# 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.

### 14 comments on “ISI 2013 B.Math and B.Stat Subjective Solutions”

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 numberNow note that all consecutive perfect squares have a difference of successive odd numberslike 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 resulthence N=51^2 or 2601 is our required solution

1. Chirantan says:

awesome solution bro. bt if we solve the problem will we get full credit??

1. Satyaki says:

surely we will! there's nothing wrong in the solution. it's artistic and well defined too!

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.

1. tanmay says:

Actually according to me this s the best solution as idid it in same way and also it has basic level and the reason a prime num. has been kept in the question.

6. Solution to problem 1Let 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^2But (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. says:

Sir,plz send the question paper and solution(at least question paper) of isi b.math/b.stat 2013.plz.......to my email - sahoosagar52@gmail.com.

1. SAGAR KUMAR SAHOO. says:

Plz send at least OBJECTIVE Question paper.PLEASE SIR ...

1. Dear Sagar,

We do not have the entire objective paper (as you know that these questions are collected from students; ISI will publish the paper in 2013.)

You may follow this blog. We will put the questions if we manage to acquire them from some source.