The ratio of the 1st gear and the 4th gear is : 3:1

So the last one will rotate: 1 * 3 = 3 ( revolutions )

Choose A

A is join into the game. UNFORTUNALY, the game is really dangerous. The game is Russian roulete, but very complex.

3 bullets are put into the 6-revolving chamber. Before put the bullets into the chamber, A is blindfold to let A don't know which chamber has the bullet. After put the bullets into the chamber, they take a shot. Luckly for A, A is survive.

After the first round, since A survives, they give A a chance. A has 3 choice:

-Choice 1: tell them spin the chamber randomly before the second shot;

-Choice 2: tell them to arange the bullets in the different position then spin the chamber randomly before the second shot ( but A has to be blindfold );

-Choice 3: tell them no spin the chamber and hope for the best.

Which choice A must choose to has the greatest chance of surviving?

Now just to find the solution! Hope you get any luck.

Hard to translate? I will translate for you.

2 , 4 , 8 FOLLOWS THE RULE . 3 , 6 , 9 FOLLOWS THE RULE . 10 , 5 , 7 FOLLOWS THE RULE . 19 , 2 , 1 FOLLOWS THE RULE. BUT 9 , 5 , 3 DON'T FOLLOWS THE RULE . WHAT IS THE RULE?

See if you can translate and solve this.

..--- / --..-- / ....- / --..-- / ---.. / ..-. --- .-.. .-.. --- .-- ... / - .... . / .-. ..- .-.. . / .-.-.- / ...-- / --..-- / -.... / --..-- / ----. / ..-. --- .-.. .-.. --- .-- ... / - .... . / .-. ..- .-.. . / .-.-.- / .---- ----- / --..-- / ..... / --..-- / --... / ..-. --- .-.. .-.. --- .-- ... / - .... . / .-. ..- .-.. . / .-.-.- / .---- ----. / --..-- / ..--- / --..-- / .---- / ..-. --- .-.. .-.. --- .-- ... / - .... . / .-. ..- .-.. . .-.-.- / -... ..- - / ----. / --..-- / ..... / --..-- / ...-- / -.. --- -. .----. - / ..-. --- .-.. .-.. --- .-- ... / - .... . / .-. ..- .-.. . / .-.-.- / .-- .... .- - / .. ... / - .... . / .-. ..- .-.. . ..--..

A rectangle has its length is 8cm longer than its width. If we double the width, its new length is still 8cm longer its new width. What is the smallest possible area of that rectangle ?

Since no box are labeled correct; box 3 must have baseballs only.So easy that box 2 contains only tennis balls.

Let  \(a@b@c=a\left(1+m\right)+a\left(2+m\right)+a\left(3+m\right)+...+a\left(10+m\right)+b\left(11+m\right)+b\left(12+m\right)+b\left(13+m\right)+...+b\left(20+m\right)+c\left(21+m\right)+c\left(22+m\right)+c\left(23+m\right)+...+c\left(30+m\right)\) ( m is some value ). If \(2@3@4=2295;9@4@2=2575;3@7@8=4420;\) will there be any solution for  \(a@b@c=1401?\) ( every value is a natural number ).

It's depends

POTT9R

- H6RRY

=H2RRY

So easily that Y = 8 and P = 1.

 1O889R

- H6RRY

=H2RRY

4th colunm : R + R = 8; but 5th colunm have R + R = 9; so Y + Y = 1R => R = 4 and Y = 7.

1O8894

- H6447

=H2447

Hmmm... maybe there is more than 1 solution... so... anyway... done I guess?

See if you can find the rules and come up with the solution.

e) 4 : 10 :: 23 : 276 :: 7 : 28 :: ? : 5050 :: 34 : ?

f) 0 : 1 :: 2 : 121 :: 5 : 15101051 :: 8 : ? :: ? : 172135352171

g) 0 : 1 :: 1 : 1 :: 2 : ? ( not 2 ) :: 3 : ? ( not 6 ) :: 4 : ? ( not 24 )

   - Bonus: What if we keep going?

Nope

Maybe -8

See if you can figure out the rule and come up with the solution.

Ex: Decagon : 10 :: Heptadecagon : 17 :: Octagon : ? Ans: 8.
a) December : 12 :: November : 11 :: June : ?

b) 6724 : 82 :: ? : 123 :: 11664 : ?

c) Twelve : 6 :: Six : 3 :: Ten : ? ( not 3 ) :: Twenty : ? ( not 10 )

533-412=121

336+312+3=651

Harry has 30 candies; 8 candies has blueberry flavor; the rest has cranberry flavor. His mother, Larry also take care of Jenny, Tom and Sala. One day; some Harry's candies are disappear. Harry told mom: " Mom, someone is stolen my candies!" Harry also knows that other child did that.

Larry says: "You must tell who stolen how much candies."

Jenny says: " Sala stole 3 blueberry candies and 3 cranberry candies, mom!"
Tom says: " It was Jenny; she took 5 cranberry candies!"

Sala says: " Um.. I don't know. Maybe Jenny took 4 blueberry candies."

If 1 statement is true:

i) Who took the candy?

ii) What is the chance that Harry pick the blueberry candies without looking, if the begining has the number of the candies is the same after someone stole the candies? Express your answer as the fraction in the simplest form.

The answer is A.

That group will have : 12 + 8 = 20 ( students )

The probability that the 1st student is girl is: \(8:20=\dfrac{2}{5}\)

The probability that the 2nd student is girl is : \(\left(8-1\right):\left(20-1\right)=\dfrac{7}{19}\)

The probability that both student are girl is : \(\dfrac{2}{5}\cdot\dfrac{7}{19}=\dfrac{14}{95}\)

We can pour water back and forth between these containers to come up with exactly 1 gallon, because we can take water of from the container to transform to outside.

This is the way to get it:

Step 1: Pour water in to the 9 gallon container until its full.

Step 2 : Take water from the 9 gallon container to the 7 gallon container until its full.

Now, the 9 gallon container has 2 gallon of water.

Step 3 : Take all the water from the 7 gallon container on to the ground.

Step 4 : Take all the water left from the 9 gallon container to the 7 gallon container.

IF WE MAKE THIS, then each times the 9 gallon container will have 2, then increases 2 more gallon until it is full.

So in step16 , the 7 gallon container is full, so The 9 gallon container have :

         8 - 7 = 1 ( gallon )

SO WE GET 1 GALLON OF WATER.

We can easily find S = 1. Because 7 + 7 = 14 ( hundred ) => 90 + 7 + 7 = 104 ( hundred ), which that means I = 0.

But NO WORD CAN BEGIN BY 0 => Wrong question  

OMG !I don't khow what do you say!

Because the number 8 is not change, but 9 is split up, so the digit number of base₉ is the remander after divde by 9, up there is the remainder after divide by 9, etc.

So we have:

\(A\left(base_{10}\right)=\left(a\equiv x\left(mod9\right)\right)\overline{10...00}+\left(a-\left(9\overline{x0...00}\right)\equiv y\left(mod9\right)\right)\overline{10...10}+...+\left(a-\left(9\overline{x0...00}-\left(...-\left(\overline{z0...00}\right)\overline{10...00}\right)...\right)\equiv t\left(mod9\right)\right)\left(base_9\right)\)

Use this formula, we get:

\(22\left(base_{10}\right)=\left(22\equiv2\left(mod9\right)\right)10+\left(22-2\times9\right)\equiv4\left(mod9\right)\left(base_9\right)\)

So when write the number 22 in Base₉, we get 24.