2018 AMC 10B Problems/Problem 21

The following problem is from both the 2018 AMC 12B #19 and 2018 AMC 10B #21, so both problems redirect to this page.

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

Since prime factorizing $323$ gives you $17 \cdot 19$ , the desired answer needs to be a multiple of $17$ or $19$ , this is because if it is not a multiple of $17$ or $19$ , $n$ will be more than a $4$ digit number. For example, if the answer were to instead be $324$ , $n$ would have to be a multiple of $2^2 * 3^4 * 17 * 19$ for both $323$ and $324$ to be a valid factor, meaning $n$ would have to be at least $104652$ , which is too big. Looking at the answer choices, $\text{(A) }324$ and $\text{(B) }330$ are both not a multiple of neither 17 nor 19, $\text{(C) }340$ is divisible by $17$ . $\text{(D) }361$ is divisible by $19$ , and $\text{(E) }646$ is divisible by both $17$ and $19$ . Since $\boxed{\text{(C) }340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$ , a four-digit number. Note that $n$ is also divisible by $2$ , one of the listed divisors of $n$ . (If $n$ was not divisible by $2$ , we would need to look for a different divisor)

-Edited by Mathandski

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 $17 \cdot 19$ is $17 \cdot 2 \cdot 10 =340$ or $19 \cdot 2 \cdot 9 =342$ $\implies \boxed{\text{(C) }340}$ .

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

~bjhhar

Solution 4 (Using the answer choices)

Note that $323$ multiplied by any of the answer choices results in a 5 or 6 digit $n$ . So, we need a choice that shares a factor(s) with 323, such that the factors we'll need to add to the prime factorization of $n$ (in result to adding the chosen divisor) won't cause our number to multiply to more than 4 digits. The prime factorization of $323$ is $17\cdot19$ , and since we know $n$ is even, our answer needs to be (1) even & (2) with a factor of $17$ or $19$ . We see 340 achieves this and is the smallest to do so (646 being the other). So, we get $\boxed{\text{(C) }340}$ -- OGBooger -- minor changes by Pearl2008

Solution 5

We see that 323 is 18^2 -1, which means it's prime factorization is (18-1)(18+1) which is 17*19. The factors that are possible is an even number with 17 as a factor or 19 as a factor. The smallest factors larger than 17*19 are 17*20 or 19*18, and we can see that 17*20 is smaller than 19*18 since 19*18 is closer to a square, so therefore our answer is $\boxed{\text{(C) }340}$ .

Video Solution 1

https://www.youtube.com/watch?v=qlHE_sAXiY8

Video Solution 2

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

Video Solution 3

https://www.youtube.com/watch?v=vc1FHO9YYKQ

~bunny1

2018 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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