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

Guest Jul 1, 2022

#1**0 **

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 \(7^2 \times 2^3 = \color{brown}\boxed{392}\)

BuilderBoi Jul 1, 2022

#3**0 **

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!

BuilderBoi Jul 1, 2022