Messages posted by kaymant

First, the prime factorization 1050:
$1050 = 2\cdot 3 \cdot 5^2\cdot 7$
Each factor of 1050 can contain only these primes. So the general form of each of the $x_i$ is
$x_i=2^{a_i}\cdot 3^{b_i}\cdot 5^{c_i}\cdot 7^{d_i}$ , where $a_i, b_i, c_i, d_i$ are non-negative integers and $i=1,\ 2,\ 3,\ 4,\ 5$ .
Accordingly,
$x_1x_2 x_3x_4 x_5 = 2^{\sum a_i}\cdot3^{\sumb_i}\cdot 5^{\sum c_i} \cdot 7^{\sum d_i}$
We must assign the $a_i$ 's, $b_i$ 's, $c_i$ 's and $d_i$ 's so that the following set of equalities hold:
$a_1+a_2+a_3+a_4+a_5=1$ ,
$b_1+b_2+b_3+b_4+b_5=1$ ,
$c_1+c_2+c_3+c_4+c_5=2$
$d_1+d_2+d_3+d_4+d_5=1$
The number of possible ordered lists for these equations are:
$a_1+a_2+a_3+a_4+a_5=1: ^5\mathrm{C}_1=5$
$b_1+b_2+b_3+b_4+b_5=1: ^5\mathrm{C}_1=5$
$c_1+c_2+c_3+c_4+c_5=2: ^6\mathrm{C}_2=15$
$d_1+d_2+d_3+d_4+d_5=1: ^5\mathrm{C}_1=5$
Since each list $(a_i,\,b_i,\,c_i,\,d_i)$ correspond to a factor, the number of solutions of the given equation
$x_1x_2x_3x_4x_5=1050$ is $5\times 5\times 15\times 5= 1875$ .

That's not correct.
First, the prime factorization 1050:
$1050 = 2\cdot 3 \cdot 5^2\cdot 7$
Each factor of 1050 can contain only these primes. So the general form of each of the $x_i$ is
$x_i=2^{a_i}\cdot 3^{b_i}\cdot 5^{c_i}\cdot 7^{d_i}$ , where $a_i$ , $b_i$ , $c_i$ , $d_i$ are non-negative integers and $i=1,\ 2,\ 3,\ 4,\ 5$ .
Accordingly,
$x_1x_2 x_3x_4 x_5 = 2^{\sum a_i}\cdot 3^{\sumb_i}\cdot 5^{\sum c_i} \cdot 7^{\sum d_i}$
We must assign the $a_i$ 's, $b_i$ 's, $c_i$ 's and $d_i$ 's so that the following set of equalities hold:
$a_1+a_2+a_3+a_4+a_5=1$ ,
$b_1+b_2+b_3+b_4+b_5=1$ ,
$c_1+c_2+c_3+c_4+c_5=2$
$d_1+d_2+d_3+d_4+d_5=1$
The number of possible ordered lists for these equations are:
$a_1+a_2+a_3+a_4+a_5=1$ : $^5\mathrm{C}_1=5$
$b_1+b_2+b_3+b_4+b_5=1$ : $^5\mathrm{C}_1=5$
$c_1+c_2+c_3+c_4+c_5=2$ : $^6\mathrm{C}_2=15$
$d_1+d_2+d_3+d_4+d_5=1$ : $^5\mathrm{C}_1=5$
Since each list $(a_i,\,b_i,\,c_i,\,d_i)$ correspond to a factor, the number of solutions of the given equation
$x_1x_2x_3x_4x_5=1050$ is $5\times 5\times 15\times 5= 1875$

Alternately,
The given equation is basically
$z^2+z+\bar{z}=2$ -----(1)
Taking conjugates
$\bar{z}^2+\bar{z}+z=2$ ---- (2)
From (1) and (2), we get $z^2=\bar{z}^2$ implying that either $z=\bar{z}$ or $z=-\bar{z}$
If $z=\bar{z}$ , then $z$ is purely real, so the given equation becomes
$z^2+2z-2=0\quad\Rightarrow \ z=-1\pm \sqrt{3}$
On the other hand, if $z=-\bar{z}$ , then $z$ is purely imaginary, i.e. $\mathrm{Re}(z)=0$ , so the original equation becomes
$z^2=2\quad\Rightarrow \ z=\pm \sqrt{2}$
But that's contradicting the fact that $z$ is purely imaginary. Hence, the only roots are $\boxed{z=-1\pm \sqrt{3}}$

An $n$ -th root of unity is of the form $e^{i\frac{2k\pi}{n}}$ where $k=0,1, \ldots, (n-1)$ .
So its $p$ -th power is $e^{i\frac{2pk\pi}{n}}$ which cannot vanish for any $p$ .
Geometrically, you can see that a root of unity lies on the unit circle. Raising it to any power rotates it by certain angle leaving the length unchanged. So no matter how many times you rotate it, it cannot end up at the origin.

(that means you should check the question.)

If the integral you gave is
$I=\int_{-\pi/4}^{\pi/3} \dfrac{2\sec x}{2+\tan x}\ \mathrm dx$ ,
then its not hyperbolic or anything like that.
One has
$\dfrac{2\sec x}{2+\tan x} = \dfrac{2}{2\cos x+\sin x}=\dfrac{2}{\sqrt{5}\sin(x+\alpha)}$
where $\tan \alpha =2$
Therefore,
$I=\int_{-\pi/4}^{\pi/3} \dfrac{2}{\sqrt{5}\sin(x+\alpha)}\ \mathrm dx=\dfrac{2}{\sqrt{5}}\left|\ln\tan\frac{x}{2}\right|_{-\p...$

If we require only the positive roots (as $\sqrt{z}\geq 0$ for a real $z$ ), then it is easy to ascertain that the only such value of $x$ will be
$x_1=\dfrac{1+\sqrt{1+4a}}{2}$

To answer both these questions, we need to know whether or not the particle is an isolated one, i.e. to say whether or not some more particles are present with whom the given particle can interact.
Generally, when one talks about a particle, it is assumed that it is isolated (until and unless, its given otherwise). Now an isolated particle can only move with a constant velocity as seen from an inertial frame. In fact, this fact is used to define an inertial reference frame.

For the first question, using the above fact, (a) must be correct.

Now, can you think of the second one?

Of course, it is integrable. Only thing is that we cannot express the integral in terms of the basic elementary functions. So in a sense, the integral of $\dfrac{\sin x}{x}$ is a higher function (or you can say that its a new function). For example, consider
$F(x)=\int_0^x \dfrac{\sin t}{t}\ \mathrm{d}t + C$ (here $C$ is an arbitrary constant)
Then, differentiating w.r.t. the upper limit $x$ , we get
$\dfrac{\mathrm{d}F(x)}{\mathrm{d}x}=\dfrac{\sin x}{x}$
As such, we can take $F(x)$ as the indefinite integral of $\dfrac{\sin x}{x}$

http://www.goiit.com/posts/list/algebra-really-good-questions-a-collection-i-will-post-all-questions-931320.htm

991 X 997 X 1009

2) Its straight forward. Assume that there is some greatest prime. Call it p_n. Let p₁, p₂, . . ., p_n-1 be the prime lesser than p_n.

The product p₁p₂p₃....p_n + 1 is the then obviously a prime and greater than p_n, which is a contradiction. Hence, there are infinitely many prime.

nugorama's explanation is correct.

Call the six colors 1, 2, 3, 4, 5, 6. Put the cube on the table so that face 1 is at the bottom. Consider face 2. If it is at the top then we can rotate the cube about a vertical axis so that face 3 is in front. Now the cube is fixed. There are 3! = 6 ways to complete the coloring. Now, suppose that face 2 is a neighbor of 1. Then we rotate the cube so that 2 is in front. Now the cube is fixed, and the coloring can be completed in 4! = 24 ways. Altogether, there are 6 + 24 = 30 distinct colorings of the cube by six colors.

@kapil: the problem says the number of zeroes, NOT ONLY THE TRAILING ZEROES.

If you want to find the number of trailing zeroes in p! (where p is a prime), it is sufficient to find the exponent of 5 in p!.

13! = 6227020800

and 101!=

9425947759838359420851623124482936749562312794702543768327889353416977599316221476503087861591808346911623490003549599583369706302603264000000000000000000000000

In both cases, Kapil's answer didn't match.

For the second case, we need to know the initial position from where it starts sliding. However the locus of the center will be simply

x² + y² = 20² .

1.) Let the fixed point be $(a,,0)$ and the fixed line be the $y$ -axis. Then, if $e$ be the eccentricity of the ellipse, its equation is
$(x-a)^2 + y^2 = e^2 x^2$
$Rightarrow (1-e^2)x^2-2ax+a^2+y^2=0$
$Rightarrow (1-e^2)left(x^2-2xdfrac{a}{1-e^2}+dfrac{a^2}{(1-e^2)^2}-dfrac{a^2}{(1-e^2)^2}ight)+a^2+y^2=0$
$(1-e^2)left(x-dfrac{a}{1-e^2}ight)^2+y^2=dfrac{a^2e^2}{1-e^2}$
$dfrac{left(x-dfrac{a}{1-e^2}ight)^2}{left(dfrac{ae}{1-e^2}ight)^2} + dfrac{y^2}{left(dfrac{ae}{sqrt{1-e^2}}ight)^2} =1$
This ellipse has its center at the point $left(dfrac{a}{1-e^2},,0ight)$ , the semi-major axis is $dfrac{ae}{1-e^2}$ and the semi-minor axis is $dfrac{ae}{sqrt{1-e^2}}$ .
The extremities of the ends of the minor axis, therefore, are $x=dfrac{a}{1-e^2}$ and $y=pm dfrac{ae}{sqrt{1-e^2}}$
From the first relation we get $e^2=1-dfrac{a}{x}$ .
Squaring the second one, we get $y^2=dfrac{a^2e^2}{1-e^2}=dfrac{a^2left(1-dfrac{a}{x}ight)}{dfrac{a}{x}}=a(x-a)$ .
Therefore, the locus of the extremities of the minor axis of the ellipses is $oxed{y^2=a(x-a)}$ , which is a parabola with vertex at the fixed point $(a,0)$ and the and the focus at the point $left(dfrac{5a}{4},,0ight)$ . Hence option (D).