A and k are integer numbers higher than 3. A is a prime number but k i

Question

The numbers  a  and  k  are integers higher than 3. The number  a  is a prime number but  k  is not. Is  k  a multiple of 6? 
(1)  a + k  is a prime number.
(2)  a + 2k  is a prime number.

mikemcgarry · Accepted Answer

Dear diota2004 , This is a fabulous question! I'm happy to respond! :-)

A few number sense facts (a) in any three consecutive integers, one is divisible by three (b) if K is not divisible by three, then in a set of three consecutive multiple of K, one is divisible by three (c) in a set of three evenly spaced integers, {p, p + d, p + 2d}, one is divisible by three The exception to the last rule is if the difference d is itself divisible by three: in that case, all three numbers in the set can miss the multiples of three. For example {13, 16, 19} None of those are divisible by 3 precisely because each one is exactly one more than a multiple of 3. Keep all this in mind. Now, let's look at the question. Statement #1 : a + k is a prime number. So a and a + k are prime, but k is not. Since all primes are odd, it must be true that k is even. If a = 37, k = 6, then 37 & 43 are prime, and 6 is not. This produces a "yes" answer. If a = 37, k = 10, then 37 & 47 are prime, and 10 is not. This produces a "no" answer. Two possible answers to the prompt. This statement, alone and by itself, is not sufficient . Statement #2 : a + 2k is a prime number. If a = 37, k = 30, then 37 & 97 are prime, and 30 is not. This produces a "yes" answer. If a = 37, k = 15, then 37 & 67 are prime, and 15 is not. This produces a "no" answer. Two possible answers to the prompt. This statement, alone and by itself, is not sufficient . Combined statements Picking numbers is good for demonstrating that something is not sufficient, but we have to use logic to show that this is sufficient. With the combined statements, we know that a set of three evenly spaced numbers are all prime {a, a + k, a + 2k} First of all, since all three are odd, we know that k must be even. Also, since all three are prime, they are not divisible by 3. Since they are not divisible by three, the spacing constant k must be divisible by 3. Any even number divisible by 3 is a multiple of 6. We can give a resounding "yes" to the prompt question. Together, the statements are sufficient . Truly wonderful question! Let me know if you have any questions. Mike :-)

vitaliyGMAT · Answer

Great explanation Mike!

Another approach with application of remainder theorem.

Prime number greater than 3 can be written in the form 6x +/- 1. In other words prime number > 3 will always leave remainders 1 or 5 when divided by 6.

(1) a + k = p, where a and p are primes (and odd as a result). Our k is even (k = p - a = odd - odd = even)

Now, we have prime number, add to it even number >2 and still have prime number. Putting this into remainders we have remainders 1 or 5 ( in mod 6), add some even number and still need to have same remainders 1 or 5 after division by 6. Even number after division by 6 will leave remainders 0, 2 and 4. Let's check them.

Let's say our prime a is in the form 6x + 1. For this we have two fitting options 0 and 4:

6x + 1 + 0 = 6x + 1 and 6x + 1 + 4 = 6x + 5 (still satisfies prime condition with remainder 5).

In case 6x + 5 option 2 will fit. 6x + 5 + 2 = 6x + 7 = 6x + 1.

So our even number k can have remainders 0 (multiple of 6), 2 and 4. Insufficient.

Ex. 5+6 = 11 or 7 + 4 = 11

(2) a + 2k = z, where a and z are primes. In this case k can be either even or odd. Using above logic we can get k = 6x + 1, k = 6x etc. k may or may not be a multiple of 6. Insufficient.

(1)&(2) Now a, a +k, a + 2k must be primes with k - even number >2.

As we already defined even number leaves remainder 0(multiple of 6), 2 and 4 upon division by 6. Let's check them.

a) a = 6x + 1.

6x + 1 + 2 = 6x + 3 (not a prime), 6x + 1 + 2*2 = 6x + 5 (possible prime)

6x + 1 + 4 = 6x + 5 (possible prime), 6x + 1 + 2*4 = 6x + 1 + 8 = 6x + 9 = 6x + 3 (not a prime)

6x + 1 + 0 in all cases give possible prime (6x + 1).

b) a = 6x + 5

6x + 5 + 2 = 6x + 7 = 6x + 1 (possible prime), 6x + 5 +2*2 = 6x + 3 (not a prime).

6x + 5 + 4 = 6x + 3 (not a prime), 6x + 5 + 8 = 6x + 1 (possible prime)

6x + 5 + 0 = 6x + 5 (possible prime), 6x + 5 + 0*2 = 6x + 5 (possible prime)

Ex. 5, 5+6=11, 5+12 = 17

Only remainder o satisfies all requirements, so our k is a multiple of 6. Sufficient.

Answer C.

glagad · Answer

Hi,

" Since they are not divisible by three, the spacing constant k must be divisible by 3." - Can you please help understand the source of this?

It made sense with few numbers, but if there is any concrete derivation/way to prove it, would love to understand

Thanks

hr1212 · Answer

One way to think around this is that, we already know a is not divisible by 3 (because a is prime) and k is even (else a + k would be even). So we can conclude,

a%3 = 1 |or| a%3 = 2

Now, a + k and a + 2k are also prime and not divisible by 3. Let's assume k is not divisible by 3, which would mean

k%3 = 1 and 2k%3 = 2 |or| k%3 = 2 and 2k%3 = 1

a%3 
 k%3 
 2k%3 
 (a + k)%3  
 (a + 2k)%3 
 
  1 
 1 
 2 
 2 
 0 
 
  1 
 2 
 1 
 0 
 2 
 
  2 
 1 
 2 
 0 
 1 
 
  2 
 2 
 1 
 1 
 0

In each of these cases, either a + k or a + 2k will be divisible by 3, if k is not divisible by 3. So, it's necessary for k to be divisible by 3, for both a + k and a + 2k to be prime.

And we already know k is even, which means both these statements can sufficiently establish that k is divisible by 6.

Bunuel · Answer

We are told that a, a + k, and a + 2k are all prime, so none are divisible by 3. We also know that k must be even.

If k is not a multiple of 3, adding k and then 2k moves you through the three possible remainder positions (0, 1, and 2) when dividing by 3. This means one of the three numbers will always be divisible by 3. For example, if a = 5 and k = 4, we get 5, 9, and 13, and 9 is divisible by 3. If a = 5 and k = 8, we get 5, 13, and 21, and 21 is divisible by 3.

If k is a multiple of 3, for example k = 12, we get 5, 17, and 29, all of which leave the same remainder when divided by 3, so none are divisible by 3.

Therefore, if a, a + k, and a + 2k are all primes (none divisible by 3), k must be a multiple of 3, and since k is even, it must be a multiple of 6.

P.S. Pure algebraic questions are no longer a part of the  DS syllabus  of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

Word Problems 
 Work Problems 
 Distance Problems 
 Mixture Problems 
 Percent and Interest Problems 
 Overlapping Sets Problems 
 Statistics Problems 
 Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check  GMAT Syllabus for Focus Edition

You can also visit the  Data Sufficiency forum  and filter questions by  OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources  to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.

nandini14 · Answer

How do we know a +k is not divisible by 3?

Bunuel · Answer

Because we are given that it's a prime, and the only prime divisible by 3 is 3, and we know that a + k is more than 3.

A and k are integer numbers higher than 3. A is a prime number but k i

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GMAT Data Sufficiency (DS) Questions

A and k are integer numbers higher than 3. A is a prime number but k i

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