Difference between revisions of "2018 AMC 10B Problems/Problem 21"
Clutch noob (talk | contribs) m (→Solution 3) |
Rockmanex3 (talk | contribs) (→See Also) |
||
| Line 14: | Line 14: | ||
{{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 16:36, 18 June 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. 
