AMC 8 1999 Problem 24
AMC8 1999 年第 24 题
余数与同余★★★☆☆
When is divided by 5 , the remainder is
当 除以 5 时，余数是多少？
- A.0
- B.1
- C.2
- D.3
- E.4
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
Note that the units digits of the powers of 9 have a pattern: , , , , and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of 9, the number ends in a 1. Since the exponent is even, the final digit is 1. Note that all natural numbers that end in 1 have a remainder of 1 when divided by 5. So, our answer is .
Write 1999 as 2000-1. We are taking . Using the binomial theorem, we see that ALL terms in this expansion are divisible by 2000, which is in turn divisible by 5, except for the very last term, which is just . This is clear because the binomial expansion is just choosing how many 2000 s and how many -1 s there are for each term. Using this, we can take the entire polynomial , which leaves just .
As , we have . Thus, the answer is .
A sum/product of any two natural numbers has the same remainder, when divided by 5, as the sum/product of their remainders. Thus, we can use the basic definition of an exponent and view the problem as .
Using the fact stated in the first sentence, we see that the remainder of 1999, when divided by 5, is 4. The problem can now be written viewed as finding the remainder of when it is divided by 5, which is already much simpler.
Now, toying with the simplified problem a little, see notice that powers of 4 alternate from ending in 4 and 6. We notice that even powers of 4 always end in 6, and also that 2000 is even. Thus must end in 6, which, when divided by 5 gives a remainder of .
By Fermat's Little Theorem, . Thus the answer is .
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