1.7 使用整数操作和再生基本理论理论

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• 选择节使用整数操作和自学基本理论

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  使用整数操作和自学基本理论

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  On the world time zone map below, the time zones are labeled with positive and negative numbers as seen on the top and bottom of the image. To find the time difference between places in different time zones, you need to find the difference between the time zone labels.
  ::在下面的世界时区地图上,时区标记为正数和负数,在图像的上部和下部可以看到。要找到不同时区地点之间的时间差,您需要找到时区标签之间的差。

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  In this section, we cover how to operate —add, subtract, multiply, and divide— with integers.
  ::在本节中,我们介绍如何用整数来增加、减、乘、除。

  

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  Operations on Integers
  ::整数业务

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  As we discussed previously, the integers  include zero , the natural numbers, and their opposites or negatives, like -1, -2, -3.... 
  
  ::正如我们以前讨论过的,整数包括零,自然数, 和它们的反面或负面,如 -1, -2, -3...

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  Addition of Integers
  
  ::增加整数

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    Addition of Integers
  
  ::增加整数

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  Like signs: Add the absolute values of the numbers and keep the common sign.
  ::类似符号:加上数字的绝对值并保留共同符号。

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  Unlike signs: Subtract the absolute values of the numbers and keep the sign of the number that is larger in absolute value.
  ::不同于符号: 减去数字的绝对值, 并保留数字的绝对值较大的符号 。

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  Note: The operation of addition is commonly referred to as finding the sum .
  ::注:增加的操作通常称为寻找总和。

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  Example 1
  
  ::例1

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  Find the following sums.
  
  ::找到以下金额。

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  a.  32 + 53
  
  ::a. 32+53

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  b. -32 + -53
  
  ::b. -32 + -53

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  c. -32 + 53
  
  ::c. -32 + 53

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  d. 32 + -53
  
  ::d. 32+-53

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  Solution:
  
  ::解决方案 :

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  a. 32 and 53 are both positive, so we add them and keep the common, positive, sign: 32 + 53 = 85.
  
  ::a. 32和53均为正数,因此我们加上它们,保持共同的正数,标志:32+53=85。

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  b. -32 and -53 are both negative, so we add the absolute values: 32 + 53 = 85. Since both of the numbers are negative, -32 + -53 = -85.
  
  ::b.-32和-53均为负数,因此我们加上绝对值:32+53=85,因为这两个数字均为负数,32+-53=85。

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  c. -32 and 53 have unlike signs, so we subtract the absolute values: 53 - 32 = 21. 53 is larger in absolute value. Since it is positive, -32 + 53 = 21.
  
  ::c.-32和53与符号不同,因此我们减去绝对值:53-32=21-53的绝对值较大。由于是正数,-32+53=21。

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  d. 32 and -53 have unlike signs, so we subtract the absolute values: 53 - 32 = 21. Now, -53 is larger in absolute value, so 32 + -53 = -21.  
  ::d. 32 和 - 53 与符号不同,因此我们减去绝对值: 53 - 32 = 21。 现在, - 53 的绝对值较大, 所以32 + - 53 = - 21。

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  by Mathispower4u demonstrates how to add integers. 
  ::通过 Mathispower4u 演示如何添加整数。

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  Subtraction of Integers
  
  ::整数减法

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  Recall the additive inverse property says that adding a number and its opposite equals 0. Subtracting a number from itself also equals 0. We can say that adding the opposite and subtracting are the same operation.
  ::回想添加的反面属性表示添加一个数字和它的对等等值为 0。 减一个数字本身也等于 0。 我们可以说, 增加反向和减是相同的操作 。

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  $$\begin{array}{rl}a+-a& =0=a-a\\ \\ a-b& =a+-b\end{array}$$

  We will use this idea for subtracting integers. 
  ::aa=0=a-aa-b=a-b我们将用这个概念来减去整数。

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    Subtracting Integers
  ::减减整数

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  To subtract integers, add the opposite.
  ::减去整数,加上相反的。

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  One way to remember this is "keep, change, change": Keep the sign of the first number, change the operation to addition, and change the sign of the second number.
  ::记住一个方法就是“ 保存, 改变, 改变 ” : 保留第一个数字的符号, 将操作更改为添加, 并更改第二个数字的符号 。

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  Note: The operation of subtraction is commonly referred to as finding the difference .
  ::注:减法的作用通常称为找出差额。

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  Example 2
  
  ::例2

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  Find  the following differences.
  
  ::找出以下差异。

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  a. 67 - 42
  
  ::a. 第67-42段

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  b. -67 - 42
  
  ::b. b. - 67 - 42

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  c. -67 - (-42)
  
  ::c. -67 - (-42)

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  d. 67 - (-42)
  
  ::d. 67 - (-42)

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  Solution:
  
  ::解决方案 :

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  a. Since these are both positive and 42 < 67, we can just subtract as we would with whole numbers , 67 - 42 = 25.
  ::a. 由于这些数值是正数和42 < 67,因此我们可以以67-42=25的整数,按我们所要的数值计算,再减去67-42=25。

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  We could do this as we would with integers, keep the sign of the first number, 67, change the operation to addition, and change the sign of the second number from positive to negative, -42: 67 + -42. Now, we follow the rules for adding numbers of unlike sign. 67 - 42 = 25 and we keep the sign of the larger in absolute value, 67, so 67 - 42 = 67 + (-42) = 25. 
  ::我们可以按整数做,保留第一个数字67的标记,将操作改为增加,将第二个数字的标记从正数改为负数, -42:67+42。现在,我们遵循规则,增加与标志不同的数字。67-42=25,我们保留最大绝对值67的标记,因此67-42=67+(-42)=25。

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  b. Keep the sign of the first number, -67, change the operation to addition, and change the sign of the second number from positive to negative, -42. We have -67 - 42 = -67 + -42.
  ::b. 保留第一个数字-67的标记,将操作改为增加,并将第二个数字的标记从正数改为负数-42。我们有-67-42=-67+-42。

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  Now, we follow the rules for addition of numbers with like sign. 67 + 42 = 109, so -67 - 42 = -67 + -42 = -109.
  
  ::现在,我们遵循增加数字的规则,加上类似符号。 67+42=109, 所以 -67 -42=67+42=109。

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  c. We follow the same process for changing this into an addition problem: -67 - (-42) = -67 + 42 = -25.
  
  ::c. 我们遵循同样的程序,将这个问题变成一个附加问题:-67-(-42)=-67+42=-25。

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  d. The same process for changing this into an addition problem yields: 67 - (-42) = 67 + 42 = 109.
  ::d. 将这一问题转化为额外问题的同样过程:67 - (-42) = 67 - 42 = 42 =109。

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  by Mathispower4u states the rules for subtracting integers and provides examples. 
  ::Mathispower4u 表示减整数的规则,并举例说明。

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  Example 3
  
  ::例3

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  What is the time difference, that is, the number of hours ahead or behind, between California and Madagascar? Is the time in California earlier or later than the time in Madagascar? A positive integer indicates that the location's time would be ahead or later in the day and a negative integer indicates that the location's time would be behind or earlier in the day. 
  
  ::加利福尼亚和马达加斯加之间的时间差是多少? 加利福尼亚和马达加斯加之间的时间差是多少? 加利福尼亚的时间早于还是晚于马达加斯加的时间? 正数整数表示该地点的时间早于或晚于一天,负数整数表示该地点的时间晚于或早于一天。

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  Solution: According to the map, California is in the -8 time zone. Madagascar is in the +3 time zone. California and Madagascar are $\begin{array}{r}-8-3=-8+-3=-11\end{array}$  hours apart. Since the result is negative, a clock in California is 11 hours behind a clock in Madagascar. The time in California is earlier than the time in Madagascar. 
  ::解答:根据地图,加利福尼亚州位于8个时区;马达加斯加位于+3时区;加利福尼亚州和马达加斯加相距8-3小时;由于结果为负数,加利福尼亚州时钟比马达加斯加时钟晚11小时;加利福尼亚州时钟比马达加斯加早。

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  Multiplication and Division of Integers
  
  ::乘数和整数司

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    Integers
  ::整整数

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  Like signs: The result is positive.
  ::类似迹象:结果是正面的。

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  Unlike signs: The result is negative.
  ::与迹象不同:结果为负。

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  Note: The operation of multiplication is commonly referred to as finding the product  and the operation of division is commonly referred to as finding the quotient .
  ::注:乘法的操作通常称为寻找产品,分法的操作通常称为寻找商数。

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  Example 4
  
  ::例4

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  Find the product or quotient for  the following.
  
  ::找到下列产品或商数。

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  a.  $\begin{array}{r}20\times 4\end{array}$  
  
  ::a. 20×4

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  b.  $\begin{array}{r}-20\times 4\end{array}$  
  
  ::b.-20×4

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  c.  $\begin{array}{r}20÷-4\end{array}$  
  ::c. 204

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  d.  $\begin{array}{r}-20÷-4\end{array}$  
  
  ::d. - 204

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  Solution:
  
  ::解决方案 :

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  a. These two numbers have the same sign, so the result is positive: $\begin{array}{r}20\times 4=80\end{array}$ .
  ::a. 这两个数字的符号相同,结果为正数:20×4=80。

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  b. These two numbers have unlike signs, so the result is negative. Multiply the numbers and then include the negative sign. $\begin{array}{r}-20\times 4=-80\end{array}$ .
  
  ::b. 这两个数字不同于符号,结果为负值。乘以数字,然后包括负值。-20x480。

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  c. Again, these two numbers have unlike signs. $\begin{array}{r}20÷-4=-(20÷4)=-5\end{array}$ .
  
  ::c. 这两个数字与信号不同。 20++4+(20++4)+5。

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  d. Similar to part a, these two numbers have the same sign, so $\begin{array}{r}-20÷-4=20÷4=5\end{array}$ .
  
  ::d. 与A部分类似,这两个数字的符号相同,所以-204=204=5。

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  by Mathispower4u demonstrates how to . 
  ::Mathispower4u 演示如何...

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    Operations With 0
  ::业务与0

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  • Multiplication: $\begin{array}{r}a\cdot 0=0\cdot a=0\end{array}$  
    ::乘数: a_0=0a=0
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  • Division: $\begin{array}{r}\frac{0}{a}=0,a\ne 0\end{array}$  
    ::分区:0a=0,a=0
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  • Division: $\begin{array}{r}\frac{a}{0}=\text{undefined}\end{array}$  
    ::项数: a0=未定义

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  Prime and Composite Numbers
  ::基数和综合数

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  A factor is a number that we multiply by another number to get a result. We can use the  number of factors to classify integers. Prime numbers are numbers whose only factors are 1 and the number itself. Composite numbers are numbers whose factors include more than 1 and the number itself.
  ::一个系数是一个数字,我们乘以另一个数字来得出结果。我们可以使用该系数数来分类整数。 位数是数字, 其唯一的因素是 1 和数字本身。 组合数字是其因素包括 1 和数字本身的数。

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  Example 5
  ::例5

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  Identify whether the following numbers are prime or composite: 2, 9, 12, and 17.
  ::确定下列数字是主要数字还是综合数字:2、9、12和17。

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  Solution:
  ::解决方案 :

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  2 is a prime number since the only factors of 2 are 1 and 2, that is, $\begin{array}{r}1\cdot 2=2\end{array}$ . 
  ::2 是一个质数, 因为唯一的2因数是 1 和 2, 即 1 和 2 = 2 。

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  The factors of 9 are 1, 3, and 9 since $\begin{array}{r}1\cdot 9=9\end{array}$ and $\begin{array}{r}3\cdot 3=9\end{array}$ . Thus, 9 is a composite number. 
  ::9的因数为1、3和9,自19=9和3=9。 因此,9是一个复合数字。

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  The factors of 12 are 1, 2, 3, 4, 6, and 12, so 12 is a composite number.
  ::12的因数是1、2、3、4、6和12,所以12是一个复合数字。

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  17 is a prime number since the only factors of 17 are 1 and 17.
  ::17是质数,因为17的唯一因素是1和17。

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  Prime Factorization and The Fundamental Theorem of Arithmetic
  ::初等化和自学基本理论

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   The Fundamental Theorem of Arithmetic
  ::论解学的基本理论

  Every positive integer is either a prime number or a composite number, which can be expressed as a unique product of prime numbers. Note this means that the of each composite number is unique.

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  Example 6
  ::例6

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  Find the prime factorization of 60.
  ::查找60的基数乘数。

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  Solution: To find a prime factorization, you first want to find two numbers that multiply to 60. Below we show two possibilities to demonstrate you can choose any pair of factors for this to work.
  ::解决方案: 要找到一个质因数化, 您首先要找到两个乘以到60的数值。 下面我们展示了两种可能性, 来证明您可以选择任何一对因素, 来使用它 。

  $$\begin{array}{rlr}60& & 60\\ \swarrow \searrow & & \swarrow \searrow \\ 610& & 230\end{array}$$

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  Next, we look at the the two factors and determine if they are prime or if they can be factored further. 2 is prime but the others are not, so we continue to factor them.
  ::接下来,我们审视这两个因素,确定它们是首要因素还是可以进一步考虑的因素。 2个因素是首要因素,而其他因素则不是,因此我们继续考虑这些因素。

  $$\begin{array}{rlr}60& & 60\\ \swarrow \searrow & & \swarrow \searrow \\ 610& & 230\\ \swarrow \searrow \swarrow \searrow & & \swarrow \searrow \\ 2325& & 215\end{array}$$

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  On the left, all of the numbers are prime. On the right, 2 is prime, but 15 is not, so we factor again. 
  ::在左边,所有的数字都是正数。在右边,2是正数,但15不是正数,所以我们再考虑一下。

  $$\begin{array}{rlr}60& & 60\\ \swarrow \searrow & & \swarrow \searrow \\ 610& & 230\\ \swarrow \searrow \swarrow \searrow & & \swarrow \searrow \\ 2325& & 215\\ & & \swarrow \searrow \\ & & 35\\ \\ & 60=2\cdot 2\cdot 3\cdot 5\end{array}$$

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  The prime numbers at the ends of the branches are the prime factorization. Notice that in both cases we get two 2's, a 3 and a 5. 
  ::树枝尾端的质数是质因数。 请注意, 在这两种情况下,我们都会得到两张2张,一张3张,一张5张。

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  Least Common Multiple
  ::最不常见

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  A multiple is a number that is divisible by another number. For example, 12 is a multiple of 6 since $\begin{array}{r}12÷6=2\end{array}$ . The least common multiple of a set of numbers is the smallest multiple that all of the numbers in the set have in common.
  ::倍数是一个可以除以另一个数字的数值。例如,12是自 12\\6=2 以来的6的倍数。一组数字中最不常见的倍数是一组数字中所有数字共有的最小的倍数。

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    How to Find the Least Common Multiple
  ::如何找到最小常见多重

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  1. Find the prime factorization of all of the numbers.
     ::查找所有数字的基因子化。
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  2. Write each prime factor with an exponent that is the largest in any of the prime factorizations.
     ::在每个质因子上写一个指数,该指数是任何质因数中最大的指数。
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  3. Multiply the prime factors from step 2.
     ::从第2步乘以主要因素。

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  Example 7
  ::例7

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  Find the least common multiple of 24, 36, and 40.
  ::找到24、36和40之间最不常见的乘数。

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  Solution: First, we find the prime factorizations of 24, 36, and 40. 
  ::解决方案:第一,我们发现 24,36和40的 首要因素。

  $$\begin{array}{r}24=3\cdot 8=3\cdot 2\cdot 4=3\cdot 2\cdot 2\cdot 2=2^3\cdot 3\\ \\ 36=3\cdot 12=3\cdot 3\cdot 4=3\cdot 3\cdot 2\cdot 2=2^2\cdot 3^2\\ \\ 40=4\cdot 10=2\cdot 2\cdot 10=2\cdot 2\cdot 2\cdot 5=2^3\cdot 5\end{array}$$

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  Now, we list each factor with its largest power and multiply. 
  ::现在,我们列出每个因素 其最大的功率和乘数。

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  $$\begin{array}{r}\text{LCM}=2^3\cdot 3^2\cdot 5=8\cdot 9\cdot 5=360\end{array}$$

  
  ::LCM=23325=895=360

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  As you can see below, 360 is a multiple of 24, 36, and 40. 
  ::如下文所示,360是24、36和40的乘数。

  $$\begin{array}{rl}& 360=2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5=24\cdot 15\\ \\ & 360=2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5=36\cdot 10\\ \\ & 360=2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5=40\cdot 9\end{array}$$

  
  

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  Feature: Below Par
  
  ::特征: 光下

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  by Denise Huey
  
  ::丹妮丝·胡伊(Denise Huey)

  

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  Golf is a popular sport known all over the world. At one point, Tiger Woods was the most popular player. If we focus on his golf game, we can apply our understanding of integers and adding integers to better understand the scoring of golf. When a golfer scores on par, the golfer’s score is zero. If a player scores below par, or scores less than the par number, the player then has a negative score. Let’s look at Tiger Woods’ scores from the 2012 Open Championship in England.
  ::高尔夫是全世界最受欢迎的运动。 在某个时刻,虎伍兹是最受欢迎的球员。 如果我们专注于他的高尔夫球赛,我们可以运用我们对整数的理解并增加整数以更好地了解高尔夫球的得分。 当高尔夫球手在平面上得分时,高尔夫球手得分为零。 如果球员在平面下得分,或者分低于平面,那么球员得分为负数。让我们看看2012年英格兰公开锦标赛的虎伍球得分。

  

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  In the first round , Tiger had a score of 2, which is one less than par. Taking the difference of these two numbers, he then has a score of -1. At the second hole, he scored on par, so his score stayed the same. We can represent this relationship as different operations on integers:
  ::在第一轮中,老虎的得分为2分,比平方差1分。用这两个数字的差数,他得分为-1分。在第二轮中,他得分为1分,因此他的得分保持不变。我们可以将这种关系作为不同的整数操作来表示:

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  For the first hole: $\begin{array}{r}2-3=-1\end{array}$
  ::第一个洞: 2- 3 @% 1

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  Total for the second hole: $\begin{array}{r}(2-3)+(4-4)=-1\end{array}$
  ::第二个洞的总数2-3)+(4-4)+1

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  Continuing with his score, his total score for the first round of golf was -3. Is this a good score, or a bad score?
  ::继续他的得分 第一轮高尔夫的总得分是3 这是好得分还是坏得分?

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  This represents a good score since the par is 70 strokes for the round. He completed 67 strokes, or three under par.
  ::这表示得分不错,因为平方为70中风。 他中了67中风,或3中了中风。

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  Summary
  
  ::摘要

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  • Addition of integers: add the absolute values and keep the common sign if like signs, subtract the absolute values and keep the sign of the larger in absolute value if unlike signs.
    ::增加整数:添加绝对值并保留共同符号,如果与符号相同,则保留共同符号,减去绝对值,如果与符号不同,则保留较大符号的绝对值。
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  • Subtraction of integers: add the opposite.
    ::整数减法:增加反差。
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  • Multiplication and division of integers: if like signs, the result is positive; if unlike signs, the result is negative.
    ::乘数和整数的划分:如果与符号一样,结果为正数;如果与符号不同,结果为负数。
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  • Prime numbers are numbers whose only factors are 1 and the number itself. Composite numbers are numbers whose factors include more than 1 and the number itself.
    ::参数数是指其唯一的因素为1和数字本身的数字。复合数字是指其因素包括1以上和数字本身的数字。
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  • The Fundamental Theorem of Arithmetic: Every positive integer is either prime or the unique product of prime numbers.    
    ::论理学的基本理论:每个正整数要么是质数的质数,要么是质数的独特产物。
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  • To find the least common multiple: Find the prime factorization of all of the numbers. Then, write each prime factor with an exponent that is the largest in any of the prime factorizations, and multiply.
    ::要找到最小常见的倍数 : 查找所有数字的质因数。 然后, 将每个质因数写成一个指数, 该指数是任何质因数中最大的指数, 然后乘以 。

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  Review
  
  ::回顾

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  Calculate the following.
  ::计算如下。

  1. $\begin{array}{r}-13+-37\end{array}$
  2. $\begin{array}{r}24+-16\end{array}$
  3. $\begin{array}{r}-3+100\end{array}$
  4. $\begin{array}{r}25-27\end{array}$
  5. $\begin{array}{r}43-(-5)\end{array}$
  6. $\begin{array}{r}-87-12\end{array}$
  7. $\begin{array}{r}-61-(-17)\end{array}$
  8. $\begin{array}{r}26\cdot 14\end{array}$
  9. $\begin{array}{r}-52\times -33\end{array}$
  10. $\begin{array}{r}-125(15)\end{array}$
  11. $\begin{array}{r}260÷-20\end{array}$
  12. $\begin{array}{r}-343÷-7\end{array}$
  13. $\begin{array}{r}\frac{-123}{3}\end{array}$

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  Explore More
  ::探索更多

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  1. While riding her bike to the ocean and back home 6 times, Cameron timed herself at 54 min, 48 min, 61 min, 44 min, 53 min, and 49 min.  She had set a goal of averaging 51 minutes for her rides. One measure of average is the mean. The mean of a set of data is calculated by first finding the sum of the data and then dividing the data by the cardinality (number of data values) of the set of data. How many minutes will she need to spend on her seventh ride to attain her goal?
  ::1. Cameron骑自行车到海洋和回家六次时,在54分钟、48分钟、61分钟、44分钟、53分钟和49分钟的时间里为自己计时。她设定了平均乘车51分钟的目标,其中一种平均值是平均值。一组数据的平均值是首先找到数据的总和,然后按照数据集的基点(数据值)将数据除以。她需要花费多少分钟才能达到目标?

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  2. Identify the error and use the distributive property to correctly solve the problem:
  ::2. 查明错误并利用分配财产正确解决问题:

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  a. - $\begin{array}{r}2(4-5)=-2(4)-2(5)=-18\end{array}$
  :a) - (2-4) - (5) - (2) - (4) - (2) - (5) - (4) - (4) - (4) - (2) - (5) - (5) - (2) - (5) - (5) - (5)

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  b.  $\begin{array}{r}-6(2-3)=-6(2)-3=-15\end{array}$
  ::b. - 6(2-3) - 6(2) - 3 15

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  3. Mt. Everest, the highest elevation in Asia, is 29,028 feet above sea level. The Dead Sea, the lowest elevation, is 1,312 feet below sea level. What is the difference between these two elevations?
  ::3. 亚洲最高海拔为29 028英尺的珠穆朗玛峰,最低海拔为1 312英尺的死海,这两个海拔有何区别?

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  4. When Ava woke up it was 85 degrees. The temperature rose 22 degrees by noon. There was an afternoon thunderstorm that caused the temperature to drop 37 degrees. What was the temperature after the thunderstorm?
  ::4. Ava醒来时,温度为85度,中午上升22度,下午暴风雨造成温度下降37度,雷暴后温度是多少?

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  5. You owe $225 on your credit card. You make a $55 payment and then purchase $87 for a new book for school. What is the integer that represents the balance owed on the credit card?
  ::5. 信用卡上欠你225美元,支付55美元,然后为学校购买新书87美元。信用卡上所欠余额的整数是多少?

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  Answers to Review and Explore More Problems
  
  ::对审查和探讨更多问题的答复

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  Please see the Appendix.
  
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try these interactives that reinforce the concepts explored in this section:
  ::尝试这些强化本节所探讨概念的交互作用 :