Sunday, 12 May 2013

ISI 2013 B.Math and B.Stat Subjective Solutions

1. For how many values of N (positive integer) N(N-101) is a square of a positive integer?

Solution:
(We will not consider the cases where N = 0 or N = 101)

\( N(N-101) =  m^2 \)

=> \( N^2 - 101N - m^2 = 0 \)

Roots of this quadratic in N is 
=> \( \frac{101 \pm \sqrt { 101^2 + 4m^2}}{2} \) 

The discriminant must be square of an odd number in order to have integer values for N.

Thus \( 101^2 + 4m^2  = (2k + 1)^2 \)
=> \( 101^2 = (2k +1)^2 - 4m^2 \)
=> \( 101^2 = (2k +2m + 1)(2k - 2m + 1) \)

Note that 101 is a prime number

Hence we have two possibilities 

Case 1:

\( 2k + 2m + 1 = 101^2; 2k - 2m + 1 = 1 \)
Subtracting this pair of equations we get  \(4m = 101^2 - 1\) or \(4m = 100 \times 102\) or m = 50*51

This gives N = 2601 (ignoring extraneous solutions)

Case 2:

\(2k + 2m + 1 = 101 ; 2k - 2m + 1 = 101 \) which gives m = 0 or N = 101. This solution we ignore as it makes N(N- 101) = 0 (a non positive square).

 Hence the only solution is N = 2601 and there are no other values of N which makes N(N-101) a perfect square. 

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8 comments:

  1. I hope nothing is wrong with this solution.
    For N(N-101) to be a perfect square ,both N and N-101 should be perfect squares , we can see it from the fact that 101 is prime number
    Now note that all consecutive perfect squares have a difference of successive odd numbers
    like
    1^2 , 2^2 , 3^2 , 4^2 ...
    respective differences are 3,5,7,9..

    Now, 101 will be the common difference for the 50th square and 51st sqaure i.e. 51^2 - 50^2

    for N-101=-50 OR N=51, we have our desired result
    hence N=51^2 or 2601 is our required solution

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  2. Problem 7: Atleast one player has won more than or equal to n(n-1)/2 matches !! :O^\infty

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  3. This comment has been removed by the author.

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  4. @Anonymous ... that typo has been corrected. (however it does not affect the solution as it has not been used elsewhere). But thanks for pointing out.

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  5. i did it ds way...let us suppose N is a multiple of 101, den it bcums 101k(101k-101)=c^2...or 101^2k(k-1)=c^2...hence k(k-1)=(c/101)^2=m^2 say...but dre s no k such dat ds holds...hence N is not a multiple of 101 and so gcd of N and N-101 s 1...therfore, deir product will b a prfct square if each of dem is a prfct square...hence N=s^2 say, and N-101 =r^2 say...den (s+r)(s-r)=101..but as 101 s a prime, hence only solution s s+r=101, s-r=1...wich gives s=51...or N=s^2=51^2.

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  6. Anirban Majumdar20 May 2013 at 22:48

    Solution to problem 1
    Let N(N-101)=(N-K)^2, for some integer k
    => N^2-101N=N^2+k^2-2*k*N
    => N(2k-101)=k^2
    => N=k^2/(2k-101)
    This is an integer if (2k-101)divides k^2
    But (2k-101) divides (2k-101)^2=4k^2-4k*101+101^2
    => (2k-101) divides (-4k*101+101^2)
    => (2k-101) divides (-4k*101+101^2)+{2*(2k-101)*101}
    => (2k-101) divides 101^2
    => (2k-101) divides 101
    => (2k-101)=1 or -1 or 101 or -101
    => k=51 or 50 or 101 or 0
    => N=51^2 or -50^2 or 101 or 0
    => N=51^2 is the only solution.(As,N=101 & 0 are rejected)

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  7. 7) Consider the player with the max number of wins say A. If A's list does not contain all others' name there exists a player B whose list contains more names than A, contradiction. Thus proved.

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  8. SAGAR KUMAR SAHOO

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