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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ISI 2013 B.Math, B.Stat Subjective Paper



Saturday, 9 February 2013

AMC 10 (2013) Solutions

12. In \(\triangle ABC, AB=AC=28\) and BC=20. Points D,E, and F are on sides \(\overline{AB}, \overline{BC}\), and \(\overline{AC}\), respectively, such that \(\overline{DE}\) and \(\overline{EF}\) are parallel to \(\overline{AC}\) and \(\overline{AB}\), respectively. What is the perimeter of parallelogram ADEF?

\(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad \)

http://cache.artofproblemsolving.com/asyforum/c/e/e/cee0c4038bfb9b14cfc95050e9b3d20bf808167d.png

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = \(2 \times 28\) = 56.

Ans. (C) 56

American Math Contest 10 (AMC 10) 2013 questions