thelizzybeth

Ok, so in that case, just replace 19 with 19/4 and calculate the answer. Sorry about that!

Where is this question from? Is it from somewhere online like AOPS' Alcumus where you can see if you got it wrong? Or is it like homework?

Yes, there are twelve edges, but four are the same length, so we can express those lengths as a, b, and c. 

Just like for a rectangle, there are four edges, but we only need two variables: L and W to express the length and width. 

Now that you asked this question though, I am wondering if "the combined length of all its edges is 19." means 4(a+b+c)=19. I assumed it meant different edges (why I only did a+b+c), but I may be wrong there. 

Let's use casework.

From 1-9 there are 9x1= 9 digits

From 10-99 there are 90x2= 180 digits

So far we have 180+9=189 digits. Now we have to start breaking it down into smaller cases. 

From 100-299 there are 200x3=600 digits

This is not enough. We need 1000-(600+189)=211 more digits

Since we are in the 3 digit number region, lets divide 211 by 3. This equals 70 with a remainder of 1. We add 70 to 299 to find the last number.

So that means from 1-369, there are 999 digits. The next digit, the 1000th digit, is 3.

I mean, you can check yourself. All of my steps are correct. I checked my arithmetic again. (last time I made the mistake of forgetting a number)

To verify, just follow the steps in my solution.

deleted

I edited my answer- is it correct now?

I'm not too sure how to do number 1 (not that great with trig) but here's number 2. 

2. Let's write out the information we have so far. Set a, b, and c as the three sides.

a+b+c=19/4

2(ab+ac+bc)=12

We are trying to figure out the space diagonal. We see that a space diagonal equals \(\sqrt{a^2+b^2+c^2}\) through the Pythagorean Theorem (to see why, draw a diagram. There is a right triangle between the height of the prism, the base diagonal, and the space diagonal.)

Now let's square both sides of a+b+c=19/4, which after expanding, simplifies to \(a^2+b^2+c^2+2ab+2ac+2bc=\frac {361} {16}\)

Substituting in 2(ab+ac+bc)=12, the second equation at the top, the equation becomes \(a^2+b^2+c^2+12=\frac{361} {16}\). Simplifying, we find\(a^2+b^2+c^2=\frac{169}{16}\) 

We are trying to find the square root of this, so our answer is \(\frac{13}{2}\)

EDIT: I have corrected my solution.

AOPS has a solution:

Let's break it down and simplify each part.

\((t-1)^3\)= \(t^3-3t^2+3t-1\)

\(6(t-1)^2\)\( = 6(t^2-2t+1) = 6t^2-12t+6\)

\(12(t-1)=12t-12\)

Combining these and the 8 together, we get our final answer: \(t^3-3t^2+3t-1\) \( + 6t^2-12t+6\) \( + 12t-12+8\) =  \(t^3+3t^2-3t+1\)