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