Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 3"
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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== | ||
{{UNCO Math Contest box|n=II|year=2006|num-b=2|num-a=4}} | {{UNCO Math Contest box|n=II|year=2006|num-b=2|num-a=4}} | ||
| − | [[Category:Introductory | + | [[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 
).
What is the 
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]](http://latex.artofproblemsolving.com/9/5/3/953b3ed19882010780e63c1f5a6644729f8a74d3.png)
Solution
The way the numbers cycle are 
for a total of 14 numbers. 
, with a remainder of 4. The fourth number in the cycle is 2, so the answer is 
See Also
| 2006 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 | ||
