前一节向我们展示了如何解决单个方程的FITB难题。如果我们有更多的方程呢?它会让谜题变得更难吗?嗯,不完全是。这和解a没什么不同据一种target="_blank" rel="nofollow" href="//www.parkandroid.com/wiki/systems-of-equations/">方程组据/a>.因为我们已经有了上一节中需要的所有基础知识,让我们从一个例子开始:据/p>
在下面的方框中填入从1到5的不同整数,使方程数组保持为真。据span class="katex-display">
0.据/span>+据/span>0.据/span>−据/span>0.据/span>÷据/span>0.据/span>=据/span>1据/span>×据/span>0.据/span>=据/span>1据/span>1据/span>
让我们从分区标志上方和下方直接开始使用2个盒子(据span class="katex">
÷据/span>),并打电话给他们据span class="katex">
X据/span>和据span class="katex">
y据/span>那据/span>分别。因为所有的方框都表示整数,据span class="katex">
X据/span>必须是可分开的据span class="katex">
y据/span>.据/span>同时,据span class="katex">
1据/span>≤据/span>X据/span>那据/span>y据/span>≤据/span>5.据/span>和据span class="katex">
X据/span>据/span>=据/span>y据/span>.据/span>因此,可能性是据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>4.据/span>那据/span>2据/span>)据/span>那据/span>(据/span>5.据/span>那据/span>1据/span>)据/span>那据/span>(据/span>4.据/span>那据/span>1据/span>)据/span>那据/span>(据/span>3.据/span>那据/span>1据/span>)据/span>那据/span>(据/span>2据/span>那据/span>1据/span>)据/span>只有。据/p>
让据span class="katex">
Z.据/span>是上面盒子里的数字,然后我们可以分析这5种可能性如下:据/p>
如果据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>4.据/span>那据/span>2据/span>)据/span>, 然后据span class="katex">
Z.据/span>−据/span>(据/span>X据/span>÷据/span>y据/span>)据/span>=据/span>1据/span>⇒据/span>Z.据/span>=据/span>3.据/span>.因此,两个未使用的整数为1和5,必须满足方程式据span class="katex">
0.据/span>+据/span>4.据/span>×据/span>0.据/span>=据/span>1据/span>1据/span>.反复试验表明没有解决方案,因此据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>4.据/span>那据/span>2据/span>)据/span>不可能是一个解决方案。据/p>
如果据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>5.据/span>那据/span>1据/span>)据/span>, 然后据span class="katex">
Z.据/span>−据/span>(据/span>X据/span>÷据/span>y据/span>)据/span>=据/span>1据/span>⇒据/span>Z.据/span>=据/span>6.据/span>.但由于据span class="katex">
Z.据/span>必须是1到5之间的整数包容性,据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>5.据/span>那据/span>1据/span>)据/span>不可能是一个解决方案。据/p>
如果据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>4.据/span>那据/span>1据/span>)据/span>, 然后据span class="katex">
Z.据/span>−据/span>(据/span>X据/span>÷据/span>y据/span>)据/span>=据/span>1据/span>⇒据/span>Z.据/span>=据/span>5.据/span>.因此,两个未使用的整数是2和3,必须满足方程式据span class="katex">
0.据/span>+据/span>4.据/span>×据/span>0.据/span>=据/span>1据/span>1据/span>.试验和错误显示据span class="katex">
3.据/span>+据/span>4.据/span>×据/span>2据/span>=据/span>1据/span>1据/span>, 因此据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>4.据/span>那据/span>1据/span>)据/span>可以是一个解决方案。据/p>
如果据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>3.据/span>那据/span>1据/span>)据/span>, 然后据span class="katex">
Z.据/span>−据/span>(据/span>X据/span>÷据/span>y据/span>)据/span>=据/span>1据/span>⇒据/span>Z.据/span>=据/span>4.据/span>.所以两个未使用的整数是2和5,它们必须满足方程据span class="katex">
0.据/span>+据/span>3.据/span>×据/span>0.据/span>=据/span>1据/span>1据/span>.试验和错误显示据span class="katex">
5.据/span>+据/span>3.据/span>×据/span>2据/span>=据/span>1据/span>1据/span>, 因此据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>3.据/span>那据/span>1据/span>)据/span>可以是一个解决方案。据/p>
如果据span class="katex">
(据/span>X据/span>那据/span>y据/span>)据/span>=据/span>(据/span>2据/span>那据/span>1据/span>)据/span>, 然后据span class="katex">
Z.据/span>−据/span>(据/span>X据/span>÷据/span>y据/span>)据/span>=据/span>1据/span>⇒据/span>Z.据/span>=据/span>3.据/span>.所以这两个未使用的整数是4和5,它们必须满足方程据span class="katex">