Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 3"
Mathisfun04 (talk | contribs) (→Solution) |
Strangeplant (talk | contribs) (Solution presented) |
||
| Line 7: | Line 7: | ||
== Solution == | == Solution == | ||
| − | + | Let the smallest number of ants in the army be x. x modulo 10 is 8, so the last digit must be 8. The result of x modulo 7, 11, and 13 is 2 and the LCM of 7, 11, and 13 is 1001, so we can consider x modulo 1001 to be 2. Any multiple of 1001 plus 2 satisfies this condition, so all that must be done is finding the first instance of this where the last digit is 8. 1001*6 is 6006 and addition of 2 yields 6008, which is the smallest number of ants that could be in the army. | |
== See also == | == See also == | ||
Latest revision as of 21:14, 23 November 2018
Problem
An army of ants is organizing a march to the Obama inauguration. If they form columns of 
ants
there are 
left over. If they form columns of 
or 
ants there are 
left over. What is the
smallest number of ants that could be in the army?
Solution
Let the smallest number of ants in the army be x. x modulo 10 is 8, so the last digit must be 8. The result of x modulo 7, 11, and 13 is 2 and the LCM of 7, 11, and 13 is 1001, so we can consider x modulo 1001 to be 2. Any multiple of 1001 plus 2 satisfies this condition, so all that must be done is finding the first instance of this where the last digit is 8. 1001*6 is 6006 and addition of 2 yields 6008, which is the smallest number of ants that could be in the army.
See also
| 2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
