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OG 11 math problem solving question

This topic has 3 member replies
nuku888 Newbie | Next Rank: 10 Posts Default Avatar
Joined
19 Apr 2007
Posted:
5 messages

OG 11 math problem solving question

Post Thu Apr 19, 2007 5:29 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Hello everyone,

    I was wondering if anyone can help me with the explanation for question 241 in the OG11.

    Question

    If the integer n has exactly three positive divisors, including 1 and n how many positive divisors does n^2 have?

    I read the explanation at the back but I don't understand if anyone can explain that would be great thanks!

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    zaffar Senior | Next Rank: 100 Posts Default Avatar
    Joined
    29 Dec 2006
    Posted:
    49 messages
    Post Thu Apr 19, 2007 9:09 pm
    i did it by picking numbers and there are only 2 numbers 4 (three divisors 1 2 4) & 9 (1 3 9). and their squares 16 (1,2,4,8,16) and 81 (1,3,9,27,81)
    so 5 is the answer.

    there may be someother method but i did it this way. i hope it helps.

    nuku888 Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    19 Apr 2007
    Posted:
    5 messages
    Post Fri Apr 20, 2007 2:09 pm
    Thanks for your reply. Sorry just got one more questions they mention something about square root n is also an integer, so does that mean square root 4 = 2 that is why they said that it is also integer?

    sosaha Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    21 Apr 2007
    Posted:
    1 messages
    Post Sat Apr 21, 2007 10:33 pm
    lets consider the number to be 'n'. since it has only one more divisor other than 1 and n, n must be a perfect sqare. so say n=x^2(x is a prime).
    => n^2 = x^4
    Now divisors of x^4 are 1, x, x^2, x^3, x^4.
    i.e 5 divisors.

    _________________
    -Wish u all the best,
    Soumyajit.

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