Gauravji21
TestPrepUnlimited
TheUltimateWinner
What is the value of prime number \(p\)?
1) \(p=n^2-1\) where \(n\) is a positive integer
2) \(p\), \(p+2\), and \(p+4\) are all prime
Statement 1:\(p = n^2 - 1 = (n - 1)*(n + 1)\) so p cannot be prime unless (n - 1) = 1. In that case we have n = 2 and p = 3. Then p = 3 is the only possible prime input. Sufficient.
Statement 2:If we look at the divide by 3 world, these values are consecutive in that world with p, p + 2 and p + 4 -> p + 1. Since we have 3 consecutive values, one of them must be divisible by 3 and cannot be prime. Then we cannot have any large p's where p, p+2 and p+4 are all prime. However we have a special case of 3, 5, 7 so p = 3 is the only possible scenario. Sufficient.
Ans: D
Posted from my mobile deviceHi..
I understand that if we have three consecutive numbers then one of them will be divisible by 3. But in the questions we have P, P+2, P+4. how come these three numbers can be considered consecutive numbers.
I was confused by this 'consecutive' thing also, but looked up some math properties.
We know already that \(P, P + 2, P + 4 \) is a small list of 3 consecutive odd integers.
If we want to see it another way, can rewrite this equation here \( 2n + 1, 2n + 3, 2n + 5 \) where n is an integer.
This shows us that either the first, second, or third number can be divisible by 3 just by nature of being 3 odd numbers apart.
Any integer n (or p in the earlier equation) you plug in will always result in being divisible by three.