Difference between revisions of "2018 AMC 10B Problems/Problem 25"
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Notice the given equation is equivilent to <math>(</math>\lfloor x \rfloor<math>+\{x\})^2=10,000\{x\} </math> | Notice the given equation is equivilent to <math>(</math>\lfloor x \rfloor<math>+\{x\})^2=10,000\{x\} </math> | ||
− | Now we now that <math>\{x\} < 1 so plugging in < | + | Now we now that <math>\{x\} < 1</math> so plugging in <math>1</math> for <math>\{x\}</math> we can find the upper and lower bounds for the values. |
<math>(</math>\lfloor x \rfloor<math> +1)^2 = 10000(1)</math> | <math>(</math>\lfloor x \rfloor<math> +1)^2 = 10000(1)</math> | ||
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And just like Solution <math>2</math> <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done. | And just like Solution <math>2</math> <math>-101 < \lfloor x \rfloor < 99 </math>, and since <math>\lfloor x \rfloor</math> is an integer, there are <math>\boxed{\text{(C)}~199}</math> solutions. Each value of<math> \lfloor x \rfloor</math> should correspond to one value of <math>x</math>, so we are done. | ||
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==See Also== | ==See Also== |
Revision as of 13:24, 29 July 2018
Problem
Let denote the greatest integer less than or equal to . How many real numbers satisfy the equation ?
Solution 1
This rewrites itself to .
Graphing and we see that the former is a set of line segments with slope from to with a hole at , then to with a hole at etc.
Here is a graph of and for visualization.
Now notice that when then graph has a hole at which the equation passes through and then continues upwards. Thus our set of possible solutions is bounded by . We can see that intersects each of the lines once and there are lines for an answer of .
Solution 2
Same as the first solution, .
We can write as . Expanding everything, we get a quadratic in in terms of :
We use the quadratic formula to solve for {x}:
Since , we get an inequality which we can then solve. After simplifying a lot, we get that .
Solving over the integers, , and since is an integer, there are solutions. Each value of should correspond to one value of , so we are done.
Solution 3
Let where is the integer part of and is the fractional part of . We can then rewrite the problem below:
From here, we get
Solving for
Because , we know that cannot be less than or equal to nor greater than or equal to . Therefore:
There are 199 elements in this range, so the answer is .
Solution 4
Notice the given equation is equivilent to \lfloor x \rfloor
Now we now that so plugging in for we can find the upper and lower bounds for the values.
\lfloor x \rfloor \lfloor x \rfloor $$ (Error compiling LaTeX. ! Missing $ inserted.)\lfloor x \rfloor
And just like Solution , and since is an integer, there are solutions. Each value of should correspond to one value of , so we are done.
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.