Question

If a chessboard (8x8) were to have pennies placed on each square such that 1 penny was placed on the first square, 2 on the second, 4 on the third, and so on (doubling the number of pennies on each subsequent square), how many pennies would be on the chessboard when finished?

Answer 1

The sum of the pennies on a chessboard is -2^64 + 1.

Step-by-step explanation:

The question asks how many pennies would be on the chessboard when finished, given that one penny is placed on the first square and the number of pennies is doubled on each subsequent square. To solve this problem, we can use the formula for the sum of a geometric series:

S = a(1 - r^n) / (1 - r)

Where S is the sum of the series, a is the first term (1 penny), r is the common ratio (2), and n is the number of terms (64 squares on a chessboard). Plugging in these values, we can find the sum:

S = 1(1 - 2^64) / (1 - 2)

Simplifying further, we get:

S = -(2^64 - 1)

Therefore, when finished, there would be -2^64 + 1 pennies on the chessboard. Please note that this number is extremely large and cannot be represented by a regular 32-bit integer.

Answer 2

64 according to e2020