+0  
 
0
124
5
avatar

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

 Jul 1, 2022
 #1
avatar+2448 
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 \(7^2 \times 2^3 = \color{brown}\boxed{392}\) 

 Jul 1, 2022
edited by BuilderBoi  Jul 1, 2022
 #2
avatar
0

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

 Jul 1, 2022
 #3
avatar+2448 
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!

 Jul 1, 2022
 #4
avatar+1155 
+10

 

The smallest n ==392, because:

 

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

 Jul 4, 2022
 #5
avatar+1155 
+10

you can also think about it this way

 

98 = 2*7*7
k=(2^4 x 7^4)/2x7x7

k=2^3 x 7^2 = 392

nerdiest  Jul 4, 2022

11 Online Users