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WholeNumbers 1(较大整数的简算1)
2019-04-09 15:11:42  吴海青- 阅读:17  关键词:双语奥数

                 WholeNumbers 1(较大整数的简算1)

   The most importantand sophisticated technique that we will learn from this chapter is to be ableto express, say, ,which denotes a 4-digit number, in the form of 1000a+100b+10c+d.

The usefulness ofthis simple expression is to help us solve and appreciate a unique set ofmathematical problems such as Example 3, Example 4 and Question 10 in thischapter.

   The other categoryof mathematical problems, namely multiplication of two extremely long string ofnumbers, uses a simple concept (10-1=9) for problem-solving. The concept isdemonstrated in Example 2.

   Last but notleast, we will learn to simplify the computation of the sum or difference oftwo sets of products through skilful factorization. Example 1 and Question 7illustrate this technique.


E.g.1   999999×222222+333333×333334=?

Solution:  999999×222222+333333×333334

        =333333×3×222222+333333×333334

        =333333×666666+333333×333334

        =333333×(666666+333334)

        =333333×1000000

        =333333000000


E.g. 2   Find the sumof all the digits of 3333…3333×6666…6666

                                 2008  3s    2008 6s

Analysis:  It is not possible to multiplytwo numbers of such magnitude. The whole trick to this question lies in asimple relationshiop:10-1=9

Solution: 

    3333…3333×6666…6666

        2008 3s     2008 6s

   =3333…3333×3×2222…2222

       2008 3s       2008 2222

   =9999…9999×  2222…2222

        2008 3s       2008 2s

   =(1000…0000—1)×  2222…2222

      2008 0s       2008 2s

   =2222…2222000…000 —2222…2222

     2008 2s   2008  0s  2008 2s

   =2222…2222 1 7777…7778

     2007 2s   2007  7s  

2+7=9    (there are 2007 pairs of 9)

1+8=9    (there is one more pair of 9)

2008×9=18072

The sum of all thedigits of 3333…3333×6666…666 is 18072.


E.g.3 The sum of all the digits of a three-digit number is 21. The digit inthe ones place is greater than the digit in the tens place. A new number, whichis 198 more than the original one, is formed by interchanging the digit in theones place with the digit in the hundreds place. What is the original number?

Analysis: It will not take us long to figure out that 876-678=198. Hence, theanswer is 678. The question lies with whether this is the only answer.

Solution:

      100a+10b+c= abc ——1

      Interchange the digits in the ones andhundreds places, it will become 100c+10b+a——2

      21

               100c+10b+a—abc =198

       100c+10b+a—100a—10b—c=198

                       99c—99a=198

                        99(c—a)=198

                          (c—a)=198÷99=2

      Since the digit in the ones place isgreater than the digit in the hundreds place by 2, we try 597. However,9 isgreater than 7.

      Next, we try 759.

      957—759=198

      Hence, the original number can be 678 or957.


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