hectictar

This might help explain why that is true:

$0 \ >\ \dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}$

So...

$\dfrac{1}{1-\frac{10}{x}}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}\ >\ 1 - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-1\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{x}{x-10}-\dfrac{x-10}{x-10}\ >\ - \dfrac{5}{y}\\~\\~\\ \dfrac{10}{x-10}\ >\ -\dfrac5y$

And we know  0 < x < 10  because that is the only way  $\dfrac{1}{1-\frac{10}{x}}$  can be less than 0.

So  x - 10  is negative, and so when we multiply both sides by  (x - 10) ,  flip the sign.

$10\ <\ -\dfrac5y(x-10)$

And we know  0 < y < 5  because that is the only way  $1-\dfrac5y$  can be less than 0.

So  y  is positive, and so when we multiply both sides by  y ,  don't flip the sign.

$10y\ <\ -5(x-10)\\~\\ y\ <\ -\frac12(x-10)\\~\\ y\ <\ -\frac12x+5$

And so we have these three inequalities:

$1.\qquad0\ {<}\ x\ {<}\ 10\\~\\ 2.\qquad0\ {<}\ y\ {<}\ 5\\~\\ 3.\qquad y\ {<}\ -\frac12x+5$

We can see on a graph that the intersection of these is a triangle:

$S\quad{=\quad}\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{(2n+1)^2-1}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n^2+4n}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n(n+1)}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n(n+1)}\bigg)\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n}-\dfrac{1}{n+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}- \dfrac{1}{4}+\dots+\dfrac{1}{m-1}-\dfrac{1}{m}+\dfrac{1}{m}-\dfrac{1}{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac11-\dfrac1{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}S_m\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\dfrac32\bigg(1-0\bigg)\\~\\~\\ S\quad=\quad\dfrac32$_

Then read the remainders from bottom to top to get  100011

So     35₁₀   =   100011₂

Here's the way I think of it if it helps:

999₁₆   =   9(16⁰)   +   9(16¹)  +  9(16²)

999₁₆   =   9(1)   +   9(16)  +  9(256)

999₁₆   =   9  +  144  +  2304

999₁₆   =   2457     that is in base 10

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Ok, awesome!!!

sin( angle )  =  opposite / hypotenuse

sin( C )  =  $\frac{20}{101}$ / $\frac{101}{101}$

sin( C  )  =  $\frac{20}{101}$

Also...

sin( C )  =  x / $\frac{99}{101}$

sin( C )  =  $\frac{101}{99}x$

So we can equate both expressions of  sin( C )  and solve for  x

$\frac{20}{101}\ =\ \frac{101}{99}x\\~\\ \frac{20}{101}\cdot\frac{99}{101}\ =\ x\\~\\ \frac{1980}{10201}\ =\ x$_

1. Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=111?

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To help see this...here is the beginning of the list of pairs:

123, 234

124, 235

125, 236

132, 243

134, 245

135, 246

142, 253

143, 254

.

.

.

2. The product of all digits of positive integer M  is 105  How many such M's are there with distinct digits?

I tried 6 for the problem above but it wasn't right... Any thoughts on why?

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I think the answer is not 6 because  1  can also be a digit of M.

105  =  3 * 5 * 7

So  5, 3, and 7 must be digits of M, while  1  might be a digit of M.

So the possiblites are:

357

375

537

573

735

753

In addition to:

the total number of possibilities   =   3! + 4!   =   6 + 6*4    =   6 + 24   =   30

Note that   (10x - 3)²   =   (10x - 3)(10x - 3)   =   100x² - 60x + 9

Now let's look at the given equation:

5x² + 4  =  3x + 9

                                Subtract  3x  from both sides of the equation.

5x² - 3x + 4  =  9

                                Subtract  4  from both sides of the equation.

5x² - 3x  =  5

                                Multiply through by  20

100x² - 60x  =  100

                                         Add  9  to both sides of the equation.

100x² - 60x + 9   =   109

                                         And remember that  100x² - 60x + 9  =  (10x - 3)²

(10x - 3)²   =   109