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Wednesday, 15 May 2013
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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)
(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.
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ISI 2013 solutions
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
\(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad \)

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
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