Question

You have a chess board as pictured, with squares numbered 1 through 64. You also have a huge change jar with an unlimited number of dimes. On the first square you place one dime. On the second square you stack 2 dimes. Then you continue, always doubling the number from the previous square. [UW]
a. How many dimes will you have stacked on the 10th square?
b. How many dimes will you have stacked on the nth square?
c. How many dimes will you have stacked on the 64th square?
d. Assuming a dime is 1 mm thick, how high will this last pile be?
e. The distance from the earth to the sun is approximately 150 million km. Relate the height of the last pile of dimes to this distance.

Answer 1

Answer:

(a) 512 dimes

(b)$D(n) = 2^{n-1}$

(c) 9.22337 x 10¹⁸ dimes

(d) 9.22337 x 10¹² km

(e) roughly 61,489.13 times longer than the distance from the earth to the sun.

Step-by-step explanation:

b. Starting with 1 dime in the first square, if the number of dimes per each square is the double of the previous square, the general equation for the number of dimes in each square 'n' can be found by:

$D(1) = 1\\D(2) = 2^1\\D(3) = 2^2\\D(4) = 2^3\\D(n) = 2^{n-1}$

a. On the 10th square:

$D(10) = 2^{10-1}\\D(10) = 512\ dimes$

c. On the 64th square:

$D(64) = 2^{64-1}\\D(10) = 9.22337 *10^{18}\ dimes$

d. If a dime is 1 mm thick, the 64th pile will be:

$h= 9.22337*10^{18}*0.0001\ m\\h= 9.22337*10^{15}\ m\\h= 9.22337*10^{12}\ km$

e. The ratio between the height of the last pile (h) to the distance from the earth to the sun is:

$R=\frac{9.22337*10^{12}\ km}{150,000,000\ km}\\ R=61,489.13$

The height of the last pile is roughly 61,489.13 times longer than the distance from the earth to the sun.