Difference between revisions of "2006 SMT/General Problems/Problem 16"
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− | The condition we are looking for is <math> | + | The condition we are looking for is <math>110(n-1)\equiv 0 \mod 360</math>, because if any other <math>A_n = A_1</math>, than <math>110(n-1)</math> must be a multiple of 360. This is because the central angle formed between two consecutive points is 110 degrees, and because we have rotated <math>(n-1)</math> times by the time we place <math>A_n</math>. |
− | <cmath> | + | <cmath>110(n-1)\equiv 0\mod 360 \Rightarrow 10(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 37\mod 360</cmath> |
− | + | We want the first <math>A_n</math> that satisfies the conditions from the problem, therefore, our answer is <math>n=37</math> or <math>\boxed{A_{37}}</math> | |
− | We |
Latest revision as of 12:43, 15 January 2020
Problem
Points are placed on a circle with center
such that
and
for all positive integers
. What is the smallest
for which
?
Solution
The condition we are looking for is , because if any other
, than
must be a multiple of 360. This is because the central angle formed between two consecutive points is 110 degrees, and because we have rotated
times by the time we place
.
We want the first
that satisfies the conditions from the problem, therefore, our answer is
or