2018 AMC 10B Problems/Problem 21

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 $17 \cdot 19$ . The desired answer needs to be a multiple of $17$ or $19$ , because if it is not a multiple of $17$ or $19$ , the LCM, or the least possible value for $n$ , will not be more than 4 digits. Looking at the answer choices, $\fbox{(C) 340}$ is the smallest number divisible by $17$ or $19$ . Checking, we can see that $n$ would be $6460$ .

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 $323+17=\boxed{\text{(C) }340}$ .

Solution 3

Again, recognize $323=17 \cdot 19$ . The 4-digit number is even, so its prime factorization must then be $17 \cdot 19 \cdot 2 \cdot n$ . Also, $1000\leq 646n \leq 9998$ , so $2 \leq n \leq 15$ . Since $15 \cdot 2=30$ , the prime factorization of the number after 323 needs to have either 17 or 19. The next highest product after 17 and 19 is $17 \cdot 2 \cdot 10 = 17 \cdot 20=340$ or $19 \cdot 2 \cdot 9 = 19 \cdot 18=342$ $\implies \boxed{\text{(C) }340}.

You can also tell by inspection that$ (Error compiling LaTeX. Unknown error_msg)19\cdot18 > 20\cdot17 $, because$ 19\cdot18$ is closer to the side lengths of a square, which maximizes the product.

~bjhhar

Video

https://www.youtube.com/watch?v=KHaLXNAkDWE

2018 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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