## 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.

2.

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

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

3. This comment has been removed by the author.

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.

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.

6. Anirban Majumdar20 May 2013 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)

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.

8. SAGAR KUMAR SAHOO