2018 AMC 10B Problems/Problem 21

Revision as of 00:52, 19 February 2018 by Pi31415926535897 (talk | contribs) (Solution 2)

Problem

Mary chose an even $4$-digit number $n$. She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,...,\dfrac{n}{2},n$. 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 $323$?

$\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646$

Solution 1

Prime factorizing $323$ gives you $17 \cdot 19$. Looking at the answer choices, $\fbox{(C) 340}$ is the smallest number divisible by $17$ or $19$.

Solution 2

Let the next largest divisor be $k$. Suppose $\gcd(k,323)=1$. Then, as $323|n, k|n$, therefore, $323\cdot k|n.$ However, because $k>323$, $323k>323\cdot 324>9999$. Therefore, $\gcd(k,323)>1$. Note that $323=17\cdot 19$. Therefore, the smallest the gcd can be is $17$ and our answer is $323+17=\boxed{\text{(C) }340}$.

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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