Bunuel wrote:
For all positive integers n and m, the function A(n) equals the following product:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 3^2)(1 + 1/5 + 5^2)…(1 + 1/p_n + 1/p_n^2), where p_n is the nth smallest prime number, while B(m) equals the sum of the reciprocals of all the positive integers from 1 through m, inclusive. The largest reciprocal of an integer in the sum that B(25) represents that is NOT present in the distributed expansion of A(5) is
A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. 1/8
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:These functions are potentially confusing. Let’s start with B(25), which equals the “sum of the reciprocals of all the positive integers” from 1 through 25, inclusive. In other words, B(25) = 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + … + 1/25.
Now the “distributed expansion of A(5)” is what we get when we expand the product shown:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)(1 + 1/7 + 1/7^2)(1 + 1/11 + 1/11^2)
It’s far too much work to actually “FOIL” this crazy product out, so how can we take a shortcut? Remind yourself of what you’re looking for: the largest reciprocal in the B(25) list that’s NOT in the expansion of this crazy product.
At this point, you might look at the choices. Can you make (A) 1/4 by expanding the product? Sure—take the bolded terms as shown:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)(1 + 1/7 + 1/7^2)(1 + 1/11 + 1/11^2)
1/4 will be 1/2^2 times all of the 1′s in the rest of the product.
How about (B) 1/5? Sure – take 1’s everywhere except the 1/5:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)(1 + 1/7 + 1/7^2)(1 + 1/11 + 1/11^2)
The same logic works for (D) 1/7.
How about (C) 1/6? Take the 1/2 and the 1/3, with 1’s everywhere else:
(1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)(1 + 1/7 + 1/7^2)(1 + 1/11 + 1/11^2)
The only reciprocal on the list that we cannot generate from the product is 1/8, since 1/8 = 1/2^3. We don’t have a 1/2^3 anywhere in the first set of parentheses, and none of the other sums have 2’s in any denominators.
The correct answer is E.Incidentally, this problem was inspired by the most important unsolved problem in mathematics: whether the so-called “Riemann Hypothesis” is true. The functions shown above are simplified versions of the Riemann “zeta” function, which lies at the heart of this problem.
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