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

(1. LaTeX adjustments to all solutions. 2. Deleted repetitive solutions and combine solutions with similar thoughts. 3. Prioritize solutions that do not rely on answer choices. CREDITS ARE RETAINED AS MUCH AS POSSIBLE.)
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==Solution 1 (Inequalities)==
 
==Solution 1 (Inequalities)==
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Let <math>d</math> be the next divisor written to the right of <math>323.</math>
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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.
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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 <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath>
 
~MRENTHUSIASM ~tdeng
 
~MRENTHUSIASM ~tdeng
  
 
==Solution 2 (Inequalities)==
 
==Solution 2 (Inequalities)==
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Again, recognize <math>323=17 \cdot 19</math>. The 4-digit number is even, so its prime factorization must then be <math>17 \cdot 19 \cdot 2 \cdot n</math>. Also, <math>1000\leq 646n \leq 9998</math>, so <math>2 \leq n \leq 15</math>. Since <math>15 \cdot 2=30</math>, the prime factorization of the number after <math>323</math> needs to have either <math>17</math> or <math>19</math>. The next highest product after <math>17 \cdot 19</math> is <math>17 \cdot 2  \cdot 10 =340</math> or <math>19 \cdot 2  \cdot 9 =342</math> <math>\implies \boxed{\textbf{(C) } 340}</math>.
 
  
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You can also tell by inspection that <math>19\cdot18 > 20\cdot17</math>, because <math>19\cdot18</math> is closer to the side lengths of a square, which maximizes the product.
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Let <math>d</math> be the next divisor written to the right of <math>323.</math>
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Since <math>n</math> is even and <math>323=17\cdot19,</math> we have <math>n=2\cdot17\cdot19\cdot k=646k</math> for some positive integer <math>k.</math> Moreover, since <math>1000\leq n\leq9998,</math> we get <math>2\leq k\leq15.</math> As <math>d>323,</math> it is clear that <math>d</math> must be divisible by <math>17</math> or <math>19</math> or both.
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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>
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~MRENTHUSIASM ~bjhhar
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==Solution 3 (Quick)==
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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>
  
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~bjhhar
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~[https://artofproblemsolving.com/wiki/index.php/User:South South]
  
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==Solution 3 (Answer Choices)==
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==Solution 4 (Answer Choices)==
  
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Since prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>, the desired answer needs to be a multiple of <math>17</math> or <math>19</math>, this is because if it is not a multiple of <math>17</math> or <math>19</math>, <math>n</math> will be more than a <math>4</math> digit number. For example, if the answer were to instead be <math>324</math>, <math>n</math> would have to be a multiple of <math>2^2\cdot3^4\cdot17\cdot19</math> for both <math>323</math> and <math>324</math> to be a valid factor, meaning <math>n</math> would have to be at least <math>104652</math>, which is too big. Looking at the answer choices, <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are both not a multiple of neither <math>17</math> nor <math>19</math>, <math>\textbf{(C)}</math> is divisible by <math>17</math>. <math>\textbf{(D)}</math> is divisible by <math>19</math>, and <math>\textbf{(E)}</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\boxed{\textbf{(C) } 340}</math> is the smallest number divisible by either <math>17</math> or <math>19</math> it is the answer. Checking, we can see that <math>n</math> would be <math>6460</math>, a four-digit number. Note that <math>n</math> is also divisible by <math>2</math>, one of the listed divisors of <math>n</math>. (If <math>n</math> was not divisible by <math>2</math>, we would need to look for a different divisor.)
+
Since prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>, the desired answer needs to be a multiple of <math>17</math> or <math>19</math>, this is because if it is not a multiple of <math>17</math> or <math>19</math>, <math>n</math> will be more than a <math>4</math> digit number. For example, if the answer were to instead be <math>324</math>, <math>n</math> would have to be a multiple of <math>2^2\cdot3^4\cdot17\cdot19</math> for both <math>323</math> and <math>324</math> to be a valid factor, meaning <math>n</math> would have to be at least <math>104652</math>, which is too big. Looking at the answer choices, <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are both not a multiple of neither <math>17</math> nor <math>19</math>, <math>\textbf{(C)}</math> is divisible by <math>17</math>. <math>\textbf{(D)}</math> is divisible by <math>19</math>, and <math>\textbf{(E)}</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\boxed{\textbf{(C) } 340}</math> is the smallest number divisible by either <math>17</math> or <math>19</math> it is the answer. Checking, we can see that <math>n</math> would be <math>6460</math>, a <math>4</math>-digit number. Note that <math>n</math> is also divisible by <math>2</math>, one of the listed divisors of <math>n</math>. (If <math>n</math> was not divisible by <math>2</math>, we would need to look for a different divisor.)
  
 
-Edited by Mathandski
 
-Edited by Mathandski
  
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==Solution 4 (Answer Choices)==
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==Solution 5 (Answer Choices)==
 
Note that <math>323</math> multiplied by any of the answer choices results in a <math>5</math> or <math>6</math>-digit <math>n</math>. So, we need a choice that shares a factor(s) with <math>323</math>, such that the factors we'll need to add to the prime factorization of <math>n</math> (in result to adding the chosen divisor) won't cause our number to multiply to more than <math>4</math> digits.  
 
Note that <math>323</math> multiplied by any of the answer choices results in a <math>5</math> or <math>6</math>-digit <math>n</math>. So, we need a choice that shares a factor(s) with <math>323</math>, such that the factors we'll need to add to the prime factorization of <math>n</math> (in result to adding the chosen divisor) won't cause our number to multiply to more than <math>4</math> digits.  
 
The prime factorization of <math>323</math> is <math>17\cdot19</math>, and since we know <math>n</math> is even, our answer needs to be  
 
The prime factorization of <math>323</math> is <math>17\cdot19</math>, and since we know <math>n</math> is even, our answer needs to be  
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https://www.youtube.com/watch?v=qlHE_sAXiY8
 
https://www.youtube.com/watch?v=qlHE_sAXiY8
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 +
https://www.youtube.com/watch?v=T94oxV8schA&ab_channel=Jay
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~Coach J
  
 
==Video Solution 2==
 
==Video Solution 2==

Latest revision as of 03:50, 6 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

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

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

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