The Least Common Multiple
最小公倍数
Objective:
学习目标:
We will learn and understand the concept of LCM, The Lowest Common Multiple.
我们将学习和理解最小公倍数的概念。
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Dozo and Mozo are given the task of enclosing a part of the ground which is 5 m wide and of an assumed length. They are told that the mats with width 5 m and length 6 m (we will call this type as type P) or the mats with width 5 m and length 8 m (we will call this type as type Q) are available. The length of the area should be the least so that in case any of the two types of mats are used, they should cover the area completely. Dozo decides that the length should be 16 m. When they start covering the enclosed area with type Q mats, they find that with 2 mats the area is completely covered. However, when they use type P, the third mat goes outside the enclosed area. Mozo suggests, to increase the length to 18 m. When they start covering the enclosed area with type P mats, they find that using 3 mats the area is completely covered. However when they use type Q, two mats are used but the enclosed area is left uncovered. They decide to call their mathematician friend Mr. Genius. He arrives and after studying the situation does a little calculation and states that 24 m, 48 m, 72 m etc. are the lengths wherein both the types of mats will fit completely. Since we require the minimum length, we will consider the length of the enclosed area as 24 m.
Dozo和Mozo接到了一个任务，圈出一块宽为5米，长度待定的地面。有人告知他们可用的有5米宽，6米长的垫子（我们称这种类型为P型）或5米宽8米长的的垫子（我们称这种类型为Q型）。这一区域的长度应为最小，以便两种类型中的任一种都能使用，且它们应完全覆盖此区域。Dozo决定地面长度应为16米。当他们开始使用Q型垫子覆盖封闭的区域时，发现使用两个Q型垫子就能将这个区域完全覆盖。
但当他们使用P型垫子时，第三个垫子超出了被圈区域。Mozo建议将地面长度增加至18米。当他们使用P型垫子覆盖封闭的区域时，用上了三个垫子后地面仍未被完全覆盖。他们决定请教他们的数学家朋友Genius先生。他到了现场，在了解了情况后做了一下计算，然后称地面长度为24米，48米，72米等时两种类型的垫子都能将其完全覆盖。因为要求的是最小长度，所以我们认为被圈区域的长度为24米。
Dozo and Mozo now see that both the types of mats cover the enclosed area completely. They are thoroughly impressed and thank Mr. Genius. They ask Mr. Genius to explain as to how he was able to find the correct length so fast? Mr. Genius explains that this is simply an application of LCM. LCM is the smallest such number which the given numbers can divide. In this case 24 is the smallest number that can be divided by 6 as well as 8. Now we proceed to learn the method of finding LCM. Take the given numbers as 24 and 36. Write the prime factors of each.
现在， Dozo 和Mozo发现两种类型的垫子都能完全覆盖被圈区域。他们十分佩服和感谢Genius先生。并询问Genius先生是怎样如此迅速的找到正确的长度的？Genius先生解释道这只是最小公倍数的简单应用。最小公倍数是给定的数能划分的数中最小的数。本例中24是能被6和8除的最小的数。现在，我们继续学习计算最小公倍数的方法。给定的数字为24和36。分别写出这两个数的质数
24 = 2 × 2 × 2 × 3
36 = 2 ×2 ×3 ×3
24 等于 2 乘以 2乘以2乘以3
36 等于2乘以2乘以3乘以3
The LCM is determined by multiplying the common factors as well as the non common factors. Thus, the
LCM = 2 × 2 × 2 × 3 × 3 = 72
最小公倍数取决于公因数和非公因数，所以，
LCM等于2 乘以 2乘以2乘以3乘以3 等于 72
Let us take another example to understand this. Let us find the LCM of 15 and 25.
15 = 3 × 5
25 = 5 × 5
LCM = 5 × 5 × 3 = 75
让我们举另一个例子来进行理解。计算15和25的最小公倍数。
15 等于 3乘以5
25等于5乘以5
LCM等于5乘以5乘以3 等于 75
Summary:
Thus in this session we have discussed the concept of LCM and understood the method of finding LCM
总结:
本节中我们探讨了最小公倍数的概念并理解了计算最小公倍数的方法。
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