diota2004 wrote:

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.

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 statementsPicking 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

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)