Difference between revisions of "2012 AIME II Problems/Problem 10"

Revision as of 12:35, 6 August 2020

Problem 10

Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$ .

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .

Solution

Solution 1

We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational.

Let $x = a + \frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \le b < c$ (essentially, $x$ is a mixed number). Then, $n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}$

Here it is sufficient for $\frac{ab}{c}$ to be an integer. We can use casework to find values of n based on the value of a:

$a = 0 \implies$ nothing because n is positive

$a = 1 \implies \frac{b}{c} = \frac{0}{1}$

$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$

The pattern continues up to $a = 31$ . Note that if $a = 32$ , then $n > 1000$ . However if $a = 31$ , the largest possible x is $31 + 30/31$ , in which $n$ is still less than $1000$ . Therefore the number of positive integers for n is equal to $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$

Solution 2

Notice that $x\lfloor x\rfloor$ is continuous over the region $x \in [k, k+1)$ for any integer $k$ . Therefore, it takes all values in the range $[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$ , the maximum value attained is $31*32 < 1000$ . It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

@@ Line 6: / Line 6: @@
 == Solution ==
+=== Solution 1 ===
 We know that <math>x</math> cannot be irrational because the product of a rational number and an irrational number is irrational (but <math>n</math> is an integer). Therefore <math>x</math> is rational.
@@ Line 23: / Line 24: @@
-The pattern continues up to <math>a = 31</math>. Note that if <math>a = 32</math>, then <math>n > 1000</math>. However if <math>a = 31</math>, the largest possible x is <math>31 + 30/31</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31*32}{2} = \boxed{496}.</math>
+The pattern continues up to <math>a = 31</math>. Note that if <math>a = 32</math>, then <math>n > 1000</math>. However if <math>a = 31</math>, the largest possible x is <math>31 + 30/31</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.</math>
-== Solution 2==
+=== Solution 2===
 Notice that <math>x\lfloor x\rfloor</math> is continuous over the region <math>x \in [k, k+1)</math> for any integer <math>k</math>. Therefore, it takes all values in the range <math>[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)</math> over that interval. Note that if <math>k>32</math> then <math>k^2 > 1000</math> and if <math>k=31</math>, the maximum value attained is <math>31*32 < 1000</math>. It follows that the answer is <math> \sum_{k=1}^{31} (k+1)k-k^2  = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.</math>

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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