HELLPPP What is the smallest positive integer n such that the fourth root of 98n is an integer?

The prime factorization of 98 is $7^2 \times 2$. 

Because we are taking the fourth root, everything must be raised to a power of a multiple of 4, (4, 8, 12, etc.)

This means that the smallest value of n that works is $7^2 \times 2^3 = \color{brown}\boxed{392}$ 

Thanks but im still confused how you got 7^2 and 2^3  

Thanks for the feedback!!

Note that because all the bases in the term $7^4 \times 2^4$ are raised to the 4th power, the 4th root of this is an integer. 

So, we basically have: ${7^4 \times 2^4 \over 7^x \times 2^y }= 7^2 \times 2^1$

Start with $7^4 \over 7^x$ . We want this to equal $7^2$. 

Recall that when you are dividing exponents with the same base, you subtract the exponents ($3^4 \div 3^2 = 3^{4 - 2} = 3^2$).

Applying that here, we have $7^4 \div 7^x = 7^2$, thus $x = 2$. ($7^4 \div 7^2 = 7 ^ {4 - 2} = 7^2$)

Now, doing the same thing to $2^4 \over 2^y$. We want this to equal $2^1$, so $y = 3$. ($2^4 \div 2^3 = 2^{4 - 3} = 2^1$).

So, the answer we are looking for is $7^2 \times 2^3 = \color{brown}\boxed{392}$

Hope this helps!

The smallest n ==392, because:

98 * 392 ==38,416^(1/4) ==14