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When I started drawing this problem out, I was really focusing on both the aspects of "smallest possible sum" and "greatest common divisor" which for some reason combined in my head to think of using the least common multiple of the 3 sides that make the vertex be the number at the vertex. I think this would also create the smallest possible sum. I'm not entirely sure, though, lmao.

Here is my solution.

Please excuse my poor handwriting, bad lighting, and dirty whiteboard hehehe

Also this is where I do all my math if anyone is curious!!

To clarify, the drawing has visible aspects of the die in purple and hidden aspects of the die in orange. So each vertex becomes the least common multiple of each side that it touches. For example, \(a = lcm(1,3,5) = 15\).

So, my answer is 153. I'm not sure if this is THE lowest answer, but this is what I came up with.

When \(f(x) = ax^3 - 6x^2 + bx - 5\) is divided by \(x - 1\), the remainder is \(-5\).
When \(f(x)\) is divided by \(x + 2\), the remainder is \(-53\).
Find the ordered pair \((a,b)\).

In general:

If a polynomial \(f(x)\) is divided by \(x-a\) , the remainder is the constant \(f(a)\) , and \(f(x)=q(x)·(x-a)+f(a)\)

\(\begin{array}{|lrcll|} \hline (x-1),\ a= 1: & \mathbf{f(1)} &=& \mathbf{-5} \\ & a\cdot 1^3 - 6\cdot 1^2 + b\cdot 1 - 5 &=& -5 \\ & a - 6 + b - 5 &=& -5 \\ &\mathbf{ a + b } &=& \mathbf{6} \qquad (1) \\ & \mathbf{b} &=& \mathbf{6-a} \\ \hline (x+2),\ a= -2: & \mathbf{f(-2)} &=& \mathbf{-53} \\ & a\cdot(-2)^3 - 6\cdot (-2)^2 + b\cdot(-2) - 5 &=& -53 \\ & -8a - 24 -2b &=& -48 \quad | \quad : (-2) \\ & 4a +12 +b &=& 24 \\ & \mathbf{4a+b} &=& \mathbf{12} \qquad (2) \\ & 4a+b &=& 12 \quad | \quad b=6-a \\ & 4a+6-a &=& 12 \\ & 3a &=& 6 \\ &\mathbf{ a } &=& \mathbf{2} \\\\ & b &=& 6-a \\ & b &=& 6-2 \\ &\mathbf{ b } &=& \mathbf{4} \\ \hline \end{array}\)

\(\mathbf{(a,b) = (2,4)}\)

Check:

deleted... had f(1)=5 not -5

I agree. The question states "She travels a certain distance on the first day, half that distance on the second day, a third that distance on the third day..."

The usage of the word 'certain' combined with a repeated 'that,' as in referring to the same certain distance over and over, implies the same setup to the problem for me.

Visualizing this problem can go a few different ways, but I'm just gonna keep it in simple geometry.

Like, really simple geometry lmao.

To explain the numbers, the horizontal 130 N is given to us. The downward force, or weight of the acrobat, is \(115 * 9.8 = 1127\).

Now, what is the magnitude of the force exerted by the floor on his hand? Well, it definitely isn't a normal force with two perpendicular forces. Oo! Then we can just use the resultant of the two forces. In other words, the magnitude of the force exerted by the floor on his hand is the hypotenuse of the above triangle in the drawing.

\(\sqrt{1127^2+130^2} = 1134.473N\)

Billie, a bushwalker, goes on a four day journey. She travels a certain distance on the first day, half that distance on the second day, a third that distance on the third day and a quarter of that distance on the fourth day. If the total distance is 50km, how far did she walk on the first day?

The meaning of the blue words is ambiguous. So there are multiple different interpretations of the problem. Each interpretation yields a different answer.

I think that the intended interpretation is this:

Billie, a bushwalker, goes on a four day journey. She travels a certain distance on the first day, half that distance on the second day, a third the distance traveled on the first day on the third day and a quarter of the distance traveled on the first day on the fourth day. If the total distance is 50km, how far did she walk on the first day?

There are 2 reasons why I think that is the intended interpretation:

1.  All instances of "that distance" in the problem statement would refer to the same thing.

2.  The solution would be an integer, and an integer that is divisible by 2, 3, and 4 at that.

I made an account to say thank you again (I'm the person who asked the question)

Also, the person that said "I have no idea what those other three stupidumb answers are" wasn't me. It was just my friend poking fun at my answer haha.

According to the question, the last answer # 6 is the accurate answer:

(x)  + 1/2( x) + 1/6( x)  + 1/24( x) =50, solve for x

x = 29 11/41 - Km - she walked on 1st day. 

   = 14 26/41 - Km on 2nd day

   = 4 36/41 - km on the 3rd day,

   =1 9/41- Km on th 4th day. 

Total = 50 Km

\(6(y + 1) \,=\, 7(x - 2)\)

                                               Divide both sides of the equation by  7 .

\(\frac67(y + 1)\, =\, x - 2\)

                                               Add  2  to both sides of the equation.

\(\frac67(y + 1)+2\, =\, x\)

                                               We could stop here, but let's simplify it.

\(x\,=\,\frac67(y + 1)+2\)

                                               Distribute  \(\frac67\)  to the terms in parenthesees.

\(x\,=\,\frac67y + \frac67+2\)

                                               Add  \(\frac67+2\)

\(x\,=\,\frac67y + \frac67+\frac{14}{7}\\~\\ x\,=\,\frac67y + \frac{20}{7}\)

Yes, thanks for the help.

Here is the solution to the problem the way you interpreted it:

Let the number of km traveled on the first day  =  x

\(x+\frac12x+\frac13\Big(\frac12x\Big)+\frac14\Big[\frac13(\frac12x)\Big]\,=\,50\\~\\ x+\frac12x+\frac16x+\frac1{24}x\,=\,50\\~\\ x(1+\frac12+\frac16+\frac1{24})\,=\,50\\~\\ x(\frac{41}{24})\,=\,50\\~\\ x(\frac{41}{24})\cdot\frac{24}{41}\,=\,50\cdot\frac{24}{41}\\~\\ x\,=\,\frac{1200}{41}\\~\\ x\,\approx\,29.268\)

The number of km she traveled on the first day  ≈  29.268  

You solved a different problem. It is kind of unclear what "that distance" refers to.

Also, we can check whether 24 km is a solution (to the problem I solved) like this:

num of km walked on first day  +  num of km walked on second day  +  num of km walked on third day  +

num of km walked of fourth day  =  50  ?

24  +  half of 24  +  a third of 24  +  a quarter of 24  =  50   ?

24  +  12  +  8  +  6  =  50   ?

50  =  50  ?

True

I have no idea what those other three stupidumb answers are, the answer was 29.2682924 km on the first day

Let the number of kilometers she traveled on the first day  =  x

When you say, "a third that distance on the third day and a quarter of that distance on the fourth day," I think "that distance" is  x .

\(x+\frac12x+\frac13x+\frac14x\,=\,50\\~\\ x(1+\frac12+\frac13+\frac14)\,=\,50\\~\\ x(\frac{25}{12})\,=\,50\\~\\ x(\frac{25}{12})\cdot\frac{12}{25}\,=\,50\cdot\frac{12}{25}\\~\\ x\,=\,24\)

She traveled 24 kilometers on the first day.

the answer is : the girl walked 50 miles 

the answer is he walked 87 km on the second day

Note that   sin( 90° - a )  =  cos( a )      and      cos( 90° - a )  =  sin( a )      and      tan( 90° - a )  =  cot( a )

This might help you see why those are true....

sin( 90° - a )  =  cos( a )   is saying that line  r  is the same length as line  s .

cos( 90° - a )  =  sin( a )   is saying that line  b  is the same length as line  c .

tan( 90° - a )  =  cot( a )   is saying that   r / b  =  s / c

So.....

     sin( 38° )  -  cos( 52° )

=   sin( 90° - 52° )  -  cos( 52° )

=   cos( 52° )  -  cos( 52° )

=   0

     tan( 12° )  -  cot( 78° )

=   tan( 90° - 78° )  -  cot( 78° )

...Do you see where this one is going? Can you finish it?

\(\phantom{=}\qquad\frac{\cos(10°)}{\sin(80°)}\\~\\ =\qquad\frac{\cos(90° - 80°)}{\sin(80°)}\\~\\ =\qquad\frac{\sin(80°)}{\sin(80°)}\\~\\ =\qquad1\)

The last one is a really similar to previous one. Do you know how to do it now?

I'm showing 2 as the first prime and 9973 as the 1229th prime.

10007 is the 1230th prime.

Oh, my bad... between 10 and 10000... blah

I get 

(1,3,5,7) = (306,309,307,303)

as you did.  Righto then.

Rom: I get slightly different count. There are 1,225 Primes between 10 and 10,000:

1 = 306, 3 = 309,
7 = 307, 9 = 303
Total =1,225 

deleted - see below

I don't know exactly what you want, but:

1 Degree Kelvin = - 272.15 Degrees Celsius.

Absolute 0 (Kelvin) =- 273.15 degrees Celsius = - 459.67 Degrees Fahrenheit.