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A is always 1 D & E can be reduced to 2^200 so either B or C. B can be reduced to 2^300 = 4^150 = 8^100 C can be reduced to 3^300 = 9^150 hence C > B. SO answer is C

I think you meant \(3^{200}\) in C. Thus C is reduced to \(9^{100}\). The answer is still C, I just wanted to make things clearer.

bangalorian2000 wrote:

A is always 1 D & E can be reduced to 2^200 so either B or C. B can be reduced to 2^300 = 4^150 = 8^100 C can be reduced to 3^300 = 9^150 hence C > B. SO answer is C

I think you meant \(3^{200}\) in C. Thus C is reduced to \(9^{100}\). The answer is still C, I just wanted to make things clearer.

bangalorian2000 wrote:

A is always 1 D & E can be reduced to 2^200 so either B or C. B can be reduced to 2^300 = 4^150 = 8^100 C can be reduced to 3^300 = 9^150 hence C > B. SO answer is C

C is correct. Here is my 2 cents. 1^999 = 1 2^300 = (2^3)^100 = 8^100 3^200 = (3^2)^100 = 9^100 4^100 = (2^2)^100 = 2^200 < 2^300 16^50 = (2^4)^50 = 2^200 < 2^300

So, 3^200 = 9^100 is greatest.
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