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

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #19]] and [[2018 AMC 10B Problems|2018 AMC 10B #21]]}}
 ==Problem==
-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>
-==Solution==
+==Solution 1 (Inequalities)==
+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 <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath>
+~MRENTHUSIASM ~tdeng
+==Solution 2 (Inequalities)==
+Let <math>d</math> be the next divisor written to the right of <math>323.</math>
+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.
+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
+==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)==
+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
+==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.
+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
+* has a factor of <math>17</math> or <math>19</math>
+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>.
+~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==
+{{AMC10 box|year=2018|ab=B|num-b=20|num-a=22}}
+{{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}}
+{{MAA Notice}}
-{{MAA Notice}}
+[[Category:Intermediate Number Theory Problems]]

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