The money spent on buying a 40% down jacket is:

$80.(100% - 40%) = $48

The money spent on buying a 55% down shirt originally is:

$40.(100% - 55%) = $18

The sum of original price of jacket and shirt originally is:

$80 + $40 = $120

The sum of jacket and shirt originally after down price is:

$48 + $18 = $66

The $ are saved after down price is:

$120 - $66 = $54

The total amount saved was percent of the total of the original prices is:

$54 : $120.100 = 45%

So choose A

The number of the burger isn't filler is:

120 - 30 = 90 [grams]

So the percent of the burger is not filler is:

90: 120.100 = 75%

So choose D

1729 = 7.13.19

So the sum of all prime factor of 1729 is:

7 + 13 + 19 = 39

\(A=1\cdot3\cdot5\cdot...\cdot99\)

\(\Rightarrow2\cdot4\cdot6\cdot...\cdot100\cdot A=1\cdot2\cdot3\cdot4\cdot...\cdot100\)

\(\Rightarrow A=\dfrac{1\cdot2\cdot3\cdot...\cdot100}{2\cdot4\cdot6\cdot...\cdot100}=\dfrac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot50\right)\cdot\left(51\cdot52\cdot...\cdot100\right)}{\left(2\cdot1\right)\left(2\cdot2\right)\left(2\cdot3\right)...\left(2\cdot50\right)}\)\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot50\right)\left(51\cdot52\cdot...\cdot100\right)}{2^{50}\cdot\left(1\cdot2\cdot3\cdot...\cdot50\right)}=\dfrac{51\cdot52\cdot...\cdot100}{2^{50}}=\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}=B\)So A = B

\(\dfrac{1}{2}\cdot\dfrac{1}{6}\cdot...\cdot\dfrac{1}{110}=\dfrac{1}{1\cdot2}\cdot\dfrac{1}{2\cdot3}\cdot...\cdot\dfrac{1}{10\cdot11}\)

\(=\dfrac{1}{1\cdot2\cdot2\cdot3\cdot...\cdot10\cdot11}=\dfrac{1}{1\cdot2^2\cdot3^2\cdot...\cdot10^2\cdot11}\)

\(=\dfrac{1}{4\cdot9\cdot16\cdot...\cdot100\cdot11}=\dfrac{1}{13168189440000\cdot11}=\dfrac{1}{144850083840000}\):V, I think the question is:

\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{110}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}=1-\dfrac{1}{11}=\dfrac{10}{11}\)You should see again and write in English

1 + 2 + 3 + 4 + ... + 198*23

\(=\left(1+2+3+4+...+197\right)+198\cdot23\)

\(=\left(197+1\right)\dfrac{197-1+1}{2}+4554\)

\(=19503+4554=24057\)

Yeah, I just find the slution of this question. 

The sign is divisible by 23:
Get the last number multiplied by 7 and add the remaining numbers, if the sum is divisible by 23, then the number is divisible by 23

So we apply for this question.

10 000 000 099 have: A = 1 000 000 009 + 9*7 = 1 000 000 072

B = 100 000 007 + 2*7 = 100 000 021

C = 10 000 002 + 1*7 = 10 000 009

D = 1 000 000 + 9*7 = 1 000 063

E = 100 006 + 3*7 = 100 027

F = 10 002 + 7*7 = 10 051

G = 1005 + 1*7 = 1012

H = 101 + 2*7 = 115

I = 11 + 5*7 = 46 divisible by 23

So 10 000 000 099 = 23.434782613

Now, I must go to school :[

10000000099 = 23.434782613

What is the expressions to calculate of question 2 ? 

a,

We have: 19x + 19y ⋮ 19 and A ⋮ 19

\(\Rightarrow19x+19y-3A=19x+19y-3\left(5x+y\right)=19x+19y-15x-3y=4x+16y⋮19\)

\(\Rightarrow4x+16y-19y⋮19\Leftrightarrow4x-3y⋮19\)

So 4x - 3y ⋮ 19

b,

- 4x + 3y ⋮ 13

=> 4x + 3y + 13y ⋮ 13

=> 4x + 16y ⋮ 13

=> \(4\left(x+4y\right)⋮13\)

=> x+4y ⋮ 19 *** Because 13 is a prime number

- 4x + 3y ⋮ 13 

=> 4x + 3y + 26x ⋮ 13

=> 30x + 3y ⋮ 13

=> \(3\left(10x+y\right)⋮13\)

=> 10x + y ⋮ 13

=> \(\left(10x+y\right)-\left(4x+3y\right)\text{⋮}13\)

=> 6x - 2y ⋮ 13

==> \(\left(6x-2y\right)+\left(x+4y\right)⋮13\)

=> 7x + 2y ⋮ 13 

So D ⋮ 13

We have:

9^2k = ...1

9^2k+1 = ...9

=> A = 1 + 9 + 9² + .... + 9⁵⁰

= 1 + 9 + ... 1 + .... + ...1

We easy see A will have sum A have 51 terms.

and 1 is first and last, so sum A have 26 last-digit terms 1 and 25 last-digit terms 9

So last-digit is: 26*1 + 25*9 = 251 = ...1

So last digits of 1 + 9 + ... + 9⁵⁰ is 1

This number is: 27 * 130% = 35,1

So answer is 35,1

a+b+c = 0

\(\Rightarrow\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\)

\(=\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)\)

\(=\left(b-c\right)\left(0-a\right)+\left(c-a\right)\left(0-b\right)+\left(a-b\right)\left(0-c\right)\)

\(=-a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)\)

\(=-ab+ac-bc+ab-ac+bc\)

\(=\left(-ab+ab\right)+\left(-bc+bc\right)+\left(-ac+ac\right)=0+0+0=0\)

\(\Rightarrow Q=\dfrac{\left(a^2+b^2+c^2\right)}{0}=????\)

Unsatisfactory, so Q is void

Because \(GCD\left(a,b\right)=7\Rightarrow\left\{{}\begin{matrix}a=7k\\b=7h\end{matrix}\right.\)with \(k,h\in N\text{*}\)

\(\Rightarrow a\cdot b=GCD\left(a,b\right)\cdot LCM\left(a,b\right)=7h\cdot7k=49hk=7\cdot70=490\)\(\Rightarrow kh=\dfrac{490}{49}=10\)

We'll have a table:

Inferred:

So \(\left(a,b\right)=\left(7,70\right);\left(70,7\right);\left(14,35\right);\left(35,14\right)\)

\(2\left(x+1\right)+\dfrac{2}{3}\cdot\left(x-1\right)=2^3\cdot4\)

\(\Rightarrow2\left(x+1\right)+\dfrac{2}{3}\left(x+1-2\right)=2^3\cdot4\)

\(\Rightarrow2\left(x+1\right)+\dfrac{2}{3}\cdot\left(x+1\right)-\dfrac{2}{3}\cdot2=2^3\cdot4\)

\(\Rightarrow2\left(x+1\right)+\dfrac{2}{3}\left(x+1\right)-\dfrac{4}{3}=32\)

\(\Rightarrow\left(x+1\right)\left(2+\dfrac{2}{3}\right)=32+\dfrac{4}{3}\)

\(\Rightarrow\left(x+1\right)\cdot\dfrac{8}{3}=\dfrac{100}{3}\)

\(\Rightarrow x+1=\dfrac{100}{3}\div\dfrac{8}{3}=\dfrac{25}{2}\)

\(\Rightarrow x=\dfrac{23}{2}\)

So the value of x is \(\dfrac{23}{2}\)

In a hour, hour needle run slowest. We counted how long after the cycle, minutes and seconds meet again.

We have:

For an hour, hour needle run 1/12 ring

For an hour, minute neddle run 1 ring

Each hour, minute needle faster at hour needle is:

1 - 1/12 = 11/12 [ring]

To meet again, minute needle, hour needle to run more than 1 ring. So the time to meet they again is:

1 : 11/12 = 12/11 [hour]

Same, we have:

For an hour, hour needle run 1/12 ring

For an hour, second needle run 60 ring

Each hour, second needle faster at hour needle is:

60 - 1/12 = 719/12 [ring]

To meet again, second needle, hour needle to run more than 1 ring. So the time to meet they again is:

1 : 719/12 = 12/719 [hour]

So, after: k*12/11 hour, hour needle and minute needle meet. 

m*12/719 hour, hour needle and second needle meet.

So, to three needles on a stright light after:

k*12/11 = m*12/719

=> k*719 = m*11

=> After law: 719*12/719 = 12 hour , three needles on a stright

\(\forall x\in\)N*, we have:\(n^4+\dfrac{1}{4}=\left(n^4+2\cdot n^2\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)-n^2=\left(n^2+\dfrac{1}{2}\right)^2-n^2=\left(n^2+\dfrac{1}{2}-n\right)\left(n^2+\dfrac{1}{2}+n\right)\)\(\Rightarrow A=\dfrac{\left(1^2+\dfrac{1}{2}+1\right)\left(1^2+\dfrac{1}{2}-1\right)...\left(29^2+\dfrac{1}{2}+29\right)\left(29^2+\dfrac{1}{2}-29\right)}{\left(2^2+\dfrac{1}{2}-2\right)\left(2^2+\dfrac{1}{2}+2\right)...\left(30^2+\dfrac{1}{2}-30\right)\left(30^2+\dfrac{1}{2}-30\right)}\)

\(=\dfrac{\dfrac{1}{2}\left(1\cdot2+\dfrac{1}{2}\right)\left(2\cdot3+\dfrac{1}{2}\right)....\left(29\cdot30+\dfrac{1}{2}\right)}{\left(2\cdot3+\dfrac{1}{2}\right)\left(1\cdot2+\dfrac{1}{2}\right)\left(3\cdot4+\dfrac{1}{2}\right)...\left(30\cdot31+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}}{30\cdot31+\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}}{\dfrac{1861}{2}}=\dfrac{1}{1861}\)

Very tired :[

Oh, you miss a condition is \(P\left(32\right)=-247\)

Then you can refer at: https://math.stackexchange.com/questions/2437594/px-an-integer-polynomial-with-p21-17-p32-247-p37-33-and-pn-n/2437614#2437614?newreg=5e33bcea067a40a3ae8a1a3d3d7ec95f

Good!

\(PutA=\dfrac{B=2+2^2+...+2^{100}}{C=2+2^2+...+2^{50}}\)

\(\Rightarrow A=\dfrac{2B-B}{2C-C}=\dfrac{\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)}{\left(2^2+2^3+...+2^{51}\right)-\left(2+2^2+...+2^{50}\right)}\)

\(\Rightarrow A=\dfrac{2^{101}-2}{2^{51}-2}=\dfrac{2^{100}-1}{2^{50}-1}\)

We have:

\(\dfrac{1}{2}=\dfrac{1}{2}\)

\(\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}\cdot2=\dfrac{1}{2}\)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{1}{8}\cdot4=\dfrac{1}{2}\)

\(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}>\dfrac{1}{16}\cdot8=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{16}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=2\)

But \(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+...+\dfrac{1}{63}>0\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)

\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{x}{3}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x}{15}-\dfrac{3}{15}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{5x-3}{15}\)

\(\Rightarrow60=y\left(5x-3\right)\)

=> 5x-3 is divisor of 60

and we see: 5x-3 \(\equiv2\left(mod5\right)\)

\(\Rightarrow5x-3\in\left\{2;12\right\}\)

\(\Rightarrow\left[{}\begin{matrix}5x-3=2\Rightarrow x=1;y=30\\5x-3=12\Rightarrow x=3;y=5\end{matrix}\right.\)

So \(\left(x,y\right)=\left(1;30\right),\left(3,5\right)\)