**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)

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

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.**

I hope nothing is wrong with this solution.

ReplyDeleteFor 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

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

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ReplyDelete@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.

ReplyDeletei 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.

ReplyDeleteSolution to problem 1

ReplyDeleteLet 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) 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.

ReplyDeleteSAGAR KUMAR SAHOO

ReplyDelete