2018 AMC 10B Problems/Problem 25
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 portion of and is the decimal portion. 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
As in Solution 1, we can write . Obviously, , so x is between . Hence, the answer is .
-Pi_3.14_Squared
Solution 5
The following code confirms that there are solutions in total.
typedef long double ld;
int main() {
// x^2=10000{x}
// (t+b)^2=10000b
// b^2+(2t-10000)b+t^2=0
FOR(t,-200,201) {
ld b = 2*t-10000, c = t*t;
ld s1 = (-b+sqrt(b*b-4*c))/2;
ld s2 = (-b-sqrt(b*b-4*c))/2;
if (0 <= s1 && s1 < 1) cout << t+s1 << "\n";
if (0 <= s2 && s2 < 1) cout << t+s2 << "\n";
}
}
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 |
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