Since the hundreds digit can't be 0, there are 9 possibilities here
The tens digit can be 0, but it can't be the same as the hundreds digit, so it's 9 possibilities here.
The ones digit can be 0,but it can neither be the same as the hundreds or the tens digits, so there are 8 posibilities here.
In total we have 9.9.8=648 integers
 

The least common multiple between 4 and 7 is 28 (seconds)
That means they blink together every 28 seconds.
10:15 p.m. - 10:00 p.m. = 15 minutes = 900 seconds
900/28 = 32 remainder 4
So they will blink together 32 more times

Call the number of boxes with 5 cars a (boxes)
        the number of boxes with 7 cars b (boxes)
For a,b\(\in\)N*
We have 5a+7b=53
The only solution to this equation is a=5 and b=4 (analyzing the ones digit of 53)
So the chances of selecting a box with 7 cars is \(\dfrac{4}{9}\)

a) 12+8#11-2=2(12-8)(11+12)=184
b) (11+8#11-2)+(13+8#11-2)+708=2(11-8)(11+11)+2(13-8)(11+13)+708=132+240+708=1080

Sorry, the last answer I posted glitched out
4# . 3# . 2# . 1#
= 4? . 3? . 2? . 1? . 3? . 2? . 1? . 2? . 1?
= 4? . (3?)² . (2?)³ . (1?)⁴
= 4! . 3! . 2! . 1! . (3! . 2! . 1!)² . (2! . 1!)³ . (1!)⁴
= 4! . (3!)³ . (2!)⁶ . (1!)¹⁰
= 24 . 6³ . 2⁶ . 1¹⁰
= 24 . 216 . 256 . 1
= 331776

\(4#\times3#\times2#\times1#=4?\times3?\times2?\times1?\times3?\times2?\times1?\times2?\times1?\times1?=4?\times\left(3?\right)^2\times\left(2?\right)^3\times\left(1?\right)^4=4!\times3!\times2!\times1!\times\left(3!\times2!\times1!\right)^2\times\left(2!\times1!\right)^3\times\left(1!\right)^4=4!\times\left(3!\right)^3\times\left(2!\right)^6\times\left(1!\right)^{10}=24\times6^3\times2^6\times1^{10}=24\times216\times64\times1=331776\)

The diagonal of the rectangle is \(\sqrt{6^2+8^2}=10\)(cm)
The diagonal of the rectangle is also the diameter of the circumscribing circle.
The radius of the circle is \(10\div2=5\)(cm)
The area of the circle is \(5^2\pi=25\pi\approx79\)(cm²)

1+2+3+4+5+6=21, so the greatest value of n is 6 (6 numbers, not the greatest number in that group is 6)

We shall call the length of the screen a (in)
                      the width of the screen b (in)
                      satisfying 37>a>b>0
Since the triangle constructed within the length, width and the diagonal line is a right triangle, we can apply the Pythagorean theorem to get the equation:
a²+b²=37²
Following the question, we can get the equation:
\(\dfrac{a}{16}=\dfrac{b}{9}\)
Since both sides are positive, we can square each of those sides to receive some equations:
\(\dfrac{a^2}{256}=\dfrac{b^2}{81}=\dfrac{a^2+b^2}{256+81}=\dfrac{37^2}{337}\)(Oh man, that fractions is not simplifiable)
\(\Rightarrow b^2=\dfrac{81\times37^2}{337}\)(I feel bad for anyone who doesn't use a calculator for this part)
\(\Rightarrow b=\sqrt{\dfrac{9^2\times37^2}{337}}=\dfrac{9\times37}{\sqrt{337}}=\dfrac{333\sqrt{337}}{337}\approx18.1\)(in)
So the width is his Tv screen is approximately 18.1 inches

Call each of the angles in an increasing order a,b,c,d (°) for 0°<a<b<c<d<180° (Assuming that this is a convex quadrilateral)
We have the equal expressions: \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{d}{6}=\dfrac{a+b+c+d}{3+4+5+6}=\dfrac{360\text{°}}{18}=20\text{°}\)
\(\Rightarrow d=20\text{°}\times6=120\text{°}\)(satisfies the condition)
So the largest of the angles is 120°

The sum of the numbers are 6.6=36
The sum of the two middle numbers are 8.2=16
Putting the numbers in a consecutively increasing order, let's say that the first number is 1, the second number is 2, the third number is 3 (these numbers are as small as possible to make the later numbers really big)
It would mean that the fourth number is 7 and the fifth number is 9, (try to make the number as close to the mean as possible while having the sum of the two numbers be 16, so that it will leave the most possibilities for the last numbers, just in case of running into any issues)
And so the largest possible number - the sixth number is 36-1-2-3-7-9=14
(Please take note of these tricks for later, similar problems)
 

Consider the original price is 100%
After the first increase, the price is 100%+150%.100%=100%+150%=250%
After the decrease, the price is 250%-75%.250%=250%-187,5%=62,5%
So it needs to increase 100%-62,5%=37,5% compared to the original
Which means it needs to increase 37,5%/62,5%=60%

Oh, sorry, accidentally left a (1) up there, sorry

First of all, analyzing P(2):
      probability of the ones digit being 2: \(\dfrac{1}{10}\)
      probability of the tens digit being 4: \(\dfrac{1}{9}\) (it can't be 0)
      probability of having both digits being 4 and 2 consecutively: \(P\left(2\right)=\dfrac{1}{10}\times\dfrac{1}{9}=\dfrac{1}{90}\)  (1)

Second, analyzing P(3)
   Case 1: 42 being on the tens and the ones:
      probability of the ones digit being 2:\(\dfrac{1}{10}\)
      probability of the tens digit being 4:\(\dfrac{1}{10}\)
      probability of having both digits being 4 and 2 consecutively:\(\dfrac{1}{10}\times\dfrac{1}{10}=\dfrac{1}{100}\)
   Case 2: 42 being on the hundreds and the tens: proven exactly P(2):\(\dfrac{1}{90}\)
   Therefore, \(P\left(3\right)=\dfrac{1}{100}\times\dfrac{1}{90}=\dfrac{1}{9000}\)

Finally, the absolute difference between P(2) and P(3) is \(\left|P\left(2\right)-P\left(3\right)\right|=\left|\dfrac{1}{90}-\dfrac{1}{9000}\right|=\left|\dfrac{99}{9000}\right|=\left|\dfrac{11}{1000}\right|=\dfrac{11}{1000}\)

Consider that the baby elephant's weight at the very beginning is 100%
1st year: 100%+10%.100%=100%+10%=110%
2nd year: 110%+10%.110%=110%+11%=121%
3rd year: 121%+10%.121%=121%+12,1%=133,1%
4th year: 133,1%+10%.133,1%=133,1%+13,31%=146,41%
5th year: 146,41%+10%.146,41%=146,41%+14,641%=161,051%
So after five years, the weight difference is 161,051%-100%=61,051%\(\approx\)61%
The answer is 61%

Method 1:
y=8.13=104
x=y+12=104+12=116
z=x+111+120=116+111+120=347
(x+1+[y-2-{z+3}+2+x]+y-1)=x+1+y-2-z-3+2+x+y-1=2x+2y-z-3=2.104+2.116-347-3=208+232-350=440-350=90

Method 2:
(x+1+[y-2-{z+3}+2+x]+y-1)=x+1+y-2-z-3+2+x+y-1=2x+2y-z-3=2(y+12)+2y-(y+12+111+120)-3=2y+24+2y-y-243-3=3y-222
Since y=8.13=104
=>3y-222=3.104-222=312-222=90

54=2.3³
Therefore, the smallest square to 2.3³ is 2².3⁴=324
So x=2.3=6

The sum of numbers in the first set is 12n
The sum of numbers in the second set is 6.3n=18n
The combined sum is 12n+18n=30n;
The number of the numbers in both sets are n+3n=4n;
So when the two sets are combined, the mean is 30n/4n=7.5

Sorry, I don't have a good way to prove it but...
x and y are negative intergers so xy will always be positive.
For xy to have the least value, x and y must have the greatest values.
Satisfying x-y=1, I believe that x=-1 and y=-2
Therefore xy=2
Please guys, if you have any more detailed way to prove this, please help Lê Quốc Trần Anh

Are you sure that your wrote the question correctly? It is a bit to inspecific and ambiguous to answer concretely.