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28 Feb 2017, 09:46
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
42% (02:37) correct
58%
(02:37)
wrong
based on 135
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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.
(1) a + k is a prime number.
(2) a + 2k is a prime number.
28 Feb 2017, 13:05
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diota2004
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.
[url]Statement #1[/url]: 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.
[url]Statement #2[/url]: 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
General Discussion
01 Mar 2017, 00:37
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diota2004
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.
17 Oct 2025, 22:35
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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
"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
vitaliyGMAT
