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

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

Jul 1, 2022

#1
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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
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Thanks but im still confused how you got 7^2 and 2^3

Jul 1, 2022
#3
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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
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The smallest n ==392, because:

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

Jul 4, 2022
#5
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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