Melody

I think 12 ... 

 \(2^x=72y\)                 where x and y are integers.

I don't think there is one....

\(72=9*8=2^3+3^2 \)

\(2^x=72y \)

the prime factors on the left side is just 2 and on the right side it is 2 and 3

I do not think any integer answers are possible.....                                                                                                                          

You can see straight off that x cannot be 2 or 6.

Now just for the moment you can ignore that.

Try solving for x in terms of k and see what happens.   The answer will present itself.

No one answer over me please. 

But asker guest can interact with me if he/she wants      (or with others if i don't answer quickly)

Here is an approximate diagram.

You can get exact figures using trigonometry.

The positive integers from 1 to 150 inclusive are placed in a 10 by 15 grid so that each cell contains exactly one integer. Then the multiples of 3 are given a red mark, the multiples of 5 are given a blue mark, and the multiples of 7 are given a green mark. How many cells have more than 1 mark?

This diagram is just to get you started.  Work out the very middle overlap section first. Then work out from there.

HINT: The numbers that have all 3 colours must have 3*5*7 =105 as a factor.  

Please no one give a full answer.

I do not understand the question

There is a 6-digit sequence of 8 positive consecutive integers that divides 2, 3, 5, 7, 9, 11, 13 and 17 EVENLY. What is the sequence in question? Thank you for help.

What does this mean....There is a 6-digit sequence of 8 positive consecutive integers     ??

No one has answered becasue it does not make any sense.

I will ask YOU a couple of  question.

What is a median?    It is the MIDDLE score. There is 7 scores.  What is the POSITION of  the middle one? 

These extra two numbers can slot in anywhere, and they can  be any integer.

So what possibilites are there for the middle score?

Please no one else answer

Could be many different reasons.

Thanks guest.

That is an interesting calculation site.

I do not think it is necessary for thei question. Many calcs, including the web2 calc witll do it easily.  

But it certainly looks like a calc site that could be worth bookmarking.