Difference between revisions of "2009 AIME I Problems/Problem 6"

Latest revision as of 15:38, 15 February 2021

Problem

How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ ?

Solution 1

First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$ .

Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$ :

For ${\lfloor x\rfloor}=0$ , $N=1$ as long as $x \neq 0$ . This gives us $1$ value of $N$ .

For ${\lfloor x\rfloor}=1$ , $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$

Therefore, $N=1$ . However, we got $N=1$ in case 1 so it got counted twice.

For ${\lfloor x\rfloor}=2$ , $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$

This gives us $3^2-2^2=5$ $N$ 's

For ${\lfloor x\rfloor}=3$ , $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$

This gives us $4^3-3^3=37$ $N$ 's

For ${\lfloor x\rfloor}=4$ , $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$

This gives us $5^4-4^4=369$ $N$ 's

Since $x$ must be less than $5$ , we can stop here and the answer is $1+5+37+369= \boxed {412}$ possible values for $N$ .

Alternatively, one could find that the values which work are $1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt[3]{28}^{\lfloor\sqrt[3]{28}\rfloor},\ \sqrt[3]{29}^{\lfloor\sqrt[3]{29}\rfloor},\ \sqrt[3]{30}^{\lfloor\sqrt[3]{30}\rfloor},\ ...,$ $\ \sqrt[3]{63}^{\lfloor\sqrt[3]{63}\rfloor},\ \sqrt[4]{257}^{\lfloor\sqrt[4]{257}\rfloor},\ \sqrt[4]{258}^{\lfloor\sqrt[4]{258}\rfloor},\ ...,\ \sqrt[4]{624}^{\lfloor\sqrt[4]{624}\rfloor}$ to get the same answer.

Solution 2

For a positive integer $k$ , we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$ . Then $x=\sqrt[k]{N}$ , and because $k \le x < k+1$ , we have $k^k \le x^k < (k+1)^k$ , and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$ . The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$ , and this is equal to $(k+1)^k-k^k$ . If $k>4$ , the value of $x^k$ exceeds $1000$ , so we only need to consider $k \le 4$ . The requested number of values of $N$ is the same as the number of values of $x^k$ , which is $\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}$ .

Video Solutions

Video Solution 1

Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s

~IceMatrix

Video Solution 2

https://youtu.be/kALrIDMR0dg

~Shreyas S

Video Solution 3

Projective Solution: https://youtu.be/fUef_tVnM5M

~Shreyas S

@@ Line 3: / Line 3: @@
 How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</math> has a solution for <math>x</math>?
-== Solution ==
+== Solution 1==
 First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>.

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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