All Answers 357289 Answers

Not really.  I only have the information in the problem so if I could find them I'm sure the smarter people on this site could too.

Opinion only: Do you understand the question well enough to sketch a diagram of the triangles and upload it? I think you will have a much better chance of getting it answered. Good luck.

Wow Ginger! I sure wasn't expecting to see this! Thank you a lot!!! I appreciate it!

Thank you, Melody

Nice answer Heureka :)

find the base of numeration in which 135/21 has a zero remainder

Let the base be x

$\frac{x^2+3x+5}{2x+1}=n \qquad n\in Z\\ $

I am going to try trial and error.

x has to be 6 or bigger because of the 5.

If x=6 this becomes

$\frac{36+16+5}{13}=\frac{57}{13}\ne n$

If x=7

$\frac{49+21+5}{14+1}=\frac{75}{15}=5=n$

So base 7 works     

When the minute hand of a clock points exactly at a full minute, the hour hand is exactly two minutes away.

Give all possible times in a 12-hour period that satisfy condition.

$\text{ angular velocity minute hand: $ \omega_m^{\circ} =\dfrac{360^{\circ}}{1~ h} $ }\\ \text{ angular velocity hour hand: $ \omega_h^{\circ} =\dfrac{360^{\circ}}{12~ h} $ }\\ \boxed{\text{angle = angular velocity }\times \text{ time}}\\ \text{ angle minute hand: $ \alpha_m^{\circ} = \omega_m^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle hour hand: $ \alpha_h^{\circ} = \omega_h^{\circ} \times t^h \qquad t^h$ time in hours } \\ \text{ angle difference minute hand-hour hand: $ \alpha_m^{\circ}-\alpha_h^{\circ} = \Delta\alpha^{\circ}$ } \\ \Delta\alpha^{\circ} = \alpha_m^{\circ}-\alpha_h^{\circ} \\ \Delta\alpha^{\circ} = \omega_m^{\circ} \times t^h - \omega_h^{\circ} \times t^h \\ \Delta\alpha^{\circ} = \left(\omega_m^{\circ} - \omega_h^{\circ} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(\dfrac{360^{\circ}}{1~ h} - \dfrac{360^{\circ}}{12~ h} \right) \times t^h \\ \Delta\alpha^{\circ} = \left(360\cdot \dfrac{11}{12} \right) \times t^h \\ \Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} \\$

$\mathbf{\boxed{\Delta\alpha^{\circ} = 330 \times t^h \pmod{360^{\circ}} }}$

Example:

$t^h = 6\ h \\ \Delta\alpha^{\circ} = 330 \times 6 \pmod{360^{\circ}} \\ \Delta\alpha^{\circ} = 1980^{\circ} \pmod{ 360^{\circ} } \\ \Delta\alpha^{\circ} = 1980^{\circ} - 5\cdot 360^{\circ} \\ \Delta\alpha^{\circ} = 180^{\circ}$

1.

$\text{The hour hand of a clock points exactly at a full minute:} \\ \begin{array}{rcll} \dfrac{360^{\circ}}{60~\min.} &=& \dfrac{x}{1~\min.} \\\\ x&=&\dfrac{1~ \min.}{60~ \min.}\cdot 360^{\circ} \\\\ \mathbf{x}&\mathbf{=}&\mathbf{6^{\circ}} \\ \end{array}$

1 minute on the clock conforms 6 degrees.

$\text{The hour hand of a clock points exactly at a full minute, so} \\ \text{ $\alpha_h^{\circ} = 6^{\circ} \cdot n \qquad | \qquad n$ is an integer}$

2. $\mathbf{t^h =\ ?}$

$\begin{array}{rcll} \alpha_h^{\circ} &=& \omega_h^{\circ} \times t^h \\\\ t^h &=& \dfrac{ \alpha_h^{\circ} } {\omega_h^{\circ}} \quad & | \quad \alpha_h^{\circ} = 6^{\circ} \cdot n \qquad \omega_h^{\circ} = \dfrac{360^{\circ}}{12~ h} \\\\ t^h &=& \dfrac{ 6^{\circ} \cdot n } { \dfrac{360^{\circ}}{12~ h}} \\\\ \mathbf{t^h} &\mathbf{=}& \mathbf{0.2n} \\ \end{array}$

3.

$\text{The hour hand is exactly two minutes away: $\Delta\alpha^{\circ} = \pm 12^{\circ}$ } \\ \begin{array}{rcll} \Delta\alpha^{\circ} &=& 330 \times t^h \pmod{360^{\circ}} \quad & | \quad \mathbf{t^h=0.2n} \qquad \Delta\alpha^{\circ} = \pm 12^{\circ} \\ \pm 12^{\circ} &=& 330 \times 0.2n \pmod{360^{\circ}} \\ \pm 12^{\circ} &=& 66n \pmod{360^{\circ}} \\ \pm 12^{\circ} -66n &=& 360^{\circ}m \qquad & n \in N,\ m \in N \\ \pm 12^{\circ} &=& 66n + 360^{\circ}m \quad & | \quad : 6 \\ \pm 2^{\circ} &=& 11n + 60^{\circ}m \\ &&\boxed{1. \text{ Diophantine equation: } 11n + 60^{\circ}m = 2} \\ &&\boxed{2. \text{ Diophantine equation: } 11n + 60^{\circ}m = -2} \\ \end{array}$

4. Solution diophantine equation 11n + 60m = 2:

$\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& 2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \\ &=& \dfrac{ 2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m + 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ 2 - 5m } {11} \\ n &=&-5m+ \dfrac{ 2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 - 5m } {11} \\ & 11a &=& 2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& 2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \\ &=& \dfrac{ 2 - 10a-a } {5} \\ &=& \dfrac{ - 10a + 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ 2 -a } {5} \\ m &=& -2a+ \dfrac{ 2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ 2 - a } {5} \\ & 5b &=& 2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2 - 5b } \\ \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ 2 - 11a } {5}} \quad & | \quad \mathbf{a = 2 - 5b }\\ & = & \dfrac{ 2 - 11 (2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{-4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = -4 + 11b }\\ & = & \dfrac{ 2 - 60(-4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{22 - 60b } \\ \end{array}$

$\boxed{ \text{1. Diophantine equation: }\\ \mathbf{n = 22 - 60b} \\\mathbf{m=-4 + 11b} \qquad b \in Z }$$\begin{array}{|lrcll|} \hline b = 0: & n &=& 22 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 22 \\ & &=& 4.4\quad (4:24) \\ \hline \end{array}$

5. Solution diophantine equation 11n + 60m = -2:

$\text{The variable with the smallest coefficient is $n$. The equation is transformed after $n$: }\\ \begin{array}{rcll} 11n + 60m &=& -2 \\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ -2 - 60m } {11}} \\ &=& \dfrac{ -2 - 55m - 5m } {11} \\ &=& \dfrac{ - 55m - 2 - 5m } {11} \\ &=& -\dfrac{55m}{11} + \dfrac{ -2 - 5m } {11} \\ n &=&-5m+ \dfrac{ -2 - 5m } {11} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ -2 - 5m } {11} \\ & 11a &=& -2 - 5m \\ \end{array} \\ \text{The variable with the smallest coefficient is $m$. The equation is transformed after $m$: }\\ \begin{array}{rcll} 5m &=& -2 - 11a \\ \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \\ &=& \dfrac{ -2 - 10a-a } {5} \\ &=& \dfrac{ - 10a - 2 -a } {5} \\ &=& -\dfrac{10a}{5} + \dfrac{ -2 -a } {5} \\ m &=& -2a+ \dfrac{ -2 - a } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} & b &=& \dfrac{ -2 - a } {5} \\ & 5b &=& -2 - a \\ \end{array}\\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ -2 - 5b } \\ \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{m} &\mathbf{=}& \mathbf{\dfrac{ -2 - 11a } {5}} \quad & | \quad \mathbf{a = -2 - 5b }\\ & = & \dfrac{ -2 - 11 (-2 - 5b) } {5} \\ \mathbf{m} & \mathbf{=} & \mathbf{4 + 11b} \\\\ \mathbf{n} &\mathbf{=}& \mathbf{\dfrac{ 2 - 60m } {11}} \quad & | \quad \mathbf{m = 4 + 11b }\\ & = & \dfrac{ -2 - 60(4 + 11b) } {11} \\ \mathbf{n} & \mathbf{=} & \mathbf{-22 - 60b } \\ \end{array}$

$\boxed{ \text{2. Diophantine equation: }\\ \mathbf{n = -22 - 60b} \\\mathbf{m=4 + 11b} \qquad b \in Z }$$\begin{array}{|lrcll|} \hline b = -1: & n &=& -22+60 \\ & &=& 38 \\ & t^h &=& 0.2n \\ & t^h &=& 0.2\cdot 38 \\ & &=& 7.6\quad (7:36) \\ \hline \end{array}$

The solutions are: 4:24 and 7:36

Thanks GingerAle for bringing Hectictar's milesone to our attention.

And to you Hectictar.

We have been greatly honoured to have you so active on our forum.

Thank you for all the wonderful contributions you have made.    

Here's another approach    ......

1 in 1st position  and then  -  C(8,3)   -   56

2 in 1st position and then - C (7,3)  -      35

3 in 1st position and then - C (6,3)  -      20

4 in 1st position and then  -  C (5,3)  -    10

5 in 1st position and then - C ( 4,3)  -       4

6 in 1st position and then  - C (3,3)  -       1

                                                              _____

                                                               126

Just as Melody found !!!

40W  =  .04 kW

8W  =      .0008  kW

So....let the hours that equalize the cost  = H......so we have

.60 +  .04 (.20) H  =  4.20  + .0008 (.20)H     simplify

.60  + .008H   =  4.20  + .00016H    

Subtract  .60 , .00016H  from both sides

.00784H  =  3.60     

Divide both sides by  .00784

H  ≈  460 hours

So....the second option is cheaper after  ≈ 460 hours

Yeah...post your solution, Melody   [ at your leisure, of course ]

I've done somethingl like this, but I don't exactly remember what I did now  !!!!

I want to see what you did......

Kinda' like the last one....we want to find

[ f(6) - f(3) ] / [ 6 - 3 ]  =

[  (6)^2 + 3(6) + 2  -  ( (3)^2 + 3(3) + 2 )  ]  /  3    = 

[ 36  + 18 +  2  -  9 -  9  - 2 ]  /  3    =

[ 36] / 3  =   12

We want  to find this :

[ -log (4*8)  + 5   -  ( - log(4*3)  + 5) ]  /  [ 8 - 3 ]  =

[ - log(4*8)  + log (4*3) ]  /  [ 8 - 3  ] =

[ log (12)  - log (32)  ] /  5 =

[ using  log a  - log b   =  log (a/b)  ]

log ( 12 / 32)  / 5   ≈   -0.085  ≈  -0.09

I based this on the fact that the little hand moves 5 minute units in an hour so the little hand is on a minute mark every 12 minutes.

I worked out that ever 12 minutes the hands move apart by a further 11 units.  Of course this is modular so 60units=0units

Then i just did some grunt work and found the pattern. 

There would be a better modular way of doing it but that is how I did it. 

I can give more info if anyone wants it. 

Twice  4:24  and  7:36

AIDS / A  =  SEER

This means the same as

S E E R

          A

______

A I D  S

Here's the logic of the solution

The last multiplication   S x A  =  A   implies that S  = 1

The first multiplication  R x A   results  in an ending value of 1

And....only the product 7 * 3   results in an ending  "1"

Thus  R = 3  and A  = 7  or vice-versa

This means that  EA + 2  =  D 

And  EA  + ??  =   I

But there is no "carry"  because  SA  could not = A  if this were true 

So...   ??  =  0 

Thus...  EA  <  10

But  if A  = 7  and EA  < 10... guess that  E  could = 1

But this is impossible because S  =1

And E  cannot = 0  because the second multiplication (EA) would have to equal E if this was true

So.....A must equal  3

And E could only   = 2

And the "carry"  from R x A  = 2

Which means that

2  + EA  =

2 + 2(3)  =  8  =  D

And since there is no "carry"  on the  next multiplication ... I  must  = EA  =  6

And R  = 7

So we have that

1 2  2  7

           3

_______

3  6  8  1

Or

AIDS / A   = SEER

3681 / 3   =   1227

How many two-digit positive integers are congruent to 1 (mod 3)?

$\begin{array}{|rcll|} \hline \text{ $x \equiv 1 \pmod 3$ $\\$ or $ \\ x-1 = n\cdot 3 $ } \\ \hline \end{array}$

$\begin{array}{lrcll} \text{If $x = 99$} & 99-1 &=& n\cdot 3 \\ & 98 &=& n\cdot 3 \\ & n &=& \dfrac{98}{3} \\ & n &=& 32.7 \\ & \boxed{ n = 0\ldots 32 \qquad x = 1\ldots 99 } \\ \end{array}$

$\begin{array}{lrcll} \text{If $x = 9$} & 9-1 &=& m\cdot 3 \\ & 8 &=& m\cdot 3 \\ & m &=& \dfrac{8}{3} \\ & m &=& 2.7 \\ & \boxed{ m = 0\ldots 2 \qquad x = 1\ldots 9 } \\ \end{array}$

$\begin{array}{|rcll|} \hline x = 10\ldots 99 \\ n-m &=& 33 -3 = 30 \\ \hline \end{array}$

30 two-digit positive integers are congruent to 1 (mod 3)

$\begin{array}{|l|rcll|} \hline 1. & 10 \\ 2. & 13 \\ 3. & 16 \\ 4. & 19 \\ 5. & 22 \\ 6. & 25 \\ 7. & 28 \\ 8. & 31 \\ 9. & 34 \\ 10. & 37 \\ 11. & 40 \\ 12. & 43 \\ 13. & 46 \\ 14. & 49 \\ 15. & 52 \\ 16. & 55 \\ 17. & 58 \\ 18. & 61 \\ 19. & 64 \\ 20. & 67 \\ 21. & 70 \\ 22. & 73 \\ 23. & 76 \\ 24. & 79 \\ 25. & 82 \\ 26. & 85 \\ 27. & 88 \\ 28. & 91 \\ 29. & 94 \\ 30. & 97 \\ \hline \end{array}$

An increasing number has its digits increase in size from left to right. For example, 24589 is an increasing number. How many four-digit increasing numbers are there? 

There are 10C4 ways to choose 4 digits that is 210 ways

Now for each of these combiations there is exactly 1 where the digits are in ascenting order.

So that is still 210 ways.

If you discount numbers beginning with 0 then that gets harder.

Lets see... if the first one is 0 how many ways can the others be chosen. 9C3=84 ways

So I think there are 210-84= 126 ways

So there are 126 numbers that meet this criterion.

y  =  0.155x^2 - 4.074x +  38.910

Sine 1 is common factor to both polunomials, therefore whatever value n takes, the GCD of the rwo polynomials will be =1

3C + 10, where C =0 to 29

3*0 + 10 mod 3 = 1

3*1 + 10 mos 3 = 1

3*2 + 10 mod 3 = 1.........and so on to C=29

3*29 + 10 mod 2 = 1 and so on . So, there are 30 2-digit positive integers that sarisfy the congruence.

The average rate of change is given by :

[  217.2   - 238. 4 ]  /  [ 4 - 2 ]  =

[ -21.2 ]  /  2  =

-10.6    ⇒  Profits decreased by an average of $10,600 per year from Year 2 to Year 4

The polynomial will have zeroes of

3, 4 + 2i, 4- 2i, 1 + √7 and 1 - √7

So we have

(x - 3) (x - (4 + 2i)  )  ( x  - (4 - 2i) )  ( x - (1 - √7)  ) ( x - (1 + √7) )  =

Breaking this down, we have

( x - (4 + 2i ) (x - ( 4 - 2i)  =  (x^2 -x(4 + 2i) - x (4 - 2i) +  (4 + 2i) (4 - 2i)  =

(x^2  -4x - 2ix - 4x + 2ix  +  16 + 4)  =  ( x^2 - 8x + 20)

And

( x - (1 - √7)) ( x - ( 1 + √7)  =  

x^2 - x(1 - √7) - x ( 1 + √7) +  (1 - √7) (1 + √7) =

x^2  - x + √7x - x - √7x + 1 - 7  =

(x^2  - 2x  - 6)

So  we have

(x - 3) ( x^2 - 8x + 20)  (x^2 - 2x - 6)  = 

This expands to :

x^5 - 13 x^4 + 60 x^3 - 82 x^2 - 144 x + 360