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

(→‎Solution 2: I will rewrite a similar solution using detailed casework. CREDITS WILL BE RETAINED TO TDENG.)
(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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==Problem==
 
==Problem==
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Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>?
+
Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,\ldots,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>?
  
 
<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)==
 +
~MRENTHUSIASM ~tdeng
  
−
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 * 3^4 * 17 * 19</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>\text{(A) }324</math> and <math>\text{(B) }330</math> are both not a multiple of neither 17 nor 19, <math>\text{(C) }340</math> is divisible by <math>17</math>. <math>\text{(D) }361</math> is divisible by <math>19</math>, and <math>\text{(E) }646</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\boxed{\text{(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)
+
==Solution 2 (Inequalities)==
 +
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>.
 +
 
 +
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.
 +
 
 +
~bjhhar
 +
 
 +
==Solution 3 (Answer Choices)==
 +
 
 +
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.)
  
 
-Edited by Mathandski
 
-Edited by Mathandski
  
−
==Solution 3==
+
==Solution 4 (Answer Choices)==
−
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{\text{(C) }340}</math>.
+
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
  
 +
* even
  
−
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.
+
* has a factor of <math>17</math> or <math>19</math>
  
−
~bjhhar
+
We see <math>340</math> achieves this and is the smallest to do so (<math>646</math> being the other). So, we get <math>\boxed{\textbf{(C) } 340}</math>.
  
−
==Solution 4 (Using the answer choices)==
+
~OGBooger (Solution)
−
Note that <math>323</math> multiplied by any of the answer choices results in a 5 or 6 digit <math>n</math>. 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 <math>n</math>(in result to adding the chosen divisor) won't cause our number to multiply to more than 4 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 (1) even & (2) with a factor of <math>17</math> or <math>19</math>.  We see 340 achieves this and is the smallest to do so (646 being the other). So, we get <math>\boxed{\text{(C) }340}</math>
 
−
-- OGBooger
 
−
-- minor changes by Pearl2008
 
  
−
==Solution 5==
+
~Pearl2008 (Minor Edits)
−
We see that <math>323</math> is <math>18^2 -1</math>, which means it's prime factorization is <math>(18-1)(18+1)</math> which is <math>17 \cdot 19.</math> The factors that are possible is an even number with <math>17</math> as a factor or <math>19</math> as a factor. The smallest factors larger than <math>17 \cdot 19</math> are <math>17 \cdot 20</math> or <math>19 \cdot 18</math>, and we can see that <math>17 \cdot 20</math> is smaller than <math>19 \cdot 18</math> since <math>19 \cdot 18</math> is closer to a square, so therefore our answer is <math>\boxed{\text{(C) }340}</math>.
 
  
 
==Video Solution 1==
 
==Video Solution 1==

Revision as of 23:58, 17 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)

~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 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 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

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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