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Sorry, I gave above the sum of numbers !!

Here are the sums of digits:

Total Evens = 400
Total Odds   = 500

Sum of {1, 3, 5, 7, .................95, 97, 99} =2,500

Yes, there is a solution! You just missed it.

a=1; b=2#(1991*a + 1);if(ceil(b)==floor(b) and isprime(a), goto3, goto4);printc, a; a++;if(a<1000000, goto1, 0)

OUTPUT =p =1993. [1991 * 1993 + 1]^1/2 =1992

Obviously the first is true, the second is also true.
The third is not because the difference between the a and c is not stated. ac is not neccesarily greater than bd because c could be negative and a could be positive but b and d could both be negative giving negative > positive which is obviously not true
The third is not true  because if the absolute value of b > absolute value a then b^2 > a^2 and for c and d as well.
a^3 + c^3  is true because absolute value doesnt matter here
To summarize A, B and F are true and the others are not

From this graph, you can see that the triangle outlined by the red, green and black is the successful region. You know this because 10>2(0) and 10,0 is a point in this region. The area of this triangle is 11x5 * 1/2 = 27.5

If you divide the first expression on the LHS(left hand side), you get (x^2+2x+4)(3y^2-6y+7) = 12
Putting the two expressions on the LHS in square form(by completing the square) you get that((x+1)^2+3)*(3(y-1)^2+4) = 12
Using this we know that the minimum for the two expressions are 3 and 4 respectively so that means that there is only 1 ordered pair satisfying this equation. Namely, (-1,1). 

The constant that works is k = 6.

If you consider each number a switch with an in or out, then there are 2^8 = 256 subsets of the set because each number is either in or out. Now we use complementary counting. If we want to count the number of subsets with 1,2 or 5, we can also count the number of subsets without any of those and subtract that count from the total. This count is 2^5 = 32 because 1,2, and 5 have to be out so there are only 5 switches. 256 - 32 = 224

Ignoring the restriction, we can position the letters in 7!/7  = 720 ways because you have to divide by 7 to get rid of the rotations. (Think of it this way, if you order the letters you can move every letter clockwise once, twice, ... 6 times for a total of 7 times(including the original case.))
To deal with the restriction, we do coplementary counting(counting what we do not want.) If you think of O and N as one letter, meaning they are together, then there are only 6 "letters" so 120 ways however, we have to multiply by two because the O can be to the right of N or the left. This gives 240 cases we do not want. Subtracting 240 from 720 is 480. 

You can subtract the second equation from the first to get (4x-6x) + (5y-5y) + (12z-(-2z)) = 54-(-4)
This turns into -2x +14z = 58. Dividing both sides by 2 gets you 7z-x = 29. Moving the x to the other side gets you x = 7z-29

Substituting into the second equation gives 4(7z-29)+5y+12z = 54
Expanding gives 28z-116 + 5y +12z. Combining like terms gives 40z +5y = 170. Then dividing by 5 gives 8z +y = 34. If you tried to substitute into the first equation would give you 0 = 0 which is obviously useless. However adding x-7z and 8z+y gives you x+y+(8z-7z) which is just x+y+z. We know x-7z+ 8z+y is -29 +34  = 5.
 

If the sum of the interior angles of a polygon with n sides is x degrees, then the sum of the interior angles of a polygon with n+1 sides is x+180 degrees and a polygon with n+2 sides is x+360 degrees. A quick proof for this is that you can draw n-2 triangles in an n-gon such that each triangle's three vertices are vertices of the polgon. In a n+1-gon there is one more triangle contributing 180 more degrees. So 1800+360 is 2160.

2²⁰⁰ < n¹⁰⁰ < (130n)⁵⁰

Take the 50^th root:     2⁴ < n² < 130n     --->     16 < n² < 130n

For  16 < n²  to be true, n >= 5

For  n² < 130n  to be true, n<= 129

So, I think that it will be true for all positive integers in the set:  { 5, 6, 7, ..., 128, 129 }

1325 m  /  530 m/hr   = .......... hr  

1+2+3+x   has to add to a factror of 64

factors   1 2   4  8  16  32 64  

for one         1 +2+3+ 2 = 8

for another   1 + 2 + 3 + 10   = 16

another        1 + 2 + 3 + 26   = 32

 and             1+ 2 + 3 + 58    = 64

In parallelogram WXYZ, WX=6cm and XY=8cm. Angle W=45 degrees. What is the area of the parallelogram?

Answer:       A = [sqrt(6²/2)] * 8  = 33.9411255 cm²  

15b /24r   *  9 r/15y  * 1000y   = 375 b

[540 / 5] / [1440 / 10] =108 / 144 = 3 / 4

You can find the ninimum 'x'   by   using  - b/ 2a   

    then use this value in the equation to find the corresponding 'y' value

It is correct!!! 

TheXSquaredFactor

I'm sorry, but...

YOU GOT IT RIGHT! Thank you so much! I was having a lot of trouble since this morning with this problem, too. You did not waste your time, congratulations.

Suppose GetThere Airlines increases their ticket price to 200+10n = 10(20+n) dollars. Then the number of tickets they sell is 40,000-1000n = 800(40-n) . Therefore, their total revenue is

10(20+n)\cdot 800(40-n) = 8000(20+n)(40-n) = 8000(800+40n-n^2).

This is maximized when n=-(\frac{40}{2\cdot(-1)})=20 . Therefore, they should charge 200+20\cdot 10 = 400 dollars per ticket.

I don't see a "zero" or a "2" in tens digit but:1, 3, 5, 7, 9

Assuming we are dealing with standard playing cards and we are ignoring jokers, then 26 out of the 52 cards are red. 12 cards (Jacks,  Queens, and Kings)  are face cards. 6 cards are both red and a face card, so we must subtract these to avoid double-counting these.

\(P(\text{red} \cup \text{face})=P(\text{red})+P(\text{face})-P(\text{red}\cap\text{face})\\ P(\text{red} \cup \text{face})=\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\\ P(\text{red} \cup \text{face})=\frac{32}{52}=\frac{8}{13} \)

The area is 1/2*6*8 = 24.