Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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==Solution 3== | ==Solution 3== | ||
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>. | 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>. | ||
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+ | 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 | ~bjhhar |
Revision as of 23:35, 7 January 2020
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
Since prime factorizing gives you , the desired answer needs to be a multiple of or , this is because if it is not a multiple of or , will be more than a digit number. For example, if the answer were to instead be , would have to be a multiple of for both and to be a valid factor, meaning would have to be at least , which is too big. Looking at the answer choices, and are both not a multiple of neither 17 nor 19, is divisible by . is divisible by , and is divisible by both and . Since is the smallest number divisible by either or it is the answer. Checking, we can see that would be , a four-digit number. Note that is also divisible by , one of the listed divisors of . (If was not divisible by , we would need to look for a different divisor)
-Edited by Shurong.ge
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 .
Solution 3
Again, recognize . The 4-digit number is even, so its prime factorization must then be . Also, , so . Since , the prime factorization of the number after needs to have either or . The next highest product after is or .
You can also tell by inspection that , because is closer to the side lengths of a square, which maximizes the product.
~bjhhar
Video
https://www.youtube.com/watch?v=KHaLXNAkDWE
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.