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

Revision as of 21:26, 11 November 2019

Problem

Mary chose an even $4$ -digit number $n$ . She wrote down all the divisors of $n$ in increasing order from left to right: $1,2,...,\dfrac{n}{2},n$ . 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

Prime factorizing $323$ gives you $17 \cdot 19$ . The desired answer needs to be a multiple of $17$ or $19$ , because if it is not a multiple of $17$ or $19$ , the LCM, or the least possible value for $n$ , will not be more than 4 digits. Looking at the answer choices, $\fbox{(C) 340}$ is the smallest number divisible by $17$ or $19$ . Checking, we can see that $n$ would be $6460$ .

Solution 2

Let the next largest divisor be $k$ . Suppose $\gcd(k,323)=1$ . Then, as $323|n, k|n$ , therefore, $323\cdot k|n.$ However, because $k>323$ , $323k>323\cdot 324>9999$ . Therefore, $\gcd(k,323)>1$ . Note that $323=17\cdot 19$ . Therefore, the smallest the gcd can be is $17$ and our answer is $323+17=\boxed{\text{(C) }340}$ .

Solution 3

Again, recognize $323=17 \cdot 19$ . The number must be even and 4 digits, so its prime factorization must then be $17 \cdot 19 \cdot 2 \cdot n$ . Then, $1000\leq 646n \leq 9998 \implies 2 \leq n \leq 15$ . 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 and 19 is $17 \cdot 2 \cdot 10 = 17 \cdot 20=340$ or $19 \cdot 2 \cdot 9 = 19 \cdot 18=342$ . Only \boxed{\text{(C) }340}, which is also lower, is an answer choice. 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

Video

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

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.

@@ Line 10: / Line 10: @@
 ==Solution 2==
 Let the next largest divisor be <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the gcd can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{(C) }340}</math>.
+==Solution 3==
+Again, recognize <math>323=17 \cdot 19</math>. The number must be even and 4 digits, so its prime factorization must then be <math>17 \cdot 19 \cdot 2 \cdot n</math>. Then, <math>1000\leq 646n \leq 9998 \implies 2 \leq n \leq 15</math>. Since <math>15 \cdot 2=30</math>, the prime factorization of the number after 323 needs to have either 17 or 19. The next highest product after 17 and 19 is <math>17 \cdot 2  \cdot 10 = 17 \cdot 20=340</math> or <math>19 \cdot 2  \cdot 9 = 19 \cdot 18=342</math>. Only \boxed{\text{(C) }340}, which is also lower, is an answer choice. 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
 ==Video==

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