Difference between revisions of "2018 AMC 10B Problems/Problem 21"

Revision as of 14:47, 18 October 2021

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: $1,2,\ldots,\dfrac{n}{2},n$ . 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 (Inequalities)

Let $d$ be the next divisor written to the right of $323.$ If $\gcd(323,d)=1,$ then $n\geq323d>323^2>100^2=10000,$ which contradicts the precondition that $n$ is a $4$ -digit number.

It follows that $\gcd(323,d)>1.$ Since $323=17\cdot19,$ the smallest possible value of $d$ is $17\cdot20=\boxed{\textbf{(C) } 340},$ from which $n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.$

~MRENTHUSIASM ~tdeng

Solution 2 (Inequalities)

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 $19 \cdot 2 \cdot 9 =342$ $\implies \boxed{\textbf{(C) } 340}$ .

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 3 (Answer Choices)

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 $2^2\cdot3^4\cdot17\cdot19$ 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, $\textbf{(A)}$ and $\textbf{(B)}$ are both not a multiple of neither $17$ nor $19$ , $\textbf{(C)}$ is divisible by $17$ . $\textbf{(D)}$ is divisible by $19$ , and $\textbf{(E)}$ is divisible by both $17$ and $19$ . Since $\boxed{\textbf{(C) } 340}$ is the smallest number divisible by either $17$ or $19$ it is the answer. Checking, we can see that $n$ would be $6460$ , a $4$ -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 4 (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

even

has 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 $\boxed{\textbf{(C) } 340}$ .

~OGBooger (Solution)

~Pearl2008 (Minor Edits)

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

@@ Line 7: / Line 7: @@
 ==Solution 1 (Inequalities)==
-<b>IN PROGRESS
-WILL BE DONE BY TOMORROW ...</b>
+Let <math>d</math> be the next divisor written to the right of <math>323.</math> If <math>\gcd(323,d)=1,</math> then <cmath>n\geq323d>323^2>100^2=10000,</cmath> which contradicts the precondition that <math>n</math> is a <math>4</math>-digit number.
+It follows that <math>\gcd(323,d)>1.</math> Since <math>323=17\cdot19,</math> the smallest possible value of <math>d</math> is <math>17\cdot20=\boxed{\textbf{(C) } 340},</math> from which <math>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</math>
 ~MRENTHUSIASM ~tdeng

2018 AMC 10B (Problems • Answer Key • Resources)
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2018 AMC 12B (Problems • Answer Key • Resources)
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