The Least Common Multiple
最小公倍數(shù)
Objective:
學(xué)習(xí)目標(biāo):
We will learn and understand the concept of LCM, The Lowest Common Multiple.
我們將學(xué)習(xí)和理解最小公倍數(shù)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接到了一個(gè)任務(wù)，圈出一塊寬為5米，長度待定的地面。有人告知他們可用的有5米寬，6米長的墊子（我們稱這種類型為P型）或5米寬8米長的的墊子（我們稱這種類型為Q型）。這一區(qū)域的長度應(yīng)為最小，以便兩種類型中的任一種都能使用，且它們應(yīng)完全覆蓋此區(qū)域。Dozo決定地面長度應(yīng)為16米。當(dāng)他們開始使用Q型墊子覆蓋封閉的區(qū)域時(shí)，發(fā)現(xiàn)使用兩個(gè)Q型墊子就能將這個(gè)區(qū)域完全覆蓋。
但當(dāng)他們使用P型墊子時(shí)，第三個(gè)墊子超出了被圈區(qū)域。Mozo建議將地面長度增加至18米。當(dāng)他們使用P型墊子覆蓋封閉的區(qū)域時(shí)，用上了三個(gè)墊子后地面仍未被完全覆蓋。他們決定請教他們的數(shù)學(xué)家朋友Genius先生。他到了現(xiàn)場，在了解了情況后做了一下計(jì)算，然后稱地面長度為24米，48米，72米等時(shí)兩種類型的墊子都能將其完全覆蓋。因?yàn)橐蟮氖亲钚￠L度，所以我們認(rèn)為被圈區(qū)域的長度為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.
現(xiàn)在， Dozo 和Mozo發(fā)現(xiàn)兩種類型的墊子都能完全覆蓋被圈區(qū)域。他們十分佩服和感謝Genius先生。并詢問Genius先生是怎樣如此迅速的找到正確的長度的？Genius先生解釋道這只是最小公倍數(shù)的簡單應(yīng)用。最小公倍數(shù)是給定的數(shù)能劃分的數(shù)中最小的數(shù)。本例中24是能被6和8除的最小的數(shù)?，F(xiàn)在，我們繼續(xù)學(xué)習(xí)計(jì)算最小公倍數(shù)的方法。給定的數(shù)字為24和36。分別寫出這兩個(gè)數(shù)的質(zhì)數(shù)
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
最小公倍數(shù)取決于公因數(shù)和非公因數(shù)，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
讓我們舉另一個(gè)例子來進(jìn)行理解。計(jì)算15和25的最小公倍數(shù)。
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
總結(jié):
本節(jié)中我們探討了最小公倍數(shù)的概念并理解了計(jì)算最小公倍數(shù)的方法。
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