What a root

Let the fifth root of 'x' be 'a'. The number 10000-a should be positive and its fourth root should be

an integer. We know that fourth root of 10000 is 10. Thus the final result after computing both the

roots will lie between 0 to 10 (inclusive) as we can find some or the other value of a (that is some

value of x) which satisfies this. Thus answer is the number of whole numbers upto 10.

Answer 11

Because the values of x can only be positive, and since we know that negative numbers can't have fourth roots, we can figure out that the number of numbers has to be all of the numbers that have fourth roots < 10000. Since the fourth root of 10000 is 10, it has to be all of the positive integers and zero going up to 10. Including zero, you have 11 numbers, so the answer is $\boxed{11}$ .

we've :

$\sqrt[4]{10^4 - \sqrt[5]{x}} = k$ where $k$ is integer, so can be rewritten as :

$\sqrt[2n]{10^4 - \sqrt[5]{x}} = k$ where $k, n$ is integers, because the root is even implies than $k$ is non-negative integer, also can be rewritten as $\sqrt{\sqrt{10^4 - \sqrt[5]{x}}} = k$ .

by add power $4$ to both sides:

$\sqrt[4]{10^4 - \sqrt[5]{x}} = k$

become:

$10^4 - \sqrt[5]{x} = k^4$

$\sqrt[5]{x} = 10^4 - k^4$

we deduce that $k \leq 10$ because $x$ must be non-negative real number.

by replacing $\sqrt[5]{x}$ by $10^4 - k^4$ , we've:

$\sqrt[4]{10^4 - \sqrt[5]{x}} = k$

become:

$\sqrt[4]{10^4 - (10^4 - k^4)} = k$

$\sqrt[4]{k^4} = k$

then number of positive number from $k = 0$ to $k = 10$ is $11$

$(10000 - x^{1/5} )^{1/4}$ is equivalent to $(10^4 - y)^{1/4}$ , where $y = x^{1/5}$ . Note that since $x^{1/5}$ can take on any real value, it can simply be replaced by another variable, in this case $y$ . For the expression to give an integer, $10^4 - y = a^4$ , for some integer value $a$ . Rearranging, we have $10^4 - a^4 = y$ . Since y cannot be negative, $a \leq 10$ . $a$ can therefore take on any integer value from 0 to 10 inclusive, giving a total of 11 possible values for $x^{1/5}$ .

O menor valor possível que "x" pode assumir, de acordo com o enunciado, é zero; com isso:

$\\ \sqrt[4]{10000 - \sqrt[5]{x}} = \\\\ \sqrt[4]{10000 - 0} = \\\\ \boxed{10}$

E, o maior valor possível que a expressão pode assumir é 10!

Para concluir o problema não se faz necessário descobrir o maior valor que "x" assume, mas sim, o menor valor que a expressão...

Ora, se a expressão tem como valores os números inteiros, o menor valor da expressão é o próprio zero, uma vez que não poderá ser negativo, afinal trata-se de números inteiros e não complexos!

Logo,

$10 - 0 + 1 = \\\\ \boxed{\boxed{11}}$

I will try to explain myself

let t= (10000 -(x)^(1/5) ) ^(1/4)).................. T can only be an integer when t^4=0,1,16,81,.........10^4......i.e. a tally of 11 numbers and hence for 11 corresponding non negative values of X.

As $x$ can be any non-negative real value and the outcome is an integer, it is clear that $x$ itself must be a non-negative integer...

We can substitute $x$ by the $5$ th power of any non-negative integer such that $y := \sqrt[5]{x}$ can be any non-negative integer...

Hence, all we have to find is the number of all $4$ th powers $n$ such that $n \leq 10000$ ...

Since, $10^4=10000$ and $x = 0$ is also a solution, there are $10+1=\fbox{11}$ non-negative real values satisfying the conditions in the question...!!!

The largest outcome is 10, then x has to be 0.

If 9 were an outcome, then $x = (10000 - 9^4)^5$ , which is an non-negative real value. If -9 were an outcome, then $x = (10000 - (-9)^4)^5$ which is the non-negative real value of x . There are 21 possible outcomes: -10, -9 ... 0, 1, 2 ... 10, so there are 11 non-negative real values of x .

Actually I thought this was pretty easy. No, I did not compute the actual values of the possible values of x, but I used my common sense.

What are the integer values of something to the fourth power that are less than or equal to ten-thousand?

$0^4 = 0$

$1^4 =1$

$2^4=16$

$3^4=81$

$4^4=256$

$5^4=625$

$6^4=1 296$

$7^4= 2 401$

$8^4=4 096$

$9^4=6 561$

$10^4=10 000$

So we can stop at $10^4$ , because and thing greater would mean that x would be negative, and anything less than zero would mean that the number would be imaginary.

So there are $\boxed{11}$ ways