Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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− | 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 323 needs to have either 17 or 19. The next highest product after 17 | + | 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 323 needs to have either 17 or 19. 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>. |
Revision as of 21:30, 11 November 2019
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 .
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 323 needs to have either 17 or 19. 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 |
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