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

Let $d$ be the next divisor written to the right of $323.$

Since $n$ is even and $323=17\cdot19,$ we have $n=2\cdot17\cdot19\cdot k=646k$ for some positive integer $k.$ Moreover, since $1000\leq n\leq9998,$ we get $2\leq k\leq15.$ As $d>323,$ it is clear that $d$ must be divisible by $17$ or $19$ or both.

Therefore, 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 ~bjhhar

Solution 3 (Quick)

The prime factorization of $323$ is $17 \cdot 19$ . Our answer must be a multiple of either $17$ or $19$ or both. Since $17 < 19$ , the next smallest divisor that is divisble by $17$ would be $323 + 17 = \boxed{\textbf{(C) } 340}$

~South

Solution 4 (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 5 (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)

Solution 6 (Very Rigorous)

This is not the fastest solution, but if i saw this question on an Olympiad/AIME, where there are no answer choices, and my work counted, this is what i would do (for the purpose of this question, a=323, b=answer, c=the four digit number);

$\newline$ Note that for any $a,b,c \in \mathbb{Z}^+$ , if $a|c$ and $b|c$ then lcm $(a,b)|c$ (this is because if something divides a or b, it must divide c, and thus the max of all of the prime factors, ie: lcm, divides c) since $a=323,b > a$ , if $gcd(a,b)=1$ then $lcm(a,b)=ab$ and thus $ab|c \implies c>a^2 \implies c>100^2 \implies$ c is not a four digit number. $\newline$ thus, $gcd(a,b)\neq1$ . This implies that either $17|b$ , or $19|b$ , or both.

$\newline$ Case 1: $17|b$ , $19\not|b$ . We let $b=17b'$ , and by Euclid's Lemma, $19\not|b'$ . Then, $lcm(323,b)|c \implies 17(lcm(19,b'))|c$ . Since we already established that, $19\not|b$ (and since 19 is prime, if it does not divide a number it is coprime to that number), $17*19*b'|c \implies 323b'|c$ . Since $b=17b'>19*17$ , $b' \geq 20$ . A quick check shows $b'=20, b=340$ suffices. $\newline$ Now, let us show that there are no such numbers less than 340. $\newline$ Presume there exists such a number, $n \in \mathbb{Z}^+$ is in the range $(323,340)$ . By hypothesis, there is a $d>1$ such that $d|323$ , $d|n$ . By properties of divisibility $d|n-323$ . the maximum possible value of $n-323$ is $17^-$ (basically an arbitrary amount smaller than 17). But, since $d>1$ and $d|323, d \in \{17,19,323\}$ . Of which, the minimum value is d=17. but, $17>17^-$ so there is no such d, and no such n. $\newline$ Thus, our answer is just $\boxed{\textbf{(C) } 340}$ . $\newline$ ~Stereotypicalmathnerd

Video Solution 1

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

https://www.youtube.com/watch?v=T94oxV8schA&ab_channel=Jay

~Coach J

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
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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
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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