hairyberry

There's a really good inequality for this, Cauchy-Schwarz inequality.

\((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\) , where a and b are a sequence of real numbers.

Applying this, we want to get xy on the right side so we do:

\((x^2+3y^2)(1+\frac{1}{3})=(x+\sqrt{3}(\frac{1}{\sqrt{3}})y)^2\)

\(18*\frac{4}{3}\ge(x+y)^2\)

\(x+y\le\sqrt{24}\).

The maximum value is \(\sqrt{24}\).

(Yes the equality condition can be satisfied).

\(\frac{1}{\frac{1(1+1)}{2}}+\frac{1}{\frac{2(2+1)}{2}}+\dots + \frac{1}{\frac{2018(2018+1)}{2}}\)

\(\frac{2}{1(2)}+\frac{2}{2(3)}+ \dots+\frac{2}{2018(2019)}\)

\(2(\frac{1}{1(2)}+ \frac{1}{2(3)} + \dots + \frac{1}{2018(2019)})\)
\(\text{Apply } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)

\(2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} +\dots -\frac{1}{2019})\)

\(2(1-\frac{1}{2019})\)

\(\frac{4036}{2019}\)

We want all numbers in this form:

_ _ 3

_ _ 4

_ _ 5

An integer can't be part of both groups, since an integer has only one ending number.

For all of these, the first two can cycle from 10 to 99 which is 90 numbers

Sum of numbers ending in 3 is (sum of numbers 10 - 99) * 10 + 90*3 = 49320

Sum of numbers ending in 4 is (sum of numbers 10 - 99) * 10 + 90*4 = 49410
Sum of numbers ending in 5 is (sum of numbers 10 - 99) * 10 + 90*5 = 49500

49320 + 49410 + 49500 = 148230.

Since it's not specified whether Magnus can have stickers left over:

This is the solution if you don't have to hand out every sticker.

a + b <= 12.

Create a dummy variable, k, for all leftovers

a + b + k = 12.

This is great because we see k has to be at least 0, like all the others.

Apply stars and bars formula:

\({12+3-1\choose 3-1}=91\).

There are 91 ways.

This is the solution id you don't have to hand out every sticker.

Give one friend any number of stickers 0 -12. 

The other gets all the rest.

There are 13 ways.

\(0>5x^2+24x\)

\(x(5x+24)<0\)

Take interval between two solutions, because quadratic points up, and you want the part below the x axis.
Therefore the interval is \((-\frac{24}{5}, 0)\).

Probability of purple orangle purple:

\(\frac{2}{5}*\frac{3}{4}*\frac{1}{3}=\frac{1}{10}\)

Probability of orange purple orange:

\(\frac{3}{5}*\frac{1}{2}*\frac{2}{3}=\frac{1}{5}\)

\(\frac{1}{5}+\frac{1}{10}=\frac{3}{10}\).

The probability that the balls alternate is 3/10.

There are 6 students in the science club, and 3 students are a member of both clubs.

The number to students that are members of only the math club is 3-3 = 0

The number of students thare are members of only the science club is 6 - 3 = 3.

There are a total of 3 students who are in the math or science club but not both.

Thanks booboo44, I realized it is possible , thanks to your work.

But I'm getting a different answer, is there anything wrong with my work?

Set the radius of the outer circle as y.

Set the radius of the inner circle as x.

We want:

\(y^2\pi-x^2\pi=\pi(y^2-x^2)\).

By the pythagorean theorem, we know

\(y^2-x^2=5^2=25\).

Therefore the area is \(25\pi\)

Note: I will use [] to denote area.

[ABCD] = AD * DC

[ADU] = [BCT] = AD*DU/2 = (AD*DC/3)/2=AD*DC/6

[ADU]+[BCT]=AD*DC/3

[ABTU] = [ABCD]-([ADU]+[BCT]) = 2/3*(AD*DC) = 120.

2)

Is this question complete? I think it might be non-unique.