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For how many prime numbers p, 2^p+p^2 is a prime?
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18 Aug 2017, 20:46
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For how many prime numbers \(p\), \(2^p+p^2\) is a prime? A. 1 B. 2 C. 3 D. 4 E. 5
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Hasan Mahmud



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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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18 Aug 2017, 21:11
Mahmud6 wrote: For how many prime numbers \(p\), \(2^p+p^2\) is a prime?
A. 1 B. 2 C. 3 D. 4 E. 5 The possibility of prime numbers(p) which yield a prime number for the expression \(2^p+p^2\) are as follows: 2,3,5,7..... Let substitute the numbers in the expression and try and find out the value for the expression. \(2^p+p^2 = 2^2 + 2^2 = 4+4 = 8\) (when p = 2)  8 is not prime! \(2^p+p^2 = 2^3 + 3^2 = 8+9 = 17\) (when p = 3)  17 is prime! \(2^p+p^2 = 2^5 + 5^2 = 32+25 = 57\) (when p = 5)  57 is not prime! \(2^p+p^2 = 2^7 + 7^2 = 128+39 = 177\) (when p = 7)  177 is not prime! Of, these numbers only 17 is prime. So, answer should be Option A. There could be more possibilities for p(2 digit prime and 3 digit primes) but the calculations will be very tedious. I have a feeling you could change the question to read single digit primes Mahmud6 or perhaps there is some shorter method to find the answer that I am missing!
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For how many prime numbers p, 2^p+p^2 is a prime?
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18 Aug 2017, 21:29
Mahmud6 wrote: For how many prime numbers \(p\), \(2^p+p^2\) is a prime?
A. 1 B. 2 C. 3 D. 4 E. 5 Let y = 2^p + p^2 The prime numbers are 2,3,5,7,11,13 Lets start putting values..from 2 At p= 2, y = 2^2 +2^2 = 8 At p=3 , y = 2^3 + 3^2 = 8 + 9 = 17 is a prime number. At p =5, y = 2^5 + 5^2 = 32 + 25 = 57 a multiple of 3 is not a prime number is not a prime number At p = 7 y = 2^7 + 7^2 = 128 +49 = 177 a multiple of 3 is not a prime number At p =11, y = 2^11 + 11^2 = 2048 + 121 = 2169 a multiple of 3 is not a prime number So, Probably only p = 3 satisfies the prime number critrea as desired by the question.. Answer A
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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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07 Oct 2017, 02:40
pushpitkc wrote: Mahmud6 wrote: For how many prime numbers \(p\), \(2^p+p^2\) is a prime?
A. 1 B. 2 C. 3 D. 4 E. 5 The possibility of prime numbers(p) which yield a prime number for the expression \(2^p+p^2\) are as follows: 2,3,5,7..... Let substitute the numbers in the expression and try and find out the value for the expression. \(2^p+p^2 = 2^2 + 2^2 = 4+4 = 8\) (when p = 2)  8 is not prime! \(2^p+p^2 = 2^3 + 3^2 = 8+9 = 17\) (when p = 3)  17 is prime! \(2^p+p^2 = 2^5 + 5^2 = 32+25 = 57\) (when p = 5)  57 is not prime! \(2^p+p^2 = 2^7 + 7^2 = 128+39 = 177\) (when p = 7)  177 is not prime! Of, these numbers only 17 is prime. So, answer should be Option A. There could be more possibilities for p(2 digit prime and 3 digit primes) but the calculations will be very tedious. I have a feeling you could change the question to read single digit primes Mahmud6 or perhaps there is some shorter method to find the answer that I am missing! pushpitkcFor primes greater than 3, \(2^p+p^2\) is always divisible by 3 and is never a prime. Do you still think, I need to change the question to read single digit primes?
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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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07 Oct 2017, 03:45
Lets look at this this way: Any prime number greater than 3 is of the form either '6K+1' or '6k1'.
If p= 6k+1, then p^2 = (6k+1)^2 = 36k^2 + 12k + 1. When p= 6k1, then p^2 = (6k1)^2 = 36k^2  12k + 1. Both these numbers leave a remainder of '1' when divided by 3.
And now look at 2^p. If you observe powers of 2, you will see that any odd power of 2 Always leaves a remainder of '2' when divided by 3 (2^1=2, 2^3=8, 2^5=32.. and so on). Both (6k+1) and (6k1) are odd: so 2 raised to any of these will always leave a remainder '2' when divided by 3.
So when we add p^2 + 2^p  we are basically adding two numbers: one of which gives a remainder '1' when divided by 3, and another which gives a remainder '2' when divided by 3. In effect, this sum will thus lead to a combined remainder of '3' or '0'. Which means divisible by 3.
So if p is any prime number greater than 3: p^2 + 2^p is always divisible by 3 for values of p which are greater than 3. We now need to check just for two prime numbers  2 and 3.
if p=2, then 2^2 + 2^2 is NOT prime if p=3, then 2^3 + 3^2 = 17, which IS Prime.
So theres only one number  3 which satisfies this. Hence A answer.



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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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09 Oct 2018, 00:22
Easiest method to solve this question: Looking at exp = 2^p+p^2, we know that p^2 needs to be odd for the expression to be prime. So our range to check starts from 3,5,7,11,13,17,19.... p=3, exp = 17  prime p=5, exp = 57  NOT prime p=7, exp = 177  NOT prime So, the correct answer is A; there is only one such number. Also, I feel there must be some other condition present in the question to make it more specific in terms of calculating answer. If you read this, give KUDOS



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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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20 Oct 2018, 19:46
Mahmud6 wrote: For how many prime numbers \(p\), \(2^p+p^2\) is a prime?
A. 1 B. 2 C. 3 D. 4 E. 5 Is there's any shortcut to this problem ?



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Re: For how many prime numbers p, 2^p+p^2 is a prime?
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26 Oct 2018, 03:55
Mahmud6 wrote: For how many prime numbers \(p\), \(2^p+p^2\) is a prime?
A. 1 B. 2 C. 3 D. 4 E. 5 The only clue of this question is that p is a prime. we are also looking for prime. Note: In this type of question we will always find pattern . Whenever we are going thorough number properties we must be careful about pattern. So, p could be 2, 3 , 5 , 7 etc. When p=2 \(2^2 + 2^2\) = even ............can't be prime. When p = 3 \(2^3 + 3^2\) = 8 + 9 = 17 ......prime when p = 5 \(2^5 + 5^2\) = 32 + 25 = 57.......( 5 + 7 = 12. so 57 is divisible by 3) when p = 7 \(2^7 + 7^2\) = 128 + 49 = 177. ( 1 + 7 + 7 = 9, 177 is divisible by 3) It means when we put any value of p greater than 3 the sum is divisible by 3. The best answer is A.




Re: For how many prime numbers p, 2^p+p^2 is a prime? &nbs
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