Two different primes may be said to “rhyme” around an integer if they

Question

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

Source: MGMAT

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

Source: MGMAT

Heaven knows what I'll do if I encounter such a question on GMAT!! It is solvable no doubt but very time consuming.. Please do post the time you take to solve this question.. I took 1.4 minutes to grasp the question, then left it as I thought it would eat away the valuable remaining time on the test.

Bunuel · Accepted Answer

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*17=34 --> 34=3+31=5+29=11+23: 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.

KarishmaB · Answer

Alternative solution:

Since we are concerned with integers between 1 and 20, write down the primes till 40.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
(you should be very comfortable with the first few primes... )

2, 3, 5, 7, 11,  12 , 13, 17, 19, 23, 29, 31, 37 - Three pairs (11,13), (7,17), (5, 19)

2, 3, 5, 7, 11, 13,  15,  17, 19, 23, 29, 31, 37 - Three pairs (13, 17), (11, 19), (7, 23)

2, 3, 5, 7, 11, 13,  17 , 19, 23, 29, 31, 37 - Three pairs (11, 23), (5, 29), (3, 31)

2, 3, 5, 7, 11, 13, 17,  18,  19, 23, 29, 31, 37 - Four pairs (17, 19), (13, 23), (7, 29), (5, 31)

2, 3, 5, 7, 11, 13, 17, 19,  20,  23, 29, 31, 37 - definitely cannot be more than 4 since there are only 4 primes more than 20. So must be less than 4 pairs. Ignore.
Answer (D).

It doesn't take too much time to look for equidistant pairs...

Bunuel · Answer

Check other  Special Numbers and Sequences  questions in our  Special Questions Directory .

carcass · Answer

Thanks Bunuel.

This is the first time I see something like that in a question. Intersting is your formula because I don't understand why if we had 20 numbers we worked with 40.

Can you give me some links to investigate a little further this concept ??'

Thnaks again for explanation.

Bunuel · Answer

There is no special concept behind it. We have that:  2n=p_1+p_2 , for some integer  n . Answer choices give different values of  n  and we should find out which  n  has the greatest number of distinct rhyming primes around it. When plugging values from answer choices for  n  in  2n=p_1+p_2 , you'll have  2n  to wok with since there is  2n  in the formula.

Hope it's clear.

royal · Answer

why are we considering till 40?? I did not get it

doe007 · Answer

As the highest integer, for which rhyming pair to be found, is 20, we need to consider equal range below the number 20 and above the number 20. In fact, we need to consider the range (2,38) as the lowest prime is 2.

anon1111 · Answer

Bunuel and Karishma,

17 has four set of rhyming primes. You both haven't considered (3,31) as a possible answer.

KarishmaB · Answer

Both Bunuel and I have considered 3 and 31 as rhyming primes for 17 in our solutions above.

2, 3, 5, 7, 11, 13,   17  , 19, 23, 29, 31, 37 - Three pairs (11, 23), (5, 29),  (3, 31)

pacifist85 · Answer

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?

GMATinsight · Answer

Just solve it by checking every option

Answer : Option D

EMPOWERgmatRichC · Answer

Hi All,

This question requires a bit of a tactical approach combined with "brute force." The answers to this question provide 5 possible values that COULD have the GREATEST number of rhyming primes, so we just have to figure out which one it is. We can't afford to stare at the problem though; to be efficient, we have to get in and throw some punches.

We're told to look for prime numbers that are equidistant from a number, but we're limited to numbers from 1 to 20, inclusive.

Let's list out the primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 (nothing above 40 is required, since there wouldn't be a matching rhyme prime on the other "side" of the number)

Logically, the correct answer will probably be one of the bigger integers, since those values allow for a greater number of primes that are "lower." We can quickly check them all though.

A: 12 - 5&19, 7&17, 11&13 
B: 15 - 7&23, 11&19, 13&17 
C: 17 - 3&31, 5&29, 11&23 
D: 18 - 5&31, 7&29, 13&23, 17&19 
E: 20 - 3&37, 11&29, 17&23

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

BrentGMATPrepNow · Answer

If two numbers are rhyming primes, then   the integer they rhyme around will be the AVERAGE of the two primes.

For example, 3 and 7 rhyme around 5. Notice that the AVERAGE of 3 and 7 is 5. 
Likewise, 5 and 23 rhyme around 14, and the AVERAGE of 5 and 23 is 14.

Now onto the solution...

List several primes:   2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,....

Now check the answer choices:

A)12
For 12 to be the integer that two primes rhyme around, we need 2 primes that have an AVERAGE of 12. In other words, we need 2 primes that ADD to 24. Now check the list of primes to find pairs that satisfy this condition. 
We get: 5 & 19, 7 & 17, 11 & 13
Total of 3 pairs.

B)15
So, we need 2  distinct  primes that ADD to 30.
We get: 7 & 23, 11 & 19, 13 & 17
Total of 3 pairs.

C)17
So, we need 2  distinct  primes that ADD to 34.
We get: 3 & 31, 5 & 29, 11 & 23
Total of 3 pairs.

D)18
So, we need 2  distinct  primes that ADD to 36.
We get: 5 & 31, 7 & 29, 13 & 23, 17 & 19
Total of 4 pairs.

E)20
So, we need 2  distinct  primes that ADD to 40.
We get: 3 & 37, 11 & 29, 17 & 23
Total of 3 pairs.

Answer: D

Cheers,
Brent

ScottTargetTestPrep · Answer

We are looking for pairs of prime numbers that are equidistant from a given integer. The approach we will use is to consider each prime number less than that integer and then pair that number with its equidistant match on the other side of the integer. If both numbers are prime, then we have a rhyme.

For example, for choice A, which is 12, we will consider the prime numbers less than 12. Since 11 is 1 less than 12, then 13 is its match because it is 1 more than 12. The pair (11, 13) is a rhyme because both numbers are prime. For the next prime, we skip 9 (not prime) and use 7. Since 7 is 5 less than 12, we see that 17 is 5 greater than 12, and the pair is (7, 17), which is a rhyme. The next pair is (5, 19), which is a rhyme. The final consideration is the pair (3, 21), but this is not a rhyme because 21 is not prime. Thus, for choice A, the integer 12 has 6 distinct rhyming primes.

Let’s use the same approach for the remaining answer choices B through E.

B. 15

(13, 17) ... Yes; (11, 19) ... Yes; (9, 21) ... No; (7, 23) ... Yes; (5, 25) ... No; 3, 27) ... No

We see that 15 has 6 distinct rhyming primes around it.

C. 17

(15, 19) ... No; (13, 21)… No; (11, 23) ... Yes; (9, 25) ... No; (7, 27) ... No; (5, 29) ... Yes; (3, 3)... Yes

We see that 17 has 6 distinct rhyming primes around it.

D. 18

(17, 19)… Yes; (15, 21) ... No; (13, 23) ... Yes; (11, 25) ... No; (9, 27) ... No; (7, 29) ... Yes; (5, 31) ... Yes; 
(3, 33) ... No

We see that 18 has 8 distinct rhyming primes around it.

E. 20

(19, 21) ... No; (17, 23) ... Yes; (15, 25) ... No; (13, 27) ... No; (11, 29) ... Yes; (9, 31) ... No; (7, 33) ... No; (5, 35) ... No; (3, 37) ... Yes

We see that 20 has 6 distinct rhyming primes around it.

Answer: D

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Two different primes may be said to “rhyme” around an integer if they

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