Rom

you're right, good catch, tired

\(\text{The set of odd integer reciprocals with a terminating decimal representation}\\ \text{is just the set of odd integers that are powers of 5.}\\ \text{ Between 1 and 2010 This is 5,25,625}\\ \text{There are a total of 1005 odd integers less than 2010 so }\\ p = \dfrac{3}{1005}\\ 3+1005=1008\)

you're wrong

\(4\sqrt{x^2+y^2+z^2} = 28\\ x^2 + y^2 + z^2 = 49\\ \text{The other two equations are correct}\)

\(z^2 = 49-40 = 9\\ z=3\\ y^2 = 45-9=36\\ y=6\\ x^2 = 49-36-9 = 4\\ x=2\\ xyz = 36\)

\(\text{$15=3\cdot 5$ so $15n$ being a perfect square means $n=15 k^2$}\\ \text{We want all of these up to 1000}\\ 15k^2 \leq 1000\\ k^2 \leq 66\frac 2 3\\ k\leq 8\\ \text{So we have $n = 15,60, 135, 240,375,540,735,960$}\)

\(\text{I assume the books of a given type are indistinguishable}\\ \text{You have 12 books. If they were all distinguishable there would be $12!$ arrangements}\\ \text{As they are not we have to reduce this number by the number of permutations of each type}\\ N = \dfrac{12!}{5!\cdot 4!\cdot 3!} = 27720\)

\(\dfrac{dy}{dx}+y = \sin(2x)\\ a=1,~b=1,~f(x)=\sin(2x)\\ y_{cf} = Ae^{-\frac b a x}=A e^{-x}\\ \text{try $y_{pt} = B\sin(2x) + C\cos(2x)$}\\ 2B\cos(2x) - 2C \sin(2x) + B\sin(2x)+C\cos(2x) = \sin(2x)\\ 2B+C = 0\\ B-2C = 1\\ 5B = 1\\ B=\dfrac 1 5,~C = -\dfrac 2 5\\ y = y_{cf}+y_{pt} = A e^{-x} + \dfrac 1 2 \left(\sin(2x)-2\cos(2x)\right)\)

\(\dfrac 1 x + \dfrac 1 y + \dfrac 1 z = 2\\ \dfrac{x+y+z}{x+y+z}\left( \dfrac 1 x + \dfrac 1 y + \dfrac 1 z\right) = 2\\ \dfrac 1 6 \left(1+\dfrac{y+z}{x} + 1 + \dfrac{x+z}{y}+1+\dfrac{x+y}{z}\right) = 2\\ 3 + \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 12\\ \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 9\)

\(b)~a_n = a_0 d^n,~n \in \{0\}\cup \mathbb{N},~a_0, d\in \mathbb{R}\\ \text{If we let $a_0 = \left(\dfrac 1 d\right)^{r-1},~\dfrac 1 d \in \mathbb{N}$}\\ \text{we end up with exactly $r$ integers in the sequence}\)

\(a)~a_n = a_0 + n d,~n \in \mathbb{Z^+},~d,a_0 \in \mathbb{R}\\ \text{Suppose there are 2 or more integers in the sequence, at say $k_1,~k_2$}\\ a_0 + k_1 d \in \mathbb{Z} \text{ and }a_0 + k_2 d \in \mathbb{Z}\\ (k_1-k_2)d \in \mathbb{Z}\\ m(k_1 - k_2)d \in \mathbb{Z},~m \in \mathbb{Z}\\ a_0 + k_1 d + m(k_1-k_2)d \in \mathbb{Z}\\ \text{This corresponds to index }\\ (m+1)k_1 - mk_2\\ \text{and thus there are infinitely many integers in the sequence as $m \in \mathbb{Z}$}\)

\(\text{On the other hand we can easily form a sequence with 1 integer by having $\\a_0 \in \mathbb{Z},~d \not \in \mathbb{Q}$}\)