AMC 2018真题分析

AMC竞赛的最后5道题是决定学生能否进入AIME的关键。今天,Neil老师和Brian老师,为各位考生带来AMC 2018年竞赛最后两道题的分析,希望能更好地帮助大家准备明年1月底的AMC 10。

2018 AMC 10A Problem 22
Let a, b, c, and d be positive integers such that gcd (a, b) = 24, gcd (b, c) = 36, gcd (c, d) =54, and 70 < gcd (d, a) < 100. Which of the following must be a divisor of a? 

(A) 5       (B) 7      (C) 11      (D) 13      (E) 17


Firstly, we need to be clear that gcd (a, b) means the greatest common divisor. – that is the basic concept we should never confuse. 

Then rewrite 24 = 23·3, and 36 = 22·32, and 54 =2·33.


Clearly there must be Ma and Mb, such that

a = 23·3·Ma and b = 23·3·Mb.


Similarly there must be Nb and Nc, such that

b = 22·32·Nb and c = 22·32·Nc.


Compose the two expression about b, then we have an equation such that

23·3·Mb = 22·32·Nb   ( Note that Mb and Nb are not equal )


And we can generate an equation such that 2Mb = 3Nb

Since both Mb and Nb are positive integers, we can simply find out that

Mb is divisible by 3 and then Ma is NOT divisible by 3.


We run the same process about Pc and Pd, such that

c = 2·33·Pc and d = 2·33·Pd.


And we directly jump to the solution such that

Pc is divisible by 2 and Pd is NOT divisible by 2.


Compare the equations a = 23·3·Ma and d = 2·33·Pd,

gcd (d, a) = 2·3·C for some C which is not divisible by 2 or 3.


(If C is divisible by 2 then Pd should be divisible by 2, which is contradicted)


Since 70 < 6C < 100, the only solution is that 6C = 78 and C =13.

(The answer is D)

Knowledge points
The main concept of this problem is the application of greatest common divisor, which is the basic concept of AMC 10.And students need to know how to solve the problem by contradiction.


2018AMC10B Problem 21

Mary shoes an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2, …, n/2, n. At some moment Mary wrote 323 as a divisor of n. What is the smallest possible value of the next divisor written to the right of 323?


(A)324     (B)330     (C)340    (D)361   (E)646


Assume the next divisor is k such that gcd (323, k) = 1,we choose 324 as the next divisor, then we need to test its correctness.

However, 323·324 > 100·101 > 9999, which means n can NOT be 4-digit number.

(Since 324 does NOT work, same as 330)


So we know gcd (323, k) > 1, and k must contain a divisor of 323.

Note that 323 = 17·19, so both 340 = 17·20 and 361 = 19·19 can be the values of k.


Thus, we choose the smaller value, the answer is (C).

Knowledge points
The main concept of this problem is basic understanding of greatest common divisor.And students need to solve the problem by supposition, or start your answer with taking a guess.









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