Rice on a chessboard- can you help?

[Highlandterrier]

It is an old problem, and I think I nearly get it. 64 squares on the chessboard, 1 on 1st square, double that on 2nd square, etc

1st 1

2nd 2 2 to power 1
3rd 4 2 to power 2
4th 8 2 to power 3
5th 16 2 to power 4

so on square number 64 there would be
2 to the power 63

But do I then need to add all these together and add one. I am tying myself in knots. I know it is a lot of rice. Does anyone know a good website which could explain it in words of one syllable because I am confused. Thanks

? its saturday - drink wine

logarithims???????

And then have some more.

1st = 1
2nd = 2
3rd = 4
4th = 8
5th = 16
6th = 32

etc so I agree that it is 2 to the power of 63 and then you need to add all these together and add one

No website. Just tired old mind

If you need to add them all up then it is just 2^64 - 1. If you look at the first few in the sequence this is easy to see:

2^2 - 1=3=2+1
2^3 -1 =7=4+2+1 etc.

So if the question is how many grains on the chalkboard the answer is 2^64 - 1.

the answer is , there's not enough rice in the world.

OP: to know the answer, you need to know the question [made up chinese proverb emoticon]. Rephrase your question in Maths jargon and you will be able to Google it.

What you are looking for is the sum of a geometric progression

Do these sites help? link1 link2

Not sure why come people want to add one after adding up the number of grains of rice on each square. If you want to know how much rice is no the last square it is 263. The total amount of rice on the board is 264 - 1. So there are 9,223,372,036,854,775,808 grains of rice on the last square and 18,446,744,073,709,551,615 on the board.

Thanks to everyone. You are great. xxxx

Oh, go on,

I have a maths head, but have no clue what you're all talking about. DD got A gcse, Ds is about to do it, so I'm not thick, go on, see if you can explain to someone who doesn't have a clue what you're talking about.

Is it a wind up?

Go on, go on, go on.

Monty, imagine you have a chessboard. On the first square there you put a grain of rice, the next square you put two grains of rice, the next square you put four grains of rice and so on. For each square you double the number of grains of rice you put on.

I imagine that the original question is either 'How many grains of rice are on the last square?' or 'How many grains of rice are on the chessboard?'.

The answer to the first question is 263 since there are 64 squares on the board and the first square has 20 grains on it.

The answer to the second question is 2^64-1.

prh47bridge, I have no idea why people want to add one either.

it's part of some apocryphal story - a pauper challenges a king, on the condition that he'll get that many grains of rice if he wins. there isn't that much rice in the world, let alone the country, so the king is impoverished when the pauper wins.

could be from the arabian nights?

And the lesson that I hope the maths teacher draws from this is that doubling each time (ie 2x) grows faster than powers (ie xa), for any finite a, given a large enough x.

Imagine a slightly different chessboard, where on the first square you put one grain, on the second 4 (2x2), on the third 9 (3x3), on the fourth 16 (4x4) and so on: the nth square gets n^2 grains.

The 64th square gets 64x64, or 4096, which is rather less than 1.86 x 10^19 (that's 18 followed by eighteen zeros).

The next board could cube the number: the first square gets 1, the second gets 8 (2x2x2), the third gets 27 (3x3x3). Then the 64th square gets 64x64x64 = ~250 thousand. Still rather less.

The next board could multiply the number by itself four times: 1, 2x2x2x2=16, 3x3x3x3=81, 4x4x4x4=256 and so on. The 64th square gets ~17 million.

How long would you have to keep doing this before the 64th square gets more than 2^64?

The answer is "the eleventh board" as 6411 (64x64x64x64x64x64x64x64x64x64x64x64) is indeed greater than 264 (2x2x2...total of 64 2s). (exercise: show this is true, without working out what 264 or 6411 are).

But all you need then is a few more squares: 267 is 1.4x1020, while 6711 is is "only" 1.2x1020, and off it goes, with the doubling board running ahead.

OK, you say, let's do another board, this time going 1, 4096, 531441, 16777216 (ie 112, 212, 312, 412...). Now we'll beat the doubling board, won't we? Well, until the 75th square, at which point doubling pulls ahead again.

It's another interesting exercise (good GCSE student, I assume/hope) to work out how many squares you have to go before the doubling board outstrips the "to the power of" board, without using a calculator (or log tables), for any given power. By square 128 you would need to be 219, 319 and so on to be ahead, but you'd be behind within a few squares. By square 256 232, 332, 432 could stay with the doubling (because 2256 = 256^32), but on square 257 the doubling gets ahead again.

Oh, I see.