I'm going to assume that a,b and c can coincide with eachother.
So a=3, b=1, c=3
a=c=3;
\(\overline {ab}\)=\(\overline {cb}\)=31;
\(\overline {abc}\)=\(\overline {cba}\)=313; 
Of which all are prime.
It feels cheating to do it this, but you have to get the smallest prime so... oh well

Applying the triangular inequality:
\(12-5< n< 12+5\)
\(\Leftrightarrow7< n< 17\)
So n varies from 8 to 16, therefore there are 9 possible integer values of n

I think that would be between any integer with the hundreds digit being 9, and the ones digit being 1, the tens digit could be anything, and the difference would always be 792.
I'm a bit unsure about this though

The area of those spheres: \(5\cdot4\pi10^2=2000\pi\)(inches square)
Since the amount of paint is proportinal to the area, that means the area of the cylinder is the same as the area of those spheres.
So the height is \(\dfrac{2000\pi-2\pi10^2}{2\pi10}=90\)(inches)

meter and then centimeter square?
The grammar seems a bit sloppy.
In the second sentence, I think you accidentally called the square "the rectangle" and called the rectangle "the rectangular".
Or maybe I'm just really confused about this question.
I think you should recheck if you had posted the question correctly, okay?

\(\sqrt{3}x-2=2\)
\(\Leftrightarrow\sqrt{3}x=4\)
\(\Leftrightarrow x=\dfrac{4}{\sqrt{3}}=\dfrac{4\sqrt{3}}{3}>\dfrac{2}{3}\)
Xin lỗi nha, mình trả lời bạn hơi chậm rùi 

I have literarily no idea how to solve this one the "effective way", so I just kept on spamming out random stuff until it works.
So... d should be a small number, and a & b should be larger numbers...
So I found out that \(54\cdot3=162\)
NEAT! c represents 3

OwO, it's a typo: it's suppossed to be width, okay?
If you use the AMGM theorem, you'll know that any rectangle achieves the maximum possible area when both of their measurements are equal, making it a square (a square is technically a rectangle too)
So we have p=q
Therefore, the equation will be q²+q=13+q²
<=>q=13(cm)
<=>p=q=13(cm)
<=>S=pq=13²=169(cm²)

(P.S: don't do this here, cuz you'll get to a dead end, oof)
\(pq+q=13+q^2\)
\(\Leftrightarrow S=q^2-q+13\)
 

The time for the express train: \(\dfrac{18}{72}=\dfrac{1}{4}\left(h\right)=15minutes\)
The time for the local train to travel (without the stops): \(\dfrac{18}{54}=\dfrac{1}{3}\left(h\right)=20minutes\)
And now counting the stops: \(20+6\cdot1.5=29\left(minutes\right)\)
So obviously, the local train takes \(29-15=14\left(minutes\right)\)more than the express train

The positive integer divisors of 20 are 1,2,4,5,10,20
The positive integer divisors of 30 are 1,2,3,5,6,10,15,30
So the chances are \(4\cdot\dfrac{1}{6}\cdot\dfrac{1}{8}=\dfrac{1}{12}\)

Time to turn all of them into decimal
1111₃=3³+3²+3¹+3⁰=27+9+3+1=40
1111₂=2³+2²+2¹+2⁰=8+4+2+1=15
So the decimal difference is 40-15=25

2 perfects squares: 1 and 4

I won't take credit for this solution, as it goes to this website: https://www.cut-the-knot.org/proofs/LinesDividePlane.shtml
Applying 2016 to the formula, we'll get 2033137 regions, which has the sum of digit be 19

So we have 1,2,3,4,6,8,9,12,18,24,36
Then we'll get 1,{1},{1},{1,2},{1,2,3},{1,2,4},{1,3},{1,2,3,4,6},{1,2,3,6,9},{1,2,3,4,6,8,12},{1,2,3,4,6,9,12,18}
Now we'll get 1,{1},{1},{1,[1]},{1,[1],[1]},{1,[1],[1,2]},{1,[1]},{1,[1],[1],[1,2],[1,2,3]},{1,[1],[1],[1,2,3],[1,3]},{1,[1],[1],[1,2],[1,2,3],[1,2,4],[1,2,3,4,6]},{1,[1],[1],[1,2],[1,2,3],[1,3],[1,2,3,4,6],[1,2,3,6,9]}
And then, 1,{1},{1},{1,[1]},{1,[1],[1]},{1,[1],[1,(1)]},{1,[1]},{1,[1],[1],[1,(1)],[1,(1),(1)]},{1,[1],[1],[1,(1),(1)],[1,(1)]},{1,[1],[1],[1,(1)],[1,(1),(1)],[1,(1),(1,2)],[1,(1),(1),(1,2),(1,2,3)]},{1,[1],[1],[1,(1)],[1,(1),(1)],[1,(1)],[1,(1),(1),(1,2),(1,2,3)],[1,(1),(1),(1,2,3),(1,3)]}
Finally, 1,{1},{1},{1,[1]},{1,[1],[1]},{1,[1],[1,(1)]},{1,[1]},{1,[1],[1],[1,(1)],[1,(1),(1)]},{1,[1],[1],[1,(1),(1)],[1,(1)]},{1,[1],[1],[1,(1)],[1,(1),(1)],[1,(1),(1,|1|)],[1,(1),(1),(1,|1|),(1,|1|,|1|)]},{1,[1],[1],[1,(1)],[1,(1),(1)],[1,(1)],[1,(1),(1),(1,|1|),(1,|1|,|1|)],[1,(1),(1),(1,|1|,|1|),(1,|1|)]}

72=2³.3²

To do this, we will calculate the area of the rectangle that has the top left of (a) as the corner, and the bottom right of the cell we are analyzing as the other corner, minus 1
1(a): number 1, so we don't discount anything
2(b): each overlapping 1(a) => 1(one)
2(c): each overlapping 1(a), 1(b) => 1+1=2(ones)
1(d): overlapping 1(a),1(b),1(c) => 1+1+2=4(ones)
1(e): overlapping 1(a),2(b) => 1+2.1=3(ones)
2(f): each overlapping 1(a),2(b),1(c),1(e) => 1+2.1+2+3=8(ones)
1(g): overlapping 1(a),2(b),1(c),1(d),1(e),1(f) => 1+2.1+2+4+3+8=20(ones)
1(h): overlapping 1(a),2(b),2(c),1(e),2(f)=> 1+2.1+2.2+3+2.8=26(ones)
1(i): ignore this guy
So in total, we have 1+2.1+2.2+4+3+2.8+20+26=76(ones)
This took 3 hours for me to figure out... Tell Becky that I hate her sadistic game

SILLY has 5 letters
S is used once
I is used once
L is used twice
Y is used once
So the number of distinct strings are \(\dfrac{5!}{1!1!2!1!}=60\)

The number of chicks is a (creatures) for a\(\in\)N* and a<27
The number of bunnies is 27-a (creatures)
We have the equation:
2a+4(27-a)=78
<=>2a+108-4a-78=0
<=>2a=30
<=>a=15(satisfies the condition)
So there are 15 chicks and 27-15=12 bunnies

There is the greatest reduction when the two sides adjacent to the right angle are equal (I mean, a square is technically also a rectangle)
Using the Pythagorean theorem, we can see that the diagonal is a\(\sqrt{2}\) and the sum of the other sides is 2a
So the greatest percent of reduction is \(\dfrac{2a-a\sqrt{2}}{2a}=1-\dfrac{\sqrt{2}}{2}\approx29\%\)

Oh, I forgot to put "(cm)" on the fourth line 

a+b+c=50
The solid has the greatest possible volume when a=b=c=\(\dfrac{50}{3}\)=16,(6)(cm) (AMGM)
Since a,b and c are integers, a=16(cm), b=c=17(cm). V₁=abc=16.17.17=4624(cm³)

The least possible volume occurs when a=b=1;c=48; V₂=abc=1.1.48=48(cm³)

So the difference is V₁-V₂=4624-48=4576(cm³)

a²-b² is smallest when a-b is smallest and a,b are some of the smallest numbers.
So, following the question's condition, the most appropriate numbers are 1000 and 1001.
So the answer is 1001²-1000²=2001