simple maths problems

[condor]

Just a couple

1. does 0.9999....=1

2. solve x^(x^(x^(x^(....))))=2

3. what is the exact number of marbles required to build a million layer pyramid ?
the number of marbles has to be such in each layer that it fits on top of the next one

1, 3, 6, 10, 15, 21, 28.....

4. Does the infinite sum of the inverses of the prime numbers converge?

1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/23......

In other words does the sum apprach infinity as the number of terms approaches infinity or does it asymtotically approach a fix value(converge)?



At this time of night...and after much wine...is too difficult for me to solve...and I want to go to bed.
Can a little elf fix it for me by morning? Please

[tuffer]

1. Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

[unlicensed]

i thought u said they were simple.......
as in simple like 1 + 2 = 17

[iria]

so 0.999999 /infinite/ = 1?

I see now why i went on to study arts and binned the maths long time ago (i discovered my vocation when i found out i could only deal with mathematical problems containing letters in them, doh)

[B0yracer]

1,3,6,10,15,21,28.......


next is 28+8=36, 36+9=45, 45+10=55, 55+11=66

[Bluebottle]

Two non-maths problems:

Find the next number

1/ 11 - 8 - 16 - 7 - ?

2/ 22 - 16 - 11 - 8 - ?

[lightningghost]

The first one, 0.99999.... does indeed = 1 at some point, but there are a bloody lot of decimal places

[ATG]

I'd guess that the answer to 4 is "yes" and that the proof is to sum a similar but definitely slower decreasing series of fractions and show that that series converges.

[samwilliams]

the answer to third one is 5000000500000 marbles

[condor]

samwilliams said:

the answer to third one is 5000000500000 marbles

Think you've only done 1000 layers there

If we see that each layer is a triangle of marbles which has to have one more row than the layer above it. These triangular numbers are connected by the series T_n = n(n+1)/2, so we want the sum from 1 to 1,000,000 of n(n+1)/2, which is equal to one half of the sum of n*n + one half the sum of n. These are both well known expressions which can be proven by induction, and the answer is
(n(n+1)(2n+1)/6+n(n+1)/2)/2, where n=1,000,000. This appears to simplify to n(n+1)(n+2)/6, so it's a little over one-sixth of a qunitillion (10^18).

I make it 166,667,166,667,000,000 marbles.

[Wacky Racer]

How can you turn two matchsticks into five without breaking them?

[Tocka]

I am new here

...and tonker its 10TO10

Did I get it right?

[All Terrain]

In reply to Wacky

Gamble with them on a game of cards

>> Edited by All Terrain on Tuesday 28th October 10:31

[M@H]

iria said:

so 0.999999 /infinite/ = 1?

I'll explain this one in more simple detail if it helps as tuffer managed to confuse me a bit, and I know the answer.. :

let X = 0.99999 (recurring)

therefore 10X = 9.99999999 (recurring)

10X - X = 9X

accordingly

9.999999 (recurring) - 0.999999 (recurring) = 9X

9.999999 (recurring) - 0.999999 (recurring) = 9

if 9X = 9, then X = 1


Cheers
Matt



>> Edited by M@H on Tuesday 28th October 10:40

[Wacky Racer]

All Terrain said:

In reply to Wacky Gamble with them on a game of cards >> Edited by All Terrain on Tuesday 28th October 10:31

[mondeoman]

I thought pyramids had four sides??

so its 1, 4, 9, 16 ......

[XM5ER]

condor said:

samwilliams said: the answer to third one is 5000000500000 marbles Think you've only done 1000 layers there If we see that each layer is a triangle of marbles which has to have one more row than the layer above it. These triangular numbers are connected by the series T_n = n(n+1)/2, so we want the sum from 1 to 1,000,000 of n(n+1)/2, which is equal to one half of the sum of n*n + one half the sum of n. These are both well known expressions which can be proven by induction, and the answer is (n(n+1)(2n+1)/6+n(n+1)/2)/2, where n=1,000,000. This appears to simplify to n(n+1)(n+2)/6, so it's a little over one-sixth of a qunitillion (10^18). I make it 166,667,166,667,000,000 marbles.

I dont doubt your maths for a minute however I do doubt your assumption that a pyramid has a triangular base, last time I looked they have a square base.

[danielson]

Wacky Racer said:

How can you turn two matchsticks into five without breaking them?

turn them into a "V" which is 5 in roman numerals..

[Spooky]

Wacky Racer said:

How can you turn two matchsticks into five without breaking them?

Er.. I reckon you turn up into the roman numeral for 5, V.

Did I win a prize?

[M@H]

Tocka said:

I am new here ...and tonker its 10TO10 Did I get it right?

Hi and welcome


yes you did..

Cheers,
Matt