Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added another quick solution)
(→‎Video Solution 1)
(2 intermediate revisions by 2 users not shown)
Line 6: Line 6:
 
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math>
 
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math>
  
−
==Solution 1==
+
==Solution 1 (Inequalities)==
−
The prime factorization of <math>323</math> is <math>17 \cdot 19</math>. Our answer must be a multiple of either <math>17</math> or <math>19</math> or both. Since <math>17 < 19</math>, the next smallest divisor that is divisble by <math>17</math> would be <math>323 + 17 = \boxed{\textbf{(C) } 340}</math>
 
−
 
 
−
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 
−
 
 
−
==Solution 2 (Inequalities)==
 
  
 
Let <math>d</math> be the next divisor written to the right of <math>323.</math>
 
Let <math>d</math> be the next divisor written to the right of <math>323.</math>
Line 20: Line 15:
 
~MRENTHUSIASM ~tdeng
 
~MRENTHUSIASM ~tdeng
  
−
==Solution 3 (Inequalities)==
+
==Solution 2 (Inequalities)==
  
 
Let <math>d</math> be the next divisor written to the right of <math>323.</math>
 
Let <math>d</math> be the next divisor written to the right of <math>323.</math>
Line 28: Line 23:
 
Therefore, the smallest possible value of <math>d</math> is <math>17\cdot20=\boxed{\textbf{(C) } 340},</math> from which <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath>
 
Therefore, the smallest possible value of <math>d</math> is <math>17\cdot20=\boxed{\textbf{(C) } 340},</math> from which <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath>
 
~MRENTHUSIASM ~bjhhar
 
~MRENTHUSIASM ~bjhhar
 +
 +
==Solution 3 (Quick)==
 +
The prime factorization of <math>323</math> is <math>17 \cdot 19</math>. Our answer must be a multiple of either <math>17</math> or <math>19</math> or both. Since <math>17 < 19</math>, the next smallest divisor that is divisble by <math>17</math> would be <math>323 + 17 = \boxed{\textbf{(C) } 340}</math>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
  
 
==Solution 4 (Answer Choices)==
 
==Solution 4 (Answer Choices)==
Line 52: Line 52:
  
 
https://www.youtube.com/watch?v=qlHE_sAXiY8
 
https://www.youtube.com/watch?v=qlHE_sAXiY8
 +
 +
hhttps://www.youtube.com/watch?v=T94oxV8schA&ab_channel=Jay
 +
 +
~ Coach J
  
 
==Video Solution 2==
 
==Video Solution 2==

Revision as of 16:11, 5 October 2024

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)

Video Solution 1

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

hhttps://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

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&oldid=229042"