Quant Question Of The Day: 42
[latexpage]
Numbers
If p is a prime such that (p – 1)/4 and (p + 1)/2 are primes, then how many values of p are possible.
i) 0
ii) 1
iii) 2
iv) 3
v) More than 3
[latexpage]
If p is a prime such that (p – 1)/4 and (p + 1)/2 are primes, then how many values of p are possible.
i) 0
ii) 1
iii) 2
iv) 3
v) More than 3
Since both (p – 1)/4 and (p + 1)/2 are integers, we can say that
p is of form 4k + 1
=> (p – 1)/4 = k
and (p + 1)/2 = 2k + 1
So, all k, 2k + 1 and 4k + 1 should be prime
If k is prime greater than 3, then
Case (i): k = 6a + 1
2k + 1 = 12a + 3 (not prime)
Case (ii): k = 6a – 1
4k + 1 = 24a – 3 (not a prime)
So, k can only be 2 or 3
k = 2
2k + 1 = 5
4k + 1 = 9 (not possible)
k = 3
2k + 1 = 7
4k + 1 = 13
So, only p = 13 satisfies
5
Hello Arushi
5 values or p = 5 ?
More than 3
Hello Priyanka,
What was your approach?
0
What approach did you use, Samyak?
0
Hello Parth
What was your approach?
1
Correct! What approach did you use?
iii) 2
Hello Manish,
Which two values?
0
Sir, I used the fact that all the prime numbers are of the form 6n+1 or 6n-1. I replaced the value of p with the above two values and I found no solution.
1
I used hit and trial method and only 13 is that value of p that gives the other two primes .
is there any other solution ?