I got D on this. 7^196 = 49^98 which is significantly less than 98^98. n could = 1, as any multiple of 98 will be divisible by 7, but it will also surely be divisible by iterations of 7^n far beyond 196, since 49^98 is so much less than 98^98.

Yes, this is correct, if n = 1, even then it can divide 98 ^ 98 and if n = 198, it can still divide it. Answer is D

\(98 = (2)(7)(7) = (2^1)(7^2)\)

So, \(98^{98}\) = \([(2^1)(7^2)]^{98}= (2^{98})(7^{196})\)

At this point we can see that....
\((2^{98})(7^{196})\) is divisible by \(7^1\)
\((2^{98})(7^{196})\) is divisible by \(7^2\)
\((2^{98})(7^{196})\) is divisible by \(7^3\)
.
.
.
\((2^{98})(7^{196})\) is divisible by \(7^{195}\)
\((2^{98})(7^{196})\) is divisible by \(7^{196}\)

In other words \(n\) can have any value from \(1\) to \(196\) inclusive

If \(n=1\), Quantity B is greater
If \(n=196\), the two quantities are equal

Answer: D

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