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# For all positive integers n and m, the function A(n) equals the follow

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For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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24 Apr 2015, 01:58
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Difficulty:

95% (hard)

Question Stats:

43% (02:41) correct 57% (02:28) wrong based on 204 sessions

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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.

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Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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25 Apr 2015, 20:50
1
For all positive integers n and m,
A(n) = (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,
note: i think there's a typo in the above function, A(n) could be (1 + 1/2 + 1/2^2)(1 + 1/3 + 1/3^2)(1 + 1/5 + 1/5^2)…(1 + 1/p_n + 1/p_n^2)

B(m) = sum of the reciprocals of all the positive integers from 1 through m, inclusive.

A(5), here 5 represents the 5th smallest prime number; the 5th smallest prime number is 11 {2, 3, 5, 7, 11, ...}
A(5) = (1 + 1/2 + 1/2^2)(1 + 1/3 + 3^2)(1 + 1/5 + 5^2)(1 + 1/7 + 1/7^2)(1 + 1/11 + 11^2)
The distributed expansion of A(5) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/9 + 1/10 + 1/11 + 1/12 + 1/14 + ...
missing numbers are 1/8, 1/13, 1/16, ....

B(25) = (1 + 1/2 + 1/3 + 1/4 + ... + 1/16 + 1/17 + ... + 1/25)
here the largest reciprocal is 1 and the reciprocals are arranged in descending order based on their values

The largest reciprocal that present in B(25) but not in A(5) is 1/8

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For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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26 Apr 2015, 02:35
1
A(5) is a product of 5 summs of set of 3 numbers (up to 11, the 5th prime) that looks like (1 + ...)*(1+...)*(1+...)*(1+...)*(1+...) which means that every number inside the brackets can be present in B(25). Quick glance through answer choices gives us an easy E coz 1/4 is present in the first bracket, 1/5 in 3rd, 1/6 is 1/2*1/3*1*1*1, 1/7 is in 4th and 1/8 is the only one left which is the correct answer, E.
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Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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26 Apr 2015, 09:41
1
1
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.

According to the problem,

A(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) .... [ as 5th smallest Prime number is 11 ]
=> A(5) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 ..... etc.

Now , the answer to the question is " The largest reciprocal of an integer in the sum that B(25) represents that is NOT present in the distributed expansion of A(5) ".

So, Scanning through the answer choices all the choices can be eliminated (as they are present in the expansion of A(5))except choice E , i.e. 1/8.

Also to be noted , for this question we need not evaluate at B(25).

Moreover , 1/8 > (1/13, 1/16, .... etc) , thus 1/8 is the largest reciprocal.

Math Expert
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Posts: 52906
Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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27 Apr 2015, 01:03
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.

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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Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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04 Aug 2015, 06:03
Hi Bunuel,

Nice question! I thought $$\frac{1}{6}$$ was correct, but, of course, you can get $$\frac{1}{6}$$ by multiplying $$\frac{1}{2}x\frac{1}{3}$$.

Can you please correct the typo that sudh described above. It can be misleading for people solving the question! Thank you in advance!
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Thank you very much for reading this post till the end! Kudos?

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Joined: 29 Dec 2014
Posts: 67
Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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11 Nov 2016, 20:19
Bunuel wrote:
Bunuel wrote:

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)

Hi:

Please can you help me understand why are we considering prime numbers only till 11 and not beyond.

Thanks
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Joined: 29 Dec 2014
Posts: 67
Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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11 Nov 2016, 20:22
WilDThiNg wrote:
Bunuel wrote:
Bunuel wrote:

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)

Hi:

Please can you help me understand why are we considering prime numbers only till 11 and not beyond.

Thanks

Ah, got it - thanks for the detailed solution
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Re: For all positive integers n and m, the function A(n) equals the follow  [#permalink]

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16 Oct 2018, 01:00
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Re: For all positive integers n and m, the function A(n) equals the follow   [#permalink] 16 Oct 2018, 01:00
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