If you wanted to cover (as nearly as possible) the floor of a 6-foot by 8-foot room with one thickness of nickels, how many nickels would it take?

Question

If you wanted to cover (as nearly as possible) the floor of a 6-foot by 8-foot room with one thickness of nickels, how many nickels would it take?

After you have done this, if you gathered up the nickels and made one stack of nickels (not edge to edge, but face to face) that reached to the ceiling of the room, 7.5 feet high, would you have enough nickels? Assume that you will not cut any coin but can only use whole coins.

A nickel, in American usage, is a five-cent coin struck by the United States Mint. Composed of 75% copper and 25% nickel, the piece has been issued since 1866. Its diameter is .835 inches (21.21mm) and its thickness is .077 inches (1.95 mm). The silver half dime, equal to five cents, had been issued since the 1790s.

Answer 1

This question doesn't actually need the area of a nickel or to consider it (oddly) as a square!

The real question is how many nickels will fit into the dimensions of the given room: 6' x 8' or 72" x 96"?

A nickel's diameter is 0.835" so along the width, we can lay 72 / 0.835 = 86 nickels (ignoring the fraction.)

Along the length, we can lay 96 / 0.835 = 114 nickels (again, ignoring fractions of a nickel)

Therefore, on this floor, we can lay down a rectangular array of 86 x 114 =9,804 nickels. That's it!

This is known as square close packing.

The nickels can also be laid down in hexagonal close packing but this will be less efficient.

The stacked 9,804 coins, each 0.077" thick, will reach a height of 9,804 x 0.077" = 754.9" = 62.9'

So, yes, we have more than enough coins to reach a 7.5' ceiling.

Answer 2

Hello, Tim,

I assumed each nickel can be modelled as a square and every nickel touches four others around it until the nickels touch the walls. The area for one nickel is divided into the total room area to yield the number of nickels that can fit in the room. Everything is converted to inches and in^2. Ignore the 0.586 of a nickel and round down to a whole number, 9913 nickels.

To get a roomful. convert the height into inches and then divide by the nickel's thickness in inches. This gives the number of layers. Multiply that times the number of nickels needed for one layer and that is the number needed to fill the room. Please send me the leftover change.

Bob

Answer 3

Assuming you placed every nickel edge to edge, without staggering every other row,

The area of the floor is 8 x 6 = 48 ft² = 48 x 144 = 6912 in². The area of the nickel if placed in a square = .835 x .835 = 0.697225 in².

As a comparison , the area of a nickel as a circle is πr²=π(.835/2)²= 0.5475 in^2.

Going with the first assumption, the number of nickels = 6912/0.697225 ≅ 9913.

The height of 9913 stacked nickels is 9913 x 0.077 = 763.3 inches = 63.6 ft.

63.6 ft > 7.5 ft, so there are enough nickels.