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S97-09

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S97-09  [#permalink]

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New post 16 Sep 2014, 00:51
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Question Stats:

35% (02:11) correct 65% (01:54) wrong based on 82 sessions

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Two different primes may be said to"rhyme" around an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it?

A. 12
B. 15
C. 17
D. 18
E. 20

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Re S97-09  [#permalink]

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New post 16 Sep 2014, 00:51
Official Solution:

Two different primes may be said to"rhyme" around an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it?

A. 12
B. 15
C. 17
D. 18
E. 20


First, make sure that you understand the new concept that the problem presents: "rhyming primes," which are the same distance on the number line from a central number. You are given an example: 3 and 7 rhyme around 5, since both are 2 units away from 5 on the number line. Don't let the new terminology confuse you. Instead, try to rephrase the concept into something you're more familiar with. Ideally, you recognize that "rhyming" is just another way to say "average (arithmetic mean)" - saying "3 and 7 rhyme around 5" is the same thing as saying "the average of 3 and 7 is 5." So, rhyming primes rhyme around their average. Alternatively, we can say that the sum of two rhyming primes (e.g., 3 and 7) is twice the central number (\(2 \times 5 = 10\)). Sums are quick operations, so it might be good to rephrase our question in terms of taking sums of two primes.

We are asked which integer between 1 and 20, inclusive, has the greatest number of rhyming primes around it. So we should list out the primes up to 40, since the larger number in any pair of rhyming primes that average to 20 would have to be below 40 (primes are restricted to positive integers).

Here are the primes less than 40:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.

Rephrasing the question in terms of sums, we can ask: what number between 1 and 20, when multiplied by 2, can be expressed as a sum of two different primes from this list in the greatest number of different ways?

We should now start from the answer choices, rather than test all 20 theoretical possibilities. Unfortunately, there is no shortcut; you actually have to check the possibilities. Primes are unevenly distributed, so there’s no way to intuit the answer.

We should start by checking the highest number, because we will probably be able to construct more valid pairs around larger numbers than around smaller numbers. Construct the pairs by inspecting your list of primes. Since you know the smaller primes better than larger primes, and since the larger primes are more spread out, put the larger prime first in the potential sum, then look for the smaller prime in the second position.

(E) \(20 \times 2 = 40\)
\(37 + 3 = 40\)
\(29 + 11 = 40\)
\(23 + 17 = 40\)

20 has 3 rhyming pairs of primes, or 6 rhyming primes.

(D) \(18 \times 2 = 36\)
\(31 + 5 = 36\)
\(29 + 7 = 36\)
\(23 + 13 = 36\)
\(19 + 17 = 36\)

18 has 4 rhyming pairs of primes, or 8 rhyming primes. If we had to pick right now, because of time pressure, we would pick D.

(C) \(17 \times 2 = 34\)
\(31 + 3 = 34\)
\(29 + 5 = 34\)
\(23 + 11 = 34\)

\(19 + 15\) doesn’t work

Also, \(17 + 17\) doesn’t work, because the definition of "rhyming" indicates that the primes must be different.

17 has 3 rhyming pairs of primes, or 6 rhyming primes. D is still our tentative answer.

(B) \(15 \times 2 = 30\)

\(29 + 1\) doesn’t work, because 1 isn’t prime.
\(23 + 7 = 30\)
\(19 + 11 = 30\)
\(17 + 13 = 30\)

15 has 3 rhyming pairs of primes, or 6 rhyming primes. D is looking better and better.

(A) \(12 \times 2 = 24\)
\(19 + 5 = 24\)
\(17 + 7 = 24\)
\(13 + 11 = 24\)

12 has 3 rhyming pairs of primes, or 6 rhyming primes.


Answer: D
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Re: S97-09  [#permalink]

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New post 30 Dec 2014, 06:32
Hello,

I wanted to share how I ended up with the correct answer. It is probably a lucky choice, but just in case I wanted to share.

So, I didn't see the connection with the mean (even though statistics is my biggest strength). What I did was to first find the primes up to 20, just to see if there is a pattern that makes sense.

So, I lined them up, smaller to larger, and tried to find a number that is between 1 and 20. For me this meant 1<x<20, so I wanted a number that is one of these: 2,3,4....,19.

Then, I realised that there is no upper limmit to the primes - so there is no reason why they should stop at 19. What I realised then, is that the number that has most primes should be the highest possible in the range we are given: one of 2,3,4,....,19. So, 19 being the highest value, it is logical that this one would have the most primes around it. I rejected 20, because of the range, so I chose 18 (D), because it was the second highest.

Does it make any sense?
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Re S97-09  [#permalink]

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New post 05 Mar 2015, 03:20
I think this question is good and helpful.
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Re: S97-09  [#permalink]

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New post 05 Mar 2015, 05:47
1
Bunuel wrote:
Two different primes may be said to"rhyme" around an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it?

A. 12
B. 15
C. 17
D. 18
E. 20


ALTERNATIVE EXPLANATION:

As per definition two different primes \(p_1\) and \(p_2\) are "rhyming primes" if \(n-p_1=p_2-n\), for some integer \(n\) --> \(2n=p_1+p_2\). So twice the number \(n\) must equal to the sum of two different primes, one less than \(n\) and another more than \(n\).

Let's test each option:

A. 12 --> 2*12=24 --> 24=5+19=7+17=11+13: 6 rhyming primes (start from the least prime and see whether we can get the sum of 24 by adding another prime more than 12 to it);
B. 15 --> 2*15=30 --> 30=7+23=11+19=13+17: 6 rhyming primes;
C. 17 --> 2*15=30 --> 34=7+23=11+19=13+17: 6 rhyming primes;
D. 18 --> 2*18=36 --> 36=5+31=7+29=13+23=17+19: 8 rhyming primes;
E. 20 --> 2*20=40 --> 40=3+37=11+29=17+23: 6 rhyming primes.

Answer: D.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S97-09  [#permalink]

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New post 13 Feb 2016, 11:37
Bunuel wrote:
Two different primes may be said to"rhyme" around an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it?

A. 12
B. 15
C. 17
D. 18
E. 20



are you **** kidding me..!!!

this question takes at-least 4 min to solve..!!
idiotic ..!!
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Re: S97-09  [#permalink]

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New post 18 Dec 2017, 00:37
pacifist85 wrote:
Hello,

I wanted to share how I ended up with the correct answer. It is probably a lucky choice, but just in case I wanted to share.

So, I didn't see the connection with the mean (even though statistics is my biggest strength). What I did was to first find the primes up to 20, just to see if there is a pattern that makes sense.

So, I lined them up, smaller to larger, and tried to find a number that is between 1 and 20. For me this meant 1<x<20, so I wanted a number that is one of these: 2,3,4....,19.

Then, I realised that there is no upper limmit to the primes - so there is no reason why they should stop at 19. What I realised then, is that the number that has most primes should be the highest possible in the range we are given: one of 2,3,4,....,19. So, 19 being the highest value, it is logical that this one would have the most primes around it. I rejected 20, because of the range, so I chose 18 (D), because it was the second highest.

Does it make any sense?


Actually, the range, according to the question stem, is an integer between 1 and 20, inclusive. So rejecting 20 based on this would be incorrect. And even though there is no highest value, there is a lowest value (2) that needs to be taken into account.
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Re: S97-09 &nbs [#permalink] 18 Dec 2017, 00:37
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