Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
| − | Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. Looking at the answer choices, <math>\fbox{(C) 340}</math> is the smallest number divisible by <math>17</math> or <math>19</math>. | + | 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>, because if it is not a multiple of <math>17</math> or <math>19</math>, the LCM, or the least possible value for <math>n</math>, will not be more than 4 digits. Looking at the answer choices, <math>\fbox{(C) 340}</math> is the smallest number divisible by <math>17</math> or <math>19</math>. Checking, we can see that <math>n</math> would be <math>6460</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 20:11, 25 February 2018
Contents
Problem
Mary chose an even 
-digit number 
. She wrote down all the divisors of in increasing order from left to right: 
. At some moment Mary wrote 
as a divisor of . What is the smallest possible value of the next divisor written to the right of ?
Solution 1
Prime factorizing gives you 
. The desired answer needs to be a multiple of 
or 
, because if it is not a multiple of or , the LCM, or the least possible value for , will not be more than 4 digits. Looking at the answer choices, 
is the smallest number divisible by or . Checking, we can see that would be 
.
Solution 2
Let the next largest divisor be 
. Suppose 
. Then, as 
, therefore, 
However, because 
, 
. Therefore, 
. Note that 
. Therefore, the smallest the gcd can be is and our answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
