2018 AMC 10B Problems/Problem 21
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 .
~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.