If a and b are both positive integers greater than 1 and a^b = a^(11b)

Analyzing the question:
Since we are given \(a^b=a^{(11b - 60)}\) while knowing \(a\) is greater than 1, the only way for this to be true is to let the exponents equate since with a base of \(a > 1\), \(a^x\) is an increasing function. \(b = 11b - 60\) and \(b = 60/10 = 6\), we would like to know the value of \(a\) now.

Statement 1:
\(a^ 2 = |a|^2 = 7|a|\)
\(|a| = 7\)
a = 7 or a = -7, however \(a\) must be positive so a = 7 is the only case we can take. Sufficient.

Statement 2:
Same as statement 1, sufficient.

Ans: D

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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since a is positive and \(a^b = a^{11b-60}\), we have \(b = 11b - 60\) or \(b = 6\).

Since we have 2 variables (\(a\) and \(b\)) and 1 equation(\(b=6\)), D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
\(a^2 = 7|a|\)
⇔ \(a^2 = 7a\), since \(a>0\)
⇔ \(a^2 - 7a = 0\)
⇔ \(a(a-7) = 0\)
⇔ \(a = 0\) or \(a = 7\)
Since \(a\) is positive, we have \(a = 7\)

Thus, we have \(a \cdot b = 7 \cdot 6 = 42\).

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)

\(|a| = 7\)
⇔ \(a = ±7\)
Since a is positive, \(a = 7\).

Thus, we have \(a \cdot b = 7 \cdot 6 = 42\).

Since condition 2) yields a unique solution, it is sufficient.

Therefore, D is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

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