If A = (1+ 13)(1+ 13^2)(1+ 13^4)(1+ 13^8), what is 1 - 23A?

If A = (1+ 13)(1+ 13^2)(1+ 13^4)(1+ 13^8), wh… | GMAT Sprint

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PS · Problem Solving

705-805

Divisibility/Multiples/Factors

If $A = (1+\frac{1}{3})(1+\frac{1}{3^2})(1+\frac{1}{3^4})(1+\frac{1}{3^8})$ , what is $1 - \frac{2}{3}A$ ?

Correct answer marked below

\frac{1}{3}

\frac{2}{3^2}

\frac{2}{3^4}

\frac{2}{3^8}

\frac{1}{3^{16}}

Correct

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Explanation

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A = (1 + \frac{1}{3})(1 + \frac{1}{3^2})(1 + \frac{1}{3^4})(1 + \frac{1}{3^8})

A = (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8})

We can write

\frac{2}{3} = \frac{(3 - 1)}{3}

. Also, remember

(a+b)(a-b) = a^2 - b^2

. We will apply this rule to simplify the problem.

\frac{2}{3}A = (\frac{3 - 1}{3}) (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8})

\frac{2}{3}A = (\frac{3^2-1}{3^2}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8})

…

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What this question tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Forgetting that 1 is not prime and that every integer divides 0.