如何求两个数的最小公倍数

方法 1

方法 1 的 4:列出数字的所有倍数

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1
评估你要计算的数字。这个方法最适用于计算两个小于10的数字的公倍数,如果你面对的是比较大或比较多的数字,最好使用其它方法。
例如,我们需要找到5和8的最小公倍数。由于这两个数字都比较小,适合使用这个方法求出它们的最小公倍数。WH.performance.mark(‘step1_rendered’);

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2
从小到大列出第一个数字的几个倍数。用第一个数字乘以不同的整数就能得到它的倍数。
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也就是说,你可以直接查看乘法表,找到一个数的倍数。
例如,第一个数字5的倍数有5、10、15、20、25、30、35和40。

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3
从小到大写下第二个数字的几个倍数。用相同的整数乘以第二个数字,得到几个倍数,来和之前的一组倍数进行比较。
在我们的示例中,数字8的倍数有8、16、24、32、40、48、56和64。

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4
比较两个数字的倍数,找到其中最小的相同倍数。你可能需要列出更多倍数,来找到相同的那个倍数。你能找到的最小的相同数字就是最小公倍数。[2]
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研究来源

例如,5和8的倍数里都有40,而且它是最小的相同倍数,所以40是5和8的最小公倍数。

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方法 2

方法 2 的 4:使用素因式分解法

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1
评估数字。这个方法最适用于计算两个大于10的数字的公倍数,如果你面对的是比较小的数字,最好使用其它方法快速求出最小公倍数。
例如,如果你要找出数字20和84的最小公倍数,你可以使用这种方法。

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2
将第一个数字进行因式分解。你可以将第一个数字因式分解成它的素数因数,得到的几个素数因数相乘,就能够得到原始数字。你可以画出因子树来将数字分解成素数。完成因式分解后,重新写出等式。等式的一边是被分解的数字,另一边是素数因数相乘。
例如, 2×10=20{displaystyle mathbf {2} times 10=20} , 2×5=10{displaystyle mathbf {2} times mathbf {5} =10},因此,20的素数因数有2、2、和5。重新写出等式,得到 20=2×2×5{displaystyle 20=2times 2times 5}。

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3
将第二个数字也进行因式分解。用相同的方式分解第二个数字,找到它的素数因数,各个素数因数相乘能够得到第二个数字。
例如, 2×42=84{displaystyle mathbf {2} times 42=84}, 7×6=42{displaystyle mathbf {7} times 6=42}, 以及 3×2=6{displaystyle mathbf {3} times mathbf {2} =6}。因此,84的素数因数有2、7、3和2。 重新写出等式,得到 84=2×7×3×2{displaystyle 84=2times 7times 3times 2} 。

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4
写下每个相同的素数因数,并将每个因数相乘,写成乘法等式。在你写下每个因数的同时,请在因式分解的等式中划掉对应的数值。
例如,两个数字拥有共同的因数2,因此,写下因数 2×{displaystyle 2times } ,并将每个因式中的2划掉。
两个数字还拥有另一个2作为共同的因数,因此,再写下第二个数字2,并写成两数相乘: 2×2{displaystyle 2times 2},然后划掉因式分解式子里的另一个2。

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5
将剩余的因数添加到乘法式子中。剩余的因数是指划掉公因数后,几个因式分解的等式中没有被划掉的因数。也就是两个数字的因数中不相同的那些。[3]
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例如,在等式 20=2×2×5{displaystyle 20=2times 2times 5}中,两个2是两个数字共同的因数,因此你会划掉两个2。还剩下一个5,将5添加到上面的乘法式子中,得到: 2×2×5{displaystyle 2times 2times 5}。
在等式84=2×7×3×2{displaystyle 84=2times 7times 3times 2}中,你也划掉了两个2,还剩下了7和3,将这两个数字也加到乘式中,变成: 2×2×5×7×3{displaystyle 2times 2times 5times 7times 3}。

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6
计算最小公倍数。将上面写下的所有因数相乘,得到最小公倍数。
在我们的例子中, 2×2×5×7×3=420{displaystyle 2times 2times 5times 7times 3=420}。因此,20和84的最小公倍数是420。

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方法 3

方法 3 的 4:使用网格法或梯形法

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1
画一个井字形的网格。井字形的网格由两组平行线交叉组成,两组平行线彼此相互垂直,形成三行三列的网格,看上去像是手机或键盘上的井字键(#)。在网格最上方中央的方格内写下你的第一个数字,在网格右上角的方格内写下第二个数字。[4]
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例如,如果你想找到数字18和30的最小公倍数,请将18写在最上方中央的方格内,在网格右上角的方格写下30。

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2
找到两个数字共有的因数。将这个数字写在网格左上角的方格内。最好使用素数因数,这会大大方便后续的计算,但是也不是必须的。
在求解18和30的最小公倍数例题中,由于18和30都是偶数,所以都能整除2,将2写在网格左上角的方格内。

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3
用例题中的两个数除以共同的因数。将除得的商写在每个数字下面的方格中。进行除法计算就能得到商。
例如,18÷2=9{displaystyle 18div 2=9},在数字18下面写下9。

30÷2=15{displaystyle 30div 2=15},在网格中30下面的格子里写下15。

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4
找到两个商的公因数。如果两个商没有公因数,可以跳过这一步直接进入下一步。如果它们有公因数,请写在网格中央偏左的格子里。
例如,9和15的公因数为3,所以将3写在网格中央偏左的格子里。

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5
用第一步得到的商除以新的公因数。将结果写在上一步结果的下面。
例如, 9÷3=3{displaystyle 9div 3=3},将3写在9下方的方格内。

15÷3=5{displaystyle 15div 3=5},将5写在15下方的方格内。

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6
如果需要的话,继续扩展井字网格,画得大一点。然后按照上面的步骤计算除法,直到两个商没有相同的因数为止。

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7
在网格第一列和最后一行的数字上画圈。圆圈连起来,就像是画出了一个大写的“L”字母。将圈出的所有数字相乘。[5]
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在我们的例题中,2和3位于网格的第一列,3和5位于网格的最后一行,写出数学式: 2×3×3×5{displaystyle 2times 3times 3times 5}。

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8
完成乘法计算。将所有因数相乘,得到的结果就是原来两数的最小公倍数。[6]
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例如: 2×3×3×5=90{displaystyle 2times 3times 3times 5=90}。因此,18和30的最小公倍数是90。

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方法 4

方法 4 的 4:使用欧几里德算法

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1
了解除法中的名词。“被除数”是除法运算中被另一个数所除的数;“除数”是被除数除以的数字;“商”是除法的最后结果;“余数”是整数被整除以后余下的数字。[7]
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例如,在方程15÷6=2余3{displaystyle 15div 6=2;{text{余}};3}:15 是被除数6 是除数2 是商3 是余数。

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2
将方程改写成“商-余数”的形式。公式是 被除数 = 除数 × 商 + 余数。[8]
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研究来源

你需要用这个公式,根据欧几里得算法求出两个数字的最大公约数。
例如,15=6×2+3{displaystyle 15=6times 2+3}。
最大公约数是两个数字公有的最大除数或因子。[9]
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研究来源

使用本方法,你需要先求出最大公约数,然后通过它来找到最小公倍数。

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3
用两个数字中较大的数字当被除数,使用较小的一个当除数。建立两个数字的“商-余数”方程。
例如,如果你要求210和45的最小公倍数,那么方程的形式是 210=45×4+30{displaystyle 210=45times 4+30}。

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4
使用原除数作为新的被除数,使用余数作为新的除数。建立两个数字的“商-余数”方程。
例如, 45=30×2+15{displaystyle 45=30times 2+15}。

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5
一直重复这个过程,直到最后的余数变成0。每一个新方程中,你都需要使用原除数作为新的被除数,使用余数作为新的除数。[10]
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例如,30=15×2+0{displaystyle 30=15times 2+0}。因为,最后的余数是0,所以你不需要再继续除下去了。

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找到最后一个方程中的除数。这个数字就是两个数字的最大公约数。[11]
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例如,因为最后一个方程30=15×2+0{displaystyle 30=15times 2+0}中,除数是15,所以15就是210和45的最大公约数。

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7
求出两个数字的乘积。用它们的乘积除以它们的最大公约数。最后的结果就是两个数字的最小公倍数。[12]
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例如,210×45=9450{displaystyle 210times 45=9450}。用乘积除以最大公约数,得到945015=630{displaystyle {frac {9450}{15}}=630}。所以,630就是210和45的最小公倍数。

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