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Let A=111111 and B=142857. Find a positive integer N with six or fewer digits such that N is the multiplicative inverse of AB modulo 1,000,000.

[111,111 x 142,857] x M mod 1,000,000 =1, solve for M

LCM[111,111, 142,857] = 999,999

GCD[111,111, 142,857] =15,873

M =999,999 / 15,873 =63 - is the multiplicative inverse of AB mod 1,000,000.

Alan seems to be using a pack where each suit has its own colour !

you can use the second equation of motion s= ut+1/2at^2

here consider that the ball was at rest before kicking . and came to rest after comming to ground

then v=u=0m/s

and a=10m/s^2  (9.8) BECAUSE IT IS UNDER THE INFLUENCE OF GRAVITY.

this is only true if the ball is kicked vertically upward .

but if the ball was in a projectile motion i.e. in a 2D motion then apply the formula

H =v02sin2 θ/2g

H = maximum height (m)

v0 = initial velocity (m/s)

g = acceleration due to gravity (9.80 m/s2)

θ = angle of the initial velocity from the horizontal plane (radians or degrees).

First calculate the probability, q, that they are all the same colour!

Probability that second card is same colour as first:  q1 = 12/51 (12 cards of same colour to choose from out of 51 cards left).

Probability that third card is also same colour: q2 = 11/50

Hence q = q1*q2 or q = 12*11/(51*50)

Probability, p, that they are not all same colour: p = 1 - q

.

A=926,510,094,425,920

B=31,381,059,608

A/B =~ 29,524.50 - This is how much A is bigger than B

Assume it starts and ends on the ground.  Assume air resistance is negligible.  Then half of the time it will be going up and half coming down.  When it reaches its peak height it will have a vertical component of velocity of v = 0 m/s.  So we can find its initial vertical component of velocity, u, from:

v = u + at   where v = 0 m/s, a = -9.8 m/s² (gravitational acceleration), t = 9/2 s

Knowing u we can now use: 

v² = u² + 2as to find the peak height, s (metres).

"1. Suppose $f$ and $g$ are polynomials, and that $h(x)=f(g(x))+g(x)$. Find the degree of $g(x)$ given that the degree of $h(x)$ is $6$ and the degree of $f(x)$ is $2$.

2.Consider the polynomials \[f(x)=1-12x+3x^2-4x^3+5x^4\] and \[g(x)=3-2x-6x^3+9x^4.\] Find $c$ such that the polynomial $f(x)+cg(x)$ has degree 3."

First calculate the probability that the larger number is n (1 <= n <= 6). This is exactly (2n-1)/36. Why? There are three possibilities:  

The first die is exactly n, the second die is smaller -- So second die can be 1 ... n-1. 
The second die is exactly n, the first die is smaller -- So the first die can be 1 ... n-1. 
Both die are the same. This happens in only one way. 

So total number of ways in which the larger number is n = 2n-1. Hence the probability that the larger number is n is (2n - 1)/36. 

Consequently, the expected value of the larger of the two numbers is 

Sum [n = 1 to 6] (2n -1)/36 * n 
= 1/36 (Sum [n = 1 to 6] 2n^2 - Sum [n = 1 to 6] n) 

You can compute this either directly or by using formulas for sequences. The answer is 161/36.

-MathCuber

I think it's B. 3^32 is 30 times larger than 3^22, the minus 1's don't really matter, and the 1/2 makes not much of a difference either, considering that the other number is 30 times larger.

-MathCuber

I think this is impossible for people like us. We would need to be a physicist to figure this out... 

I have many questions, like how fast the ball goes, when it is at its peak point. Also, we would need to know the force of gravity, and the angle in which it is kicked. This is science...

That was right thanks

Hint: both have to be even.

First, we do 5 choose 2, which gives us 10. Then, we find the total number of possibilities, which is 10 choose 2 which equals 45. Thus the answer is 10/45=2/9

I think this is the golden ratio (sqrt(5) + 1) / 2 represented as "continued fraction" as follows:

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/a(n)))))))))))))))))))))))=Sqrt(5).

The constant times the constant = the new constant

so the original constants have to be 4

[Am I missing something? ]

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The 77th term is =daecb, and the 100th term =eacdb

Simplify the following:
3 (x - 4) + 2 (x^2 - x + 7) - 5 (x - 1)

3 (x - 4) = 3 x - 12:
3 x - 12 + 2 (x^2 - x + 7) - 5 (x - 1)

2 (x^2 - x + 7) = 2 x^2 - 2 x + 14:
-12 + 3 x + 2 x^2 - 2 x + 14 - 5 (x - 1)

-5 (x - 1) = 5 - 5 x:
2 x^2 + 3 x - 2 x + 5 - 5 x - 12 + 14

Grouping like terms, 2 x^2 + 3 x - 2 x - 5 x - 12 + 5 + 14 = 2 x^2 + (3 x - 2 x - 5 x) + (-12 + 14 + 5):
2 x^2 + (3 x - 2 x - 5 x) + (-12 + 14 + 5)

3 x - 2 x - 5 x = -4 x:
2 x^2 + -4 x + (-12 + 14 + 5)

-12 + 14 + 5 = 7:

 2x^2 - 4x + 7 - The "Constant Coefficient" is 7.

Since I'm too lazy to slug it through counting the zeros manually, I had the computer do the slugging and got the following results!!:

1- Number of zeros between 1 and 50,000 inclusive =18,893 zeros.

2- Number of zeros between 50,000 and 100,000 inclusive =20,005 zeros.

So, the ratio is= 18,893 : 20,005. Or, as a percentage =18,893 / 20,005 =94.44%

Perfect, except if there are 5 females, that is also an option.

there are six of these, meaning the answer is 186!

Yes, it was answered here: https://web2.0calc.com/questions/oof-stuck

First, put the house key and car key next to each other on the keychain. It doesn't matter where they are put, because the keychain can be rotated and reflected to move them to any other pair of adjacent locations. The remaining three keys can be put on the keychain in \(3!=\boxed6\) ways.

I will give this a go!.

1 - First scenario: You want a committee of 5 people 4 of which must be female and 1 male:

6C4 x 4C1 =15 x 4 = 60 ways of choosing in this scenario.

2- Second scenario: You want a committee of 5 people 3 of which must be female and 2 male:

6C3 x 4C2 =20 x  6 =120 ways of choosing in this scenario.

So, the total of the two scenarios =60 + 120 = 180 ways of choosing a committee of 5 people.

Note: Somebody should check this.