Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
| − | We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s | + | We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>. |
| + | By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>. | ||
| + | By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. | ||
| + | We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards. | ||
| + | After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9. | ||
| + | So our answer is <math>1239</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:04, 9 February 2017
Contents
Problem
How many triangles with positive area have all their vertices at points 
in the coordinate plane, where 
and 
are integers between 
and 
, inclusive?
Solution 1
There are a total of 
sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes 
degenerate triangles, 4 lines that go through exactly 4 points, which contributes 
degenerate triangles, and 16 lines that go through exactly three points, which contributes 
degenerate triangles. Subtracting these degenerate triangles, we get an answer of 
.
Solution 2
We can find out that the least number of digits the number 
is 
, with 
's and one .
By randomly mixing the digits up, we are likely to get: 
...
....
By adding 1 to this number, we get: ...
...
.
We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.
After subtracting 6 from every number, we can conclude that 
(originally 
) is the only number divisible by 9.
So our answer is
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
