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Oh! I thought the buttons added that number of channels to the channel you were already on. I had no idea!

math help

Consider parallelogram ABCD with points S and T chosen such that CS:SD = BT:TC = 2:1, as in the picture.
Let\( \vec{AB} = v\) and \(\vec{AD} = w\).
Then there exist constants r, s, t, u such that
\(\vec{AT} = r v + s w\),
\(\vec{BS} = t v + u w\).

2.

Also, what is\( \mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}}\) equal to?

\(\begin{array}{|rcll|} \hline \left(\vec{AT}\right)^2 &=& \left(v +\dfrac{2}{3}w \right)^2 \\ \mathbf{\left(\vec{AT}\right)^2} &=& \mathbf{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2 } \\\\ \left(\vec{BS}\right)^2 &=& \left(w -\dfrac{2}{3}v \right)^2 \\ \mathbf{\left(\vec{BS}\right)^2} &=& \mathbf{w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2 } \\\\ \left(\vec{AC}\right)^2 &=& \left(v+w \right)^2 \quad | \quad \vec{AC}=v+w \\ \mathbf{\left(\vec{AC}\right)^2} &=& \mathbf{v^2 + 2vw + w^2 } \\\\ \left(\vec{BD}\right)^2 &=& \left(w-v \right)^2 \quad | \quad \vec{BD}=w-v \\ \mathbf{\left(\vec{BD}\right)^2} &=& \mathbf{w^2 - 2vw + v^2 } \\ \hline \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}w^2+w^2 - 2\cdot \dfrac{2}{3}vw +\dfrac{4}{9}v^2} {v^2 + 2vw + w^2+w^2 - 2vw + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2+\dfrac{4}{3}vw +\dfrac{4}{9}w^2+w^2 - \dfrac{4}{3}vw +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}w^2+w^2 +\dfrac{4}{9}v^2} {v^2 + w^2+w^2 + v^2} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{v^2 +\dfrac{4}{9}v^2+w^2 +\dfrac{4}{9}w^2} {2(v^2 + w^2)} \\\\ \dfrac{AT^2+BS^2}{AC^2+BD^2} &=& \dfrac{\dfrac{13}{9}(v^2+w^2)} {2(v^2 + w^2)} \\\\ \mathbf{\dfrac{AT^2+BS^2}{AC^2+BD^2}} &=& \mathbf{\dfrac{13}{18}} \\ \hline \end{array}\)

math help

Consider parallelogram ABCD with points S and T chosen such that \(CS:SD = BT:TC = 2:1\), as in the picture.
Let \( \vec{AB} = v\) and \(\vec{AD} = w\).

1.
Then there exist constants r, s, t, u such that
\(\vec{AT} = r v + s w\),
\(\vec{BS} = t v + u w\).

\(\text{Let $\vec{AB} = \vec{DC}$ } \\ \text{Let $\vec{AD} = \vec{BC}$ }\)

\(\begin{array}{|rcll|} \hline \vec{AT} &=& v +\dfrac{2}{3}w \quad | \quad \vec{AT} = r v + s w \\ \mathbf{r} &=& \mathbf{1} \\ \mathbf{s} &=& \mathbf{\dfrac{2}{3}} \\\\ \vec{BS} &=& w -\dfrac{2}{3}v \quad | \quad \vec{BS} = t v + u w \\ \mathbf{t} &=& \mathbf{-\dfrac{2}{3}} \\ \mathbf{u} &=& \mathbf{1} \\ \hline \end{array}\)

The sum has to be positive

3.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,
and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

We can also write \(\vec{U} = x \vec{A} + y \vec{B} + z \vec{C}\)
for some real values of x, y, z. Find x, y, z.

\(\begin{array}{|rcll|} \hline \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \vec{u} = \vec{A}-\vec{C},\ \vec{v} = \vec{B}-\vec{C} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\left(\vec{B}-\vec{C}\right)+ \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\left(\vec{B}-\vec{C}\right)\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\vec{C}-\dfrac{2}{3}\vec{B}+\dfrac{2}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{US}{AS}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \quad | \quad \dfrac{US}{AS}=\dfrac{1}{5} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\left( \vec{A}-\dfrac{2}{3}\vec{B}-\dfrac{1}{3}\vec{C}\right) \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \vec{C}+\dfrac{2}{3}\vec{B}-\dfrac{2}{3}\vec{C} + \dfrac{1}{5}\vec{A}-\dfrac{2}{15}\vec{B}-\dfrac{1}{15}\vec{C} \\ \vec{U} &=& \dfrac{1}{5}\vec{A} +\dfrac{8}{15}\vec{B}+\dfrac{4}{15}\vec{C} \quad |\quad \vec{U} = x \vec{A} + y \vec{B} + z \vec{C} \\ \hline \mathbf{x}&=& \mathbf{\dfrac{1}{5}} \\ \mathbf{y}&=& \mathbf{\dfrac{8}{15}} \\ \mathbf{z}&=& \mathbf{\dfrac{4}{15}} \\ \hline \end{array}\)

ok, thanks 

The computer automatically arranges them in numerical order. Here is the list:

102695  104765  104972  105386  106835  107249  107456  109526  120497  123809  124637  128570  129605  130847  132089  134780  135608  136850  137264  138092  138506  138920  140369  140576  140783  142853  143267  146372  148235  149063  149270  150926  153824  154238  156308  159206  159620  165209  165830  170384  178043  178250  180527  180734  182390  183425  184253  185702  186530  190256  190463  192740  194603  196052  203918  204539  204953  205367  205781  206195  207851  209714  210749  210956  213647  213854  215096  217580  219650  234761  236417  238901  239108  240971  241385  245318  248630  250493  253184  256703  256910  259601  260843  263741  265190  267053  270158  270365  273056  274091  280715  281543  281750  283406  286304  290651  291065  294170  298103  298310  301829  302657  302864  305762  307418  307625  308246  309281  314870  315284  316940  320459  321908  324185  324806  325841  326048  340952  341780  342608  342815  345092  345920  346127  347162  349025  350267  350681  354821  356270  360824  362480  367241  369104  372416  376142  381524  382145  384215  390218  390425  391046  391460  392081  396014  401396  402638  403259  405329  406157  406571  407192  407813  410297  412367  413609  416093  416507  418370  428306  428513  430169  431825  432860  436172  437621  451076  451283  452318  458321  460391  463082  463910  467015  471362  472190  473018  476123  479021  480263  485231  487301  490613  491027  502619  503861  506138  509243  510278  510692  512348  514832  516074  516902  521870  523940  528701  531806  532841  534290  537602  538016  540293  541328  543812  546710  560372  561407  563270  567203  567410  568031  573206  573620  576104  581072  582107  582314  583142  592043  594320  601358  603428  603842  605291  605912  607154  609431  610259  612743  612950  617504  620195  621437  623507  623714  625370  629510  630752  631580  632408  634271  635720  638204  641309  641723  641930  642137  643172  647105  647312  650417  650831  651038  652073  652901  653108  657041  671324  674015  680432  681053  691403  694301  701546  702581  704651  706514  709412  710654  714380  715208  716243  716450  718520  724109  724316  730526  741083  741290  745016  746051  750812  751640  756401  760541  763025  765302  780413  781034  783104  790142  801527  801734  804632  810635  812705  813740  815603  817052  820157  820364  820571  830714  830921  831542  834026  835061  836510  839201  840236  842306  842513  846032  850172  851207  854312  857210  860315  861350  863420  870251  875012  890123  893021  902543  903164  904613  905234  908132  910823  912065  916205  917240  920138  920345  921380  923450  924071  925106  930281  935042  940217  940631  942701  946013  951602  952016  952430  963401  981203  982031  Total =  352

Thanks,

It doesn't really matter, I was trying to work out how to get them without a computer program.

Can you print them in groups so that the ones with the same digits are together?

Do you know if you got 13 combinations of 6 digits?

Melody: I have the entire list. If you want it, I will print it.

I know 0123467 is not divisable by 23.

Its digits do however add to a 23.  Maybe if those digits are rearranged then you will get a number that IS divisable by 23.

I have found 13 sets of 6 numbers.  Each set comply with the first and 3rd conditions.

0123467 is one of those sets.

Now it is a matter or rearranging the digits  in each set to work out which permutations of digits will give a number divisable by 23.

Notice that each number must meet 3 conditions:

That its 6 digits add up to 23, and must be divisible by 23, and all 6 digits different from each other. 123467 is NOT divisible by 23. 123467 mod 23 = 3.

2.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

Find \(\dfrac{AU}{US}\)

\(\begin{array}{|rcll|} \hline \vec{u} &=& \vec{A}-\vec{C} \\ \vec{v} &=& \vec{B}-\vec{C} \\ \hline \vec{U}-\vec{C} &=& \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) \quad | \quad \lambda \text{ is a real value} \\ \vec{U}-\vec{C} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \quad | \quad \dfrac{US}{AS} \text{ is a real value} \\ \hline \vec{U}-\vec{C} = \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda\left( \vec{v}-\dfrac{3}{7}\vec{u} \right) &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\left( \vec{u}-\dfrac{2}{3}\vec{v} \right) \\ \dfrac{3}{7}\vec{u}+ \lambda \vec{v}-\dfrac{3}{7}\lambda\vec{u} &=& \dfrac{2}{3}\vec{v}+ \dfrac{US}{AS}\vec{u}-\dfrac{2}{3}\dfrac{US}{AS}\vec{v} \\ \vec{u}\left(\underbrace{\dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} }_{=0}\right) &=& \vec{v}\left(\underbrace{\dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda }_{=0} \right) \quad | \quad \vec{u} \text{ and } \vec{v}\text{ are independent} \\ \hline \dfrac{3}{7}-\dfrac{3}{7}\lambda-\dfrac{US}{AS} &=& 0 \\ \quad |\quad \cdot \dfrac{7}{3} \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} &=& 0 \qquad (1) \\\\ \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda &=& 0 \qquad (2) \\ \hline (1)-(2): \qquad 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \left( \dfrac{2}{3}-\dfrac{2}{3}\dfrac{US}{AS}-\lambda \right) &=& 0 \\ 1- \lambda-\dfrac{7}{3}\dfrac{US}{AS} - \dfrac{2}{3}+\dfrac{2}{3}\dfrac{US}{AS}+\lambda &=& 0 \\ \dfrac{1}{3} -\dfrac{5}{3}\dfrac{US}{AS} &=& 0 \\ \dfrac{5}{3}\dfrac{US}{AS} &=& \dfrac{1}{3} \quad | \quad \cdot \dfrac{3}{5} \\ \mathbf{ \dfrac{US}{AS} } &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{AU}{AS} &=& \left( 1-\dfrac{US}{AS} \right) \\ \dfrac{AU}{AS} &=& \left( 1-\dfrac{1}{5} \right) \\ \mathbf{ \dfrac{AU}{AS} } &=& \mathbf{\dfrac{4}{5}} \\ \hline \dfrac{AU}{US} &=& \dfrac{ \dfrac{AU}{AS} }{ \dfrac{US}{AS} } \\ \dfrac{AU}{US} &=& \dfrac{ \dfrac{4}{5} }{ \dfrac{1}{5} } \\ \mathbf{ \dfrac{AU}{US} } &=& \mathbf{4} \\ \hline \end{array}\)

I dont THINK it's the answer, I am POSITIVE that it is.

1.

In triangle ABC, S is a point on side BC such that BS:SC = 1:2,

and T is a point on side AC such that AT:TC = 4:3.
Let U be the intersection of AS and BT.

We can write \( \vec{T} = w \vec{A} + x \vec{C}\), \(\vec{S} = y \vec{B} + z \vec{C}\) for some real values of w, x, y, and z.
Find w, x, y, and z.

\(\begin{array}{|rcll|} \hline \vec{T} &=& \vec{A}+ \dfrac{4}{7}\left(\vec{C}-\vec{A} \right) \\ \vec{T} &=& \vec{A}+ \dfrac{4}{7}\vec{C} -\dfrac{4}{7}\vec{A} \\ \vec{T} &=& \dfrac{3}{7}\vec{A}+ \dfrac{4}{7}\vec{C} \quad &|\quad \vec{T} = w \vec{A} + x \vec{C}\\ \hline \mathbf{w}&=& \mathbf{\dfrac{3}{7}} \\ \mathbf{x}&=& \mathbf{\dfrac{4}{7}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \vec{S} &=& \vec{C}+ \dfrac{2}{3}\left(\vec{B}-\vec{C} \right) \\ \vec{S} &=& \vec{C}+ \dfrac{2}{3}\vec{B} -\dfrac{2}{3}\vec{C} \\ \vec{S} &=& \dfrac{2}{3}\vec{B}+ \dfrac{1}{3}\vec{C} \quad &|\quad \vec{S} = y \vec{B} + z \vec{C}\\ \hline \mathbf{y}&=& \mathbf{\dfrac{2}{3}} \\ \mathbf{z}&=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

I have just been playing with the numbers. I do not know how to do this.

But these are the possible combinations of 6 digits that add up to 23.

0 1 2 3 8 9

0 1 2 4 7 9

0 1 2 5 6 9 

0 1 3 4 6 9 

0 2 3 4 5 9

0 1 2 5 7 8

0 1 3 4 7 8

0 1 3 5 6 8

0 2 3 4 6 8

1 2 3 4 5 8

1 2 3 4 6 7     (added on)

0 2 3 5 6 7 

0 1 4 5 6 7 

So that is 12 groups of 6 digits.

Now it canot begin with 0 becasue that would not be a proper number.

The digits add to 23 so none of these will be divisable by 3 or 6 or 9.  I am not sure what that means, I am getting in a mess.

So I have a lot  LESS than

5*5!*11+6! =6600+720 = 7320 ways   

plus another  720 =  8050 

352 ways certainly sounds possible

Edit

oh dear, I just realize that 1 2 3 4 6 7 is missing.    I wonder how many others are missing.  I think not many, hopefully none.

oh.. I took it as 3 separate problems

\(b=1-a\\ \dfrac 1 a + \dfrac 1 b = \\ \dfrac 1 a + \dfrac{1}{1-a} = \\ \dfrac{1-a+a}{a(1-a)} = \\ \dfrac{1}{a(1-a)} \in (-\infty,0) \cup [4,\infty)\)

Find the value of \(k\) for which the binomial \((2x – 1)\) is a factor of \(2x^3 + kx^2 – 2x + 1\)

\(\begin{array}{|rcll|} \hline (2x – 1)&=& 2\left(x – \mathbf{\dfrac12} \right) \\ \hline \end{array}\)

The root is at \(\mathbf{x=\dfrac12}\)

\(\begin{array}{|rcll|} \hline 2x^3 + kx^2 – 2x + 1 \quad &| \quad \mathbf{x=\dfrac12} \\ 2\left(\dfrac12\right)^3 + k\left(\dfrac12\right)^2 – 2\left(\dfrac12\right) + 1 &=& 0 \\ \dfrac{2}{8} + \dfrac{k}{4} – \dfrac{2}{2} + 1 &=& 0 \\ \dfrac{1}{4} + \dfrac{k}{4} – 1 + 1 &=& 0 \\ \dfrac{1}{4} + \dfrac{k}{4} &=& 0 \quad &| \quad \cdot 4\\ 1+k &=& 0 \quad &| \quad -1 \\ \mathbf{k} &=& \mathbf{-1} \\ \hline \end{array}\)

AGREED !  UNfortunately....we get a lot of folks posting stuf which is UNclear!   Parentheses and brackets are NECESSARY to get correct answers to postings (and not make an engineering mistake that kills people !)          ~EP

Rom: I get a completely different answer, way way smaller than yours! Here is the code:

n=0;p=0;cycle:a(100000+n);b=int(a/100000);c=int(a/10000)%10;d=int(a/1000)%10;e=int(a/100)%10;f=int(a/10)%10;g=a%10;n=n+1;if(b+c+d+e+f+g==23 and a%23==0 and b!=c and b!=d and b!=e and b!=f and b!=g and c!=d and c!=e and c!=f and c!=g and d!=e and d!=f and d!=g and e!=f and e!=g and f!=g, goto loop,goto cycle);loop:p=p+1;printa," ",;if(n<=882031, goto cycle, 0);print"Total = ",p

1 - They all sum up to 23. Are multiples of 23. And all 6 digits different from each other.

They total = 352. They begin with: 102695  104765  104972  105386  106835  107249  107456.......and end in 

946013  951602  952016  952430  963401  981203  982031  Total =  352

It cannot equal 0.

It must be bigger than 0

Any more takes?

ok, still I think when there is possible confusion (over and above knowing the very basic rules) people should just use brackets.

That indice one gets me every time!

2) 

999993 = 23 * 43478

100004 = 23 * 4348

43478 - 4348 + 1 = 39131

3)

9 choices for the 1st digit, 9 for the 2nd, then 8 7 6 5

9 * 9!/4! = 136080

How did you get 165?