MaxWong

OK, so the word "SMART" starts with S. We first want to know the approximate position of words that start with "S".

Arranging the letters in alphabetical order, we get "AMRST". That means words start with "S" is on a comparatively low position.

In reality, if we fix the first letter, there are 24 ways to arrange the other letters. So there are 24 words which start with "A", 24 which start with "M", so on and so on.

At the time we reach "S" finally, there would have been at least 72 words in front of the current word.

So the list of permutation looks like this:

AMRST

AMRTS

...

...

SAMRT

SAMTR

...

...

STRMA

...

...

So we need to know where does words which starts with "SM" lies.

When we fix the first two letters, there are 6 ways to arrange the other three.

So there are 72 + 6 words before the first word which starts with "SM".

And we also know that the first word which starts with "SM" is SMART, because "ART" is in alphabetical order.

Therefore, the position of the word is 72 + 6 + 1 = 79. 

Edited: Sorry, I did basic arithmetic wrongly at the very last step.

I'm not CPhill, but I will try to help.

We can consider the total number of ways that they can stand in a line (without any extra constraints) and consider the total number of ways that they stand in a line, but no girls are standing next to each other(This is the opposite of what the question asked for).

The difference of the two is the answer we want according to Inclusion-Exclusion principle.

Also, note that the problem says "Everyone is distinguishable", that means "order matters".

Total number of ways that they can stand in a line is the permutation of 9 people, which is \(P^9_9 = 9! = 362880\).

Next, we consider the number of ways that they stand in a line, but no girls are standing next to each other.

We will first consider the boys first. We have \(P^4_4 = 4! = 24\) ways to arrange the boys into a line.

But we haven't considered the girls yet. Now, we can't put two girls between two boys, and we can't put two girls at the front, or at the end. We will consider all the possible spots for the girls to fit inside the queue.

In the following diagram, B means boy, and X means an unoccupied space.

After we have considered the boy, the queue looks like this:

X B X B X B X B X

No two girls can fit into the same X spot, so each girl must go to an unoccupied space.

The queue will look like this. (B means boy and G means girl)

G B G B G B G B G

But we haven't considered the order of the girls. We can arrange their order in \(P^5_5 = 5! = 120\) ways.

Multiplying the number of ways of arranging boys and the number of ways of arranging girls:

The number of ways of arranging all of them in a line, with no girls standing next to each other is \(24\times 120 = 2880\) ways.

Subtract that from the total number of ways, our required answer is \(P^9_9 - P^4_4 \times P^5_5 = 362880 - 2880 = \boxed{360000}\)

There are much more ways to arrange the order of the boys and girls than you think. 
I hope this is helpful and easy-to-understand.

We can consider the prime factorisation of the product of 10, 20, 30, ... 90.

\(10 \times 20 \times 30 \times \cdots \times 90 = 10^9 \times 2\times 3\times 2^2 \times 5 \times 2 \times 3 \times 7\times 2^3\times 3^2=2^{16}\times 3^4\times 5^{10}\times 7\)

We can't leave that 7 there, because everything in the product must be a perfect square. By a greedy approach, we remove 70 first and see what happens.

\(\dfrac{10 \times 20 \times 30 \times \cdots \times 90 }{70} = 2^{15} \times 3^4 \times 5^9 \)

But we can't leave 2¹⁵ and 5⁹ there. Now without affecting the exponent of 3, we remove 10.

\(\dfrac{10 \times 20 \times 30 \times \cdots \times 90 }{70\times } = 2^{14} \times 3^4 \times 5^8 = \left(2^7\times 3^2\times 5^4\right)^2\)

And there we go, a perfect square, and it only took us 2 removals.

P.S. Notice that 1 removal doesn't work because you have to take away that 7, so you must first remove 70, but that affects the exponent of 2 and 5 so it doesn't work and we have to take an extra move.

By performing column operations and factoring out common factors whenever there is,

\(\begin{vmatrix} p & 2q & 5r + 4p \\ s & 2t & 5u + 4s \\ v & 2w & 5x + 4v \end{vmatrix} = 2 \begin{vmatrix} p & q & 5r \\ s & t & 5u \\ v & w & 5x \end{vmatrix} = 10 \begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix} = 10(-3) = -30\)

Multiplying both the numerator and the denominator by \((x^2 - 1)\), we have

\(\dfrac{\dfrac{x}{x - 1} - \dfrac{x}{x + 1}}{\dfrac{x}{x^2 - 1}} = \dfrac{x(x + 1) - x(x - 1)}{x} = \dfrac{x((x + 1) - (x - 1))}{x} = \dfrac{2x}{x} = 2\)

This holds for \(x \neq 0 \)

Notice that \(\begin{bmatrix} a&d&g\\b&e&h\\c&f&i \end{bmatrix} = \begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix}^{\mathbf{T}}\)

So their determinants are the same.

\(\begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix} = \begin{vmatrix} a&b&c\\d&e&f\\g&h&i\end{vmatrix} = -24\)

We can use the formula \(A = \dfrac12 ab \sin C\) directly.

\(A = \dfrac12 \cdot 12^2 \cdot \sin 120^\circ = 72 \cdot \dfrac{\sqrt 3}2 = 36\sqrt 3\)

Let's start by listing composite numbers:

4, 6, 8, 9, 10, 12, 14, 15, 16, ...

4 + 6 + 8 cannot be prime, but 4 + 6 + 9 can.

Also, 19 = 4 + 6 + 9, so the sum is indeed a prime, and the numbers are composite.

The required answer is 19.

That means 5228X is divisible by 2 and 3.

So X must be an even number, and \(5 + 2 + 2 + 8 + X\) is divisible by 3.

But that means \(2 + 2 + 2 + 2 + X \) is divisible by 3. Taking mod 3 again on both sides,

\(2 + X\) is divisible by 3, and don't forget that X is an even digit.

\(X = 4\) is the only possible solution.

OA, OB, OC divide the circle into three congruent parts \(\Longleftrightarrow\) \(\angle AOB = \angle BOC = \angle COA = 120^\circ\)

D is a point on arc AB such that arc AD has half the measure of arc DB \(\Longleftrightarrow\) \(m\stackrel{\mbox{$\large\frown$}}{AD} = 40^\circ\), \(m\stackrel{\mbox{$\large\frown$}}{DB} = 80^\circ\)

\(\text{reflex }\angle COD = m\stackrel{\mbox{$\large\frown$}}{DB} + m\stackrel{\mbox{$\large\frown$}}{BC} = 200^\circ\)

\(m\angle COD = 360^\circ - 200^\circ = 160^\circ\)

Because OC = OD,

\(m\angle OCD = m\angle ODC\)

So \(160^\circ + 2 \left(m\angle OCD\right) = 180^\circ\)

I believe you can continue from here. It's just one step away from the answer.