Q:

What is the maximum number of pencil that can be stood in this stand?

I am working on the following problem asked in an examination:
"What is the maximum number of cylindrical pencils of 0.5 cm diameter that can be stood in a square shaped stand of 5cm × 5cm inner cross section?
1. 99    2. 121   3. 100    4. 105."
A simple calculation yielded me the answer 100.  But the answer in the key is 105. I don't how such an answer is possible.
Any hint please.....Thanks in advance!

A:

Whoever constructed the question, had a path planned for you.
Because the pen diameter is one tenth of the square edge, you should immediately know that you can put at least $10\times 10 = 100$ pens within the square. However, you should also know that regular rectangular packing is not the densest one for circles; hexagonal packing is even denser.

By calculation or approximation of the equilateral triangles formed by the centers of each triplet of pens, or the right triangle halves of each such triangle (yellow triangles on the right), you were to determine that using a simple hexagonal packing, you should be able to fit six rows of ten pens each, plus five rows of nine pens each, within the same square. Because $6\times 10 + 5 \times 9 = 60 + 45 = 105$, you are supposed to choose 105.
Unfortunately, the hexagonal packing (show in blue on the right, above) is not the best known circle packing for this case, and the correct answer happens to be 106, as Seyed already pointed out. So, whoever constructed the question, happened to pick an incorrect path..

A:

The correct answer is 106 pencils.
You can use this packing calculator:
http://www.packomania.com/

