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.

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?

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

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