A king in acient times agreed to reward the inventor of chess with one grain of
ID: 3199472 • Letter: A
Question
A king in acient 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 th third square, four grains of wheat, on the fourth square eight grains of wheat. if the amount of wheat is douled in this way on each of the remaining squares. how many grains of wheat should be placed on 21? 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 weights 1/7000 pound.
How many grains of wheat should be placed on sqyare 21?
How many total grains of wheat should be on the board after the grains of wheat have been placed on square 21?
what is the total weight of all the grains of wheat on the board after the grains of wheat have been placed on square 21? (round to the nearest tenth as neede)
Explanation / Answer
Answer.
Q1. ) How many grains of wheat should be placed on square 21?
Ans. We have,
20 = 1 grain on 1st square, 201 = 2 grains on 2nd square, 202 = 4 grain on 3rd square and so on.
Hence, for every n, we have 2n-1 grains.
Therefore,
Grains on 21st square will be = 221-1 = 220.
Q 2.)How many total grains of wheat should be on the board after the grains of wheat have been placed on square 21?
Ans. Sum of a GP whose first term is 20 and 21st term is 220.
Therefore,
Sum = 20 + 201 + ..... 220 = 1*(221 - 1)/(2-1) = 221 - 1.
NOTE: - Sum of a GP a,ar,ar2,....arn = a(rn - 1)/(r-1).
Q 3.) what is the total weight of all the grains of wheat on the board after the grains of wheat have been placed on square 21? (round to the nearest tenth as neede)
Ans. We have,
Sum = 221 - 1
and, weight of 1 grain (w) = 1/7000 pounds
Hence,
Total Weight = sum * w = 299.59300 pounds ~ 300 pounds.
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