2018 AMC 10B Problems/Problem 21
- The following problem is from both the 2018 AMC 12B #19 and 2018 AMC 10B #21, so both problems redirect to this page.
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 (Inequalities)
Let be the next divisor written to the right of
If then which contradicts the precondition that is a -digit number.
It follows that Since the smallest possible value of is from which ~MRENTHUSIASM ~tdeng
Solution 2 (Inequalities)
Let be the next divisor written to the right of
Since is even and we have for some positive integer Moreover, since we get As it is clear that must be divisible by or or both.
Therefore, the smallest possible value of is from which ~MRENTHUSIASM ~bjhhar
Solution 3 (Answer Choices)
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 nor , 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 -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 Mathandski
Solution 4 (Answer Choices)
Note that multiplied by any of the answer choices results in a or -digit . So, we need a choice that shares a factor(s) with , such that the factors we'll need to add to the prime factorization of (in result to adding the chosen divisor) won't cause our number to multiply to more than digits. The prime factorization of is , and since we know is even, our answer needs to be
- even
- has a factor of or
We see achieves this and is the smallest to do so ( being the other). So, we get .
~OGBooger (Solution)
~Pearl2008 (Minor Edits)
Video Solution 1
https://www.youtube.com/watch?v=qlHE_sAXiY8
Video Solution 2
https://www.youtube.com/watch?v=KHaLXNAkDWE
Video Solution 3
https://www.youtube.com/watch?v=vc1FHO9YYKQ
~bunny1
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.