G’day folks, greetings from Perth, Western Australia.
This is my first post, so please be gentle with me.
While playing a game of Chess with my nephew the other nite, a thought occured to me:
If I was to place 1 grain of sand on the first square of a Chess-Board, 2 on the second, 4 on the third, 256 on the fourth, 65,536 on the fifth, how many grains of sand would there be on the last (64th) square?
I’ve tried to work this out on using calculators, but they are not able to calculate such a seemingly large number.
If you’ve made a mistake in your calculating and just want to find 2^64, that is, have double the number of grains of sand as the previous square (represented by the pattern 1, 2, 4, 8, 16, 32, etc) , then your answer is 18446744073709551616 grains of sand.
If you’re looking for a different pattern than that, you’re going to need to define it further, because I really have no idea what you’re looking for.
cobberone
The answer would be 2x10[sup]63[/sup] or 9.22 x 10[sup]18[/sup] grains os sand.
The total grains of sand on the board would be twice this number minus 1.
Total = 1.844 x 10[sup]19[/sup] (and don’t for get to subtract 1 grain.)
What I meant to ask is if I started with 1 grain of sand, ‘powered’ (is that the right term?) it to 2, then 4 (2x2), 16 (4x4), 256 (16x16), 65,536 (256x256), 4,294,967,296 (65,536x65,536), and so-on.
Your pattern is a little odd, but you could use Excel to do it once your explain the pattern.
Why is your pattern 1 2 4 256 65536?
You are not doubling or squaring each square. What is the pattern you want to achieve?
You took one and doubled it. Then you either took 2 to the second or 2x2. Then you took 4[sup]4[/sup]. Then you simply squared 256 to get 65536. Your pattern does not make a lot of sense and the previous answers appear wrong for the information provided by you.
Welcome to the SDMB. Enjoy the stay.
Jim (I see you update the question, give me a few minutes)
The OP doesn’t appear to be using a progression of powers of two, but rather, the next number is the square of the previous one (except in the first instance, or we’d never get past 1)
This is squaring the total of square n to get the total on square n+1, and since there are 2 grains on square n=2, the formula for any square n is 2^(2^(n-1)). Thus the 64th square has 2^(2^63) = 2^9223372036854775808.
The number is too large for my calculator to compute, but I can estimate it; since 2^10 = 1,024 ~= 1,000, the number is equal to 1024^9007199254740992, so it’s in the neighborhood of 1 followed by 10^16 = 10 quadrillion zeros.
It looks like starting with square 2, there are 2^2^(n - 2) grains on the nth square. Based on that, there should be 2^2^62 grains on the 64th square, which has about 1.4 * 10[sup]19[/sup] digits.
It’s definitely clearer. You’re saying that the number of grains on square n is the square on the number on n - 1. Unfortunately, when you start this at 1, it never grows - square 2 has 1[sup]2[/sup], which is again 1.
The simple way to escape this difficulty is to express the sequence as Colophon has done. The number of grains on the final square would then be far greater than the number of subatomic particles in the known universe.
That still doesn’t match for square 1: That formula would give sqrt(2) grains on the first square.
You can have it in tons if you’d like, but it won’t make things any easier. These numbers are easily huge enough that, the way we’re writing them, even if you used neutrinos instead of grains of sand and used the mass of the Universe for our mass units, we still wouldn’t be able to write the answer any differently.
