This post contains problems from the first relay round of the Duke Math Meet 2009. Try to solve these problems.

1A. Find the lowest positive angle \theta that satisfies the equation \sqrt {1+\cos \theta} = \sin \theta + \cos\theta expressed in degrees.


\sqrt {1 +\cos\theta} = \cos\theta + \sin \theta \Rightarrow \sqrt{2\cos^2 \frac{\theta}{2} } = \sqrt2{\frac{1}{\sqrt2} \cos\theta + \frac{1}{\sqrt2} \sin\theta }

Now this gives

\sqrt2 \cos\frac{\theta}{2} = \sqrt2\cos(\theta - \frac{\pi}{4}) \Rightarrow \frac{\theta}{2} = \theta - \frac{\pi}{4} or \frac{\theta}{2} = -\theta + \frac{\pi}{4}

Thus the possible values of \theta are 90^o or 30^o .

Since we require the smallest positive angle hence the answer is 30^o .

1B Let n be two times the tens digit of TNYWR. Find the coefficient of the x^{n-1}y^{n+1} term in the expansion of (2x + \frac{y}{2} + 3)^{2n}


TNYWR is 3. Hence n = 6 Thus we are required to find coefficient of x^5 y^7 term in the expansion of (2x + \frac{y}{2} + 3 )^{12}

This can be easily found from trinomial expansion. The required term is {{12}\choose {5}}(2x)^5 {{7}\choose{7}} (\frac{y}{2})^7 = 792 \times 32 \times \frac{1}{128} = 198

1C Let k be TNYWR, and let n = k/2. Find the smallest integer m greater than n such that 15
divides m and 12 divides the number of positive integer factors of m.


k = 198, hence n = 99.

So we have to look at multiples of 15 greater than 99. We want 12 to divide the number of positive divisors of m.

Suppose m = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k} . The number positive divisors of k is (\alpha_1 +1 )... (\alpha_k + 1)

The first multiple of 15 greater than 99 is 105 = 15 \times 7 . By inspection we see that m = 150.

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