Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 1"
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== Problem == | == Problem == | ||
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| − | == Solution 1== | + | == Solutions == |
| + | === Solution 1 === | ||
We can try combinations of numbers that add to <math>100</math>. We soon find that one must shoot at the target <math>6</math> times and hit the <math>16</math> ring <math>2</math> times and hit the <math>17</math> ring <math>4</math> times. | We can try combinations of numbers that add to <math>100</math>. We soon find that one must shoot at the target <math>6</math> times and hit the <math>16</math> ring <math>2</math> times and hit the <math>17</math> ring <math>4</math> times. | ||
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| + | === Solution 2 === | ||
| + | Since <math>100=2^2\cdot5^2</math>, we can take <math>\pmod25</math> and <math>\pmod4</math> of the <math>6</math> numbers. From <math>\pmod25</math> we have the set (in order) <math>{15,14,-1,-2,-8,16}</math>, and from <math>\pmod4</math> we have the set <math>{0,-1,0,-1,1,0}</math>. We notice that as <math>8</math> is <math>1/2</math> times <math>16</math>, we probably have two times more <math>17</math>s than <math>16</math>s, and it works perfectly, so one has to shoot the <math>16</math> ring <math>2</math> times and the <math>17</math> ring <math>2</math> times. | ||
== See also == | == See also == | ||
Latest revision as of 23:17, 19 January 2023
Problem
How many times must one shoot at this target, and which rings must one hit in order to score exactly 
points?
![[asy] draw(circle((0,0),1),black); draw(circle((0,0),2),black); draw(circle((0,0),3),black); draw(circle((0,0),4),black); draw(circle((0,0),5),black); draw(circle((0,0),6),black); MP("40",(0,0-.3),N); MP("39",(0,1-.1),N); MP("24",(0,2-.1),N); MP("23",(0,3-.1),N); MP("17",(0,4-.1),N); MP("16",(0,5-.1),N); [/asy]](http://latex.artofproblemsolving.com/c/a/9/ca92e43d47d5fa16d0bd1cd84cb92dee8d52e9d8.png)
Solutions
Solution 1
We can try combinations of numbers that add to . We soon find that one must shoot at the target 
times and hit the 
ring 
times and hit the 
ring 
times.
Solution 2
Since 
, we can take 
and 
of the numbers. From we have the set (in order) 
, and from we have the set 
. We notice that as 
is 
times , we probably have two times more s than s, and it works perfectly, so one has to shoot the ring times and the ring times.
See also
| 1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
