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Thank for the kind words, Mike! (This is one of mine.)

Bunuel, those weren't the inspiration, at least on a conscious level, but it's great to see the convergent evolution; they are certainly excellent and similar questions.

I'm very surprised this is currently listed as sub-600 level. I debated how difficult to go with this one; I could have done something like p - 8, p, p + 20 and gotten the same answer since p + 20 = p + 8 + 12 will be the same remainder w.r.t. 3 as p + 8.
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What is the value of prime number \(p\)?

(1) \(p+8\) is prime.
(2) \(p−8\) is prime.
Dear ziyuenlau,

I'm happy to respond. :-)

This is a devilishly clever and difficult problem. Let's jump to the case in which we are considering both statements together.

Here are a few number sense rules.
1) In a set of three consecutive integers, one is always divisible by 3.
2) In a set of three evenly spaced integers, one is always divisible by 3.
(These rules generalize from 3 to n.)

The numbers (p - 8), p, and (p + 8) are three evenly spaced numbers, so it absolutely must be true that one of them is divisible by 3. The only way all three of them can be prime is if the one that is divisible by 3 is 3 itself! Thus:
p - 8 = 3
p = 11
p + 8 = 19
If p takes the value p = 11, this is the only way that all three numbers can be prime simultaneously. Thus, the value of p is unique determined. Combined the statements are sufficient. OA = (C).

Veritas wrote a truly beautiful problem here!

Does all this make sense?
Mike :-)


Hi mikemcgarry,

I did not understand the rule well.

When you say-- the numbers (p - 8), p, and (p + 8) are three evenly spaced numbers, so it absolutely must be true that one of them is divisible by 3


even spaced means they are at the equal distance from preceding and succeeding term, or distance should be even as well?

For instance--
31 34 37

3 3

but none of them is divisible by 3; I think I misunderstood the concept. Please also explain how this rule is generalized from 3 to n. sorry for the silly question.
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Good catch, AR15J! To clarify, the rule in question only applies to an evenly spaced set where the spacing isn't itself a multiple of 3.

More broadly, the general rule is this: Any list of N evenly spaced integers will contain a multiple of N as long as the spacing has no factors besides 1 in common with N. (In fact, it will contain exactly one such multiple.)

Try it!
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Good catch, AR15J! To clarify, the rule in question only applies to an evenly spaced set where the spacing isn't itself a multiple of 3.

More broadly, the general rule is this: Any list of N evenly spaced integers will contain a multiple of N as long as the spacing has no factors besides 1 in common with N. (In fact, it will contain exactly one such multiple.)

Try it!

Thanks AnthonyRitz for clearing the doubt

In GMAT also, evenly spaced means equal distance, below set is also evenly spaced?Please confirm

31,34,37
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AR15J
AnthonyRitz
Good catch, AR15J! To clarify, the rule in question only applies to an evenly spaced set where the spacing isn't itself a multiple of 3.

More broadly, the general rule is this: Any list of N evenly spaced integers will contain a multiple of N as long as the spacing has no factors besides 1 in common with N. (In fact, it will contain exactly one such multiple.)

Try it!

Thanks AnthonyRitz for clearing the doubt

In GMAT also, evenly spaced means equal distance, below set is also evenly spaced?Please confirm

31,34,37

AR15J,

Yes, that is correct. 31, 34, 37 is evenly spaced. (But it will never contain a multiple of 3 no matter how long you continue, since the spacing is 3 and since at least one element is not a multiple of 3.) (But if, for instance, you include at least one more element in either direction, you will be guaranteed a multiple of 4.)
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Any list of N evenly spaced integers will contain a multiple of N as long as the spacing has no factors besides 1 in common with N. (In fact, it will contain exactly one such multiple.)

Hello Bunuel could you verify this information, because the example below challenges what you AnthonyRitz just said.

For N: 4, Set: {2,4,6,8}, Spacing/Difference: 2; We have a spacing (2) that is a factor of 4 and there is at least one multiple of 4 in the set.

I think the rule is that for any set/list of N evenly spaced integers there is at least one multiple of N:
(1) if any term is a multiple of N; or,
(2) if the common difference (spacing) doesn't share any factors with N but 1.
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AnthonyRitz
Any list of N evenly spaced integers will contain a multiple of N as long as the spacing has no factors besides 1 in common with N. (In fact, it will contain exactly one such multiple.)
AnthonyRitz

Hello, could you verify the information below, because the example below challenges what you just said.

For N: 4, Set: {2,4,6,8}, Spacing/Difference: 2; We have a spacing (2) that is a factor of 4 and there is at least one multiple of 4 in the set.


What Anthony wrote is correct (though you wouldn't need to know it for the GMAT) - if you create an equally spaced list of N numbers that are d apart, then if the GCD of d and N is 1, the list will contain one multiple of N.

The GCD of N and d in your list is not 1, so it's not a counterexample to that. I think you're trying to establish whether the logical converse of what Anthony wrote is also true, but in math, if you have a true statement, sometimes the converse is also true, and sometimes it's not true in general. And here, the converse is not true in general: when the GCD of N and d is not 1, then we need some more info to say how many multiples of N we'll have in our list. We can't take the converse (reverse) of the true fact above to conclude we should have zero multiples of N - that's only sometimes true.

None of these facts are worth memorizing for the GMAT, though.
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IanStewart

What Anthony wrote is correct (though you wouldn't need to know it for the GMAT) - if you create an equally spaced list of N numbers that are d apart, then if the GCD of d and N is 1, the list will contain one multiple of N.

The GCD of N and d in your list is not 1, so it's not a counterexample to that. I think you're trying to establish whether the logical converse of what Anthony wrote is also true, but in math, if you have a true statement, sometimes the converse is also true, and sometimes it's not true in general. And here, the converse is not true in general: when the GCD of N and d is not 1, then we need some more info to say how many multiples of N we'll have in our list. We can't take the converse (reverse) of the true fact above to conclude we should have zero multiples of N - that's only sometimes true.

None of these facts are worth memorizing for the GMAT, though.

Hi Ian,

Indeed, I was trying to establish the converse. Now I understand.

Thks :)
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Hello, this is my first time posting a question here. I really need to understand why 5 is not a solution here, because in that case, it will satisfy the conditions of (p-8) = -3, and (p+8) = 13, or is it that negative numbers are not considered as prime numbers?
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Msfollie
Hello, this is my first time posting a question here. I really need to understand why 5 is not a solution here, because in that case, it will satisfy the conditions of (p-8) = -3, and (p+8) = 13, or is it that negative numbers are not considered as prime numbers?

hi , prime no.'s are only "positive"by definition . so -3 is not a prime no .
hope it helps :-)
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