+据/span>7.据/span>+据/span>3.据/span>+据/span>3.据/span>+据/span>9.据/span>+据/span>0.据/span>=据/span>4.据/span>2据/span>,它是可被3所以3,所以65973390可被3所以3。据/li>
- 由于65973390的最后一位数字为0,因此它可被5可分开。据/li>
- 要通过7查看可分配性,作为最初的步骤,我们计算据span class="katex">
6.据/span>5.据/span>9.据/span>7.据/span>3.据/span>3.据/span>9.据/span>-据/span>2据/span>(据/span>0.据/span>)据/span>=据/span>6.据/span>5.据/span>9.据/span>7.据/span>3.据/span>3.据/span>9.据/span>。然而,这个号码仍然有点太大,告诉我们是否被7.在这些情况下,我们一直在又一次地又一次地应用可分配规则,直到我们有足够小的数字来处理:据span class="katex-display">
6.据/span>5.据/span>9.据/span>7.据/span>3.据/span>3.据/span>-据/span>2据/span>(据/span>9.据/span>)据/span>6.据/span>5.据/span>9.据/span>7.据/span>1据/span>-据/span>2据/span>(据/span>5.据/span>)据/span>6.据/span>5.据/span>9.据/span>6.据/span>-据/span>2据/span>(据/span>1据/span>)据/span>6.据/span>5.据/span>9.据/span>-据/span>2据/span>(据/span>4.据/span>)据/span>6.据/span>5.据/span>-据/span>2据/span>(据/span>1据/span>)据/span>=据/span>6.据/span>5.据/span>9.据/span>7.据/span>1据/span>5.据/span>=据/span>6.据/span>5.据/span>9.据/span>6.据/span>1据/span>=据/span>6.据/span>5.据/span>9.据/span>4.据/span>=据/span>6.据/span>5.据/span>1据/span>=据/span>6.据/span>3.据/span>。据/span>现在我们可以看到我们留下来了据span class="katex">
6.据/span>3.据/span>那据/span>我们可以容易地识别为7.因此,65973390也是7的倍数。据/li>
由于65973390以2,3,5,7全部可分开,因此可通过据span class="katex">
2据/span>×据/span>3.据/span>×据/span>5.据/span>×据/span>7.据/span>=据/span>2据/span>1据/span>0.据/span>。据/span>□据/span>
尝试一些问题让自己看看你是否了解这个话题:据/p>
如果我们知道整数是5的倍数,则整数的最后两位数字有多少种可能性?据/p>