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

Revision as of 15:46, 16 February 2018

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

Prime factorizing $323$ gives you $17 \cdot 19$ . Looking at the answer choices, $\fbox{(B) 340}$ is the smallest number divisible by $17$ or $19$ .

2018 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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

@@ Line 6: / Line 6: @@
 ==Solution==
-Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. Looking at the answer choices, only <math>\fbox{(B) 340}</math> is divisible by <math>17</math>.
+Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. Looking at the answer choices, <math>\fbox{(B) 340}</math> is the smallest number divisible by <math>17</math> or <math>19</math>.
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Difference between revisions of "2018 AMC 10B Problems/Problem 21"

Revision as of 15:46, 16 February 2018

Problem

Solution

See Also