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Math Help - Birthday Problem; Prime Number Problem

  1. #1
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    Birthday Problem; Prime Number Problem

    I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

    The questions are:

    1.A man whose birthday was January 1st died on Feb 9th 1992. He was x years of age in the year x^2 AD for some interger x. In what year was he born?

    2. If p and p^2+14 are both prime numbers, find, with justification, all possible values of p.

    I think that question 1 is to do with quadratics and you can tell that if x is 44, then x^2 would be 1936 which looks to be the most sensible, although I do not know where to go. Question 2 I don't really know where to start.
    These are both really nagging me and I cannot get the answers.

    Thanks so much for any help given!
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  2. #2
    MHF Contributor alexmahone's Avatar
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    re: Birthday Problem; Prime Number Problem

    Quote Originally Posted by BobtheBob View Post
    I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

    The questions are:

    1.A man whose birthday was January 1st died on Feb 9th 1992. He was x years of age in the year x^2 AD for some interger x. In what year was he born?

    2. If p and p^2+14 are both prime numbers, find, with justification, all possible values of p.

    I think that question 1 is to do with quadratics and you can tell that if x is 44, then x^2 would be 1936 which looks to be the most sensible, although I do not know where to go. Question 2 I don't really know where to start.
    These are both really nagging me and I cannot get the answers.

    Thanks so much for any help given!
    2. Let p be a prime other than 3.

    p\equiv\pm 1\pmod{3}

    p^2\equiv 1\pmod{3}

    p^2+14\equiv 1+14\equiv 0\pmod{3}

    p^2+14 is not prime.

    When p=3, p^2+14=23 is a prime.

    So, the only possible value of p is 3.
    Last edited by alexmahone; January 5th 2012 at 09:31 AM.
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  3. #3
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    re: Birthday Problem; Prime Number Problem

    Quote Originally Posted by BobtheBob View Post
    I am currently teaching myself some maths but I have found these two questions on the internet, and I'm completely stuck.

    The questions are:

    1.A man whose birthday was January 1st died on Feb 9th 1992. He was x years of age in the year x^2 AD for some interger x. In what year was he born?

    2. If p and p^2+14 are both prime numbers, find, witjustification, all possible values of p.

    I think that question 1 is to do with quadratics and you can tell that if x is 44, then x^2 would be 1936 which looks to be the most sensible, although I do not know where to go.
    Good. You might also note that if x= 45, then x^2= 2025 which cannot be right because he died before that. If x= 43, then x^2= 1849 which would mean that when he died, in 1992, he was 43+ (1992- 1849)= 43+ 143= 186 years old! Now, if he was 44 in 1936, when was he born?

    Question 2 I don't really know where to start.
    p must not be 2 or 7 since, in that case, p^2+ 14 would be divisible by 2 or 7 respectively.

    These are both really nagging me and I cannot get the answers.

    Thanks so much for any help given!
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  4. #4
    Member kalyanram's Avatar
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    re: Birthday Problem; Prime Number Problem

    Question 1:

    Observation 1: \sqrt{1992} = 44.xxx hence maximum possible integers whose squares are less than 1992 are atmost 44. Now lets consider the simplest case of 1 for a man to be 1 year old on 1 AD he as to be born on 1 Jan 0 AD the he would be 1992 year of age when he died which would be absurd.

    So for any x^2 AD the man has to be born x^2 - x AD to be x years old and given that he died on 1992 only practical solution can be obtained if choose an year closer to 1992 so choosing the year 1936 we have x^2 = 1936 \implies x = 44. So the man should be born on 1936 - 44 = 1892 AD. So he was just over 100 years old when he died which is practically possible.

    Kalyan.
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  5. #5
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    re: Birthday Problem; Prime Number Problem

    Ok thanks so much! I understand question 1. Is there a more elegant answer for question 1 or would trial and error be a good answer?

    I'm still not all that sure on question 2 though. I don't quite understand why some steps are being done? for example:

    p\equiv\pm 1\pmod{3}
    surely 7 --1=8 and therefore not divisable by 3?

    Thanks very much!
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  6. #6
    MHF Contributor alexmahone's Avatar
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    re: Birthday Problem; Prime Number Problem

    Quote Originally Posted by BobtheBob View Post
    Ok thanks so much! I understand question 1. Is there a more elegant answer for question 1 or would trial and error be a good answer?

    I'm still not all that sure on question 2 though. I don't quite understand why some steps are being done? for example:



    surely 7 --1=8 and therefore not divisable by 3?

    Thanks very much!
    7\equiv 1\pmod{3}
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  7. #7
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    re: Birthday Problem; Prime Number Problem

    Ok thanks very much for the help. I'm understanding it better now!

    Thanks!
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  8. #8
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    re: Birthday Problem; Prime Number Problem

    Actually, are you able to explain this step for me because I don't quite understand what you are doing?

    p^2+14\equiv 1+14\equiv 0\pmod{3}
    Thanks very much.
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  9. #9
    Member kalyanram's Avatar
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    re: Birthday Problem; Prime Number Problem

    For any prime p \not = 3,
    p = 3k+1 \implies p^2 = (3k+1)^2 = 9k^2+6k+1 = 3K+1
    or
    p = 3k+2 \implies p^2 = (3k+2)^2 = 9k^2+12k+4 = 3K'+1
    So either ways
    p^2 = 3L+1 for some k,K,K',L \in Z^+
    so p^2  \equiv 1 \mod(3) \implies p^2+ 14 \equiv 1+14 \equiv 0 \mod(3)

    Kalyan
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  10. #10
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    re: Birthday Problem; Prime Number Problem

    Ok thanks very much for all the help given!
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