MATH SOLVE

4 months ago

Q:
# A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess board. On the second square the King would place two grains of wheat, on the third square, four grains of wheat, and on the fourth square eight grains of wheat. If the amount of wheat is doubled in this way on each of the remaining squares, how many grains of wheat should be placed on square 18? Also find the total number of grains of wheat on the board at this time and their total weight in pounds. (Assume that each grain of wheat weighs 1/7000 pound.)

Accepted Solution

A:

Answer:Square 15: 16,384Sum of all grains up to and including square 15: 32,767Total weight up to and including square 15: 4.681 lbStep-by-step explanation:Square 1: 1 = 2^0Square 2: 2 = 2^1Square 3: 4 = 2^2Square 4: 8 = 2^3Notice that the exponent of the 2 is 1 less than the number of the square, so for square n, the number of grains is 2^(n - 1)Square n: 2^(n - 1)Square 15: 2^(15 - 1) = 2^14 = 16,384Now let's look at the sums.Square 1 sum: 1Square 2 sum: 1 + 2 = 3Square 3 sum: 3 + 4 = 7Square 4 sum: 7 + 8 = 15Notice that each sum is one grain less than 2 raised to the number of the square.Square 1 sum: 1 = 2^1 - 1Square 2 sum: 1 + 2 = 3 = 2^2 - 1Square 3 sum: 3 + 4 = 7 = 2^3 - 1Square 4 sum: 7 + 8 = 15 = 2^4 - 1Square n sum: 2^n - 1Square 15 sum: 2^15 - 1 = 32,767Weight:32,767 × 1/7000 lb = 4.681 lb