Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 3"

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== Solution ==
 
== Solution ==
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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 ==
{{UNC Math Contest box|year=2009|n=II|num-b=2|num-a=4}}
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{{UNCO Math Contest box|year=2009|n=II|num-b=2|num-a=4}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 22:14, 23 November 2018

Problem

An army of ants is organizing a march to the Obama inauguration. If they form columns of $10$ ants there are $8$ left over. If they form columns of $7, 11$ or $13$ ants there are $2$ 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 (ProblemsAnswer KeyResources)
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