Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 3"

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==Solution==
 
==Solution==
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The way the numbers cycle are <math>1, 6, 11, 2, 7, 12, 3, 8, 13, 4, 9, 14, 5, 10</math> for a total of 14 numbers. <math>2006\div14=143R4</math>, with a remainder of 4. The fourth number in the cycle is 2, so the answer is <math>\boxed2</math>
  
 
==See Also==
 
==See Also==
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{{UNCO Math Contest box|n=II|year=2006|num-b=2|num-a=4}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:47, 4 December 2016

Problem

The first 14 integers are written in order around a circle.

Starting with 1, every fifth integer is underlined. (That is $1,6,11,2,7,\ldots$). What is the $2006^{th}$ number underlined?

[asy] draw(unitcircle,black); pair A; for(int j=1;j<15;++j){ A=dir(90-(j-1)*(360/14)); MP(string(j),A,A); } [/asy]

Solution

The way the numbers cycle are $1, 6, 11, 2, 7, 12, 3, 8, 13, 4, 9, 14, 5, 10$ for a total of 14 numbers. $2006\div14=143R4$, with a remainder of 4. The fourth number in the cycle is 2, so the answer is $\boxed2$

See Also

2006 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