# 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 Replies to “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

2. @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.

3. 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. 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.

4. 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)

5. 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.

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.