Maths

[mandylee01]

As a New Year Resolution a man decides to save 1p on 1st Jan., 2p on 2nd Jan., 4p on 3rd Jan., 8p on 4th Jan., and so on - each day saving twice the amount saved on the previous day. What would his total be on the last day of January?

[smouse]

10737418.24

[Fibonacci]

1st Jan = 1p : 1 = 2^0 (^ = to the power of)
2nd Jan = 2p: 2 = 2^1
3rd Jan = 4p: 4 =2^2
4th Jan = 8p: 8 =2^3
so on
31st Jan 2^(31-1) = 2^30 pence
= 1,073,741,824p
= �10,737,418.24

[Janads]

Surely if he saved the different amounts every day, all the days of the month need to be added up including the last day entry to get the total?
Which would be �21474836.47

[Aquagility]

I think that this figure, �10,737,418.24 is the man's saving on Jan 31st. His total on the last day would be much more!

[Aquagility]

Well done, Jonads. I was not going to bother adding it up.

[Janads]

You don't need to add it all up. If you look at any entry along the dates all the previous entries add up to 1p less eg. on day 7 = 64 the previous 6 add up to 63. So you just have to take a penny off the final sum and add those two.

[Aquagility]

On which day, I wonder, did he decide to give up!