tertre

The diagram should look like this:

Let the smaller circle have radius of length y units and that means that the bigger circle has a radius of length 2y units.

Thus, the smaller circle's area is $\pi(y)^2$ and the larger circle's area is $\pi(2y)^2=\pi4y^2$ units.

Thus, the answer is 1:4.

3.4 repeating at the four... to base ten....

3+4/13+4/169+...

3+4/12

3+1/3

10/3..

2dd56 _ 6 to base ten is...

(2*6^3)+(d*6^2)+(d*6^1)+(5*6^0)=432+36d+6d+5=437+42d...

437 + 42d has to be divisible by eleven...

437 mod 11 = 8

Thus, the least number is d=3..

Yes, @below, I forgot. Thank you!

Four(4) Identical balls and Four(4) identical boxes mean that the balls are indistinguishable and the boxes are also indistinguishable...

Since there is no restriction, let's try number separation.

1+1+1+1

1+1+2+0

2+2+0+0

1+3+0+0

4+0+0+0

Five(5) ways are possilbe.

Yes, it is 13 C 4.... 0 and 12 are included...

(a): This seems complicated, yet I think 13 C 4 = 715 ways if a is the least, b is the second to least, etc.

Yes, you're right...

The keywords noted here are different=distingusiable balls and boxes(you can tell the balls apart)...

Let's say the for balls have color white, yellow, red, and blue... and the boxes are labeled from box 1, box 2, box 3, and box 4.

Pick up the first white ball, I can place it in any of the four boxes, for 4 ways.

Next, pick up the yellow ball, I can place it in any of the four boxes again(since there is no restriction one ball per box)..

Similarly, the same flow for the red and blue balls..

Thus, the answer is 4*4*4*4=4^4=256 ways.

The choices are independent of each other...

Hello Melody,

Are there restrictions that there is one ball per box?

If there is a restriction, then yes, 3!=six(6) ways.

If there are no restrictions, ball one can go in the red, white, or blue box...this also happens for ball two and ball three...

Then, there are 3^3=27 ways.

Remember that the choices are independent.

We need at least two three's and three two's...

We have to add one more three and two more two..

 However, 2*3*2*2 is not a perfect cube but 3*3*2*2

We need three  3's and two 2's..

3^3*2^2=27*4=108-smallest positive integer.