JPF wrote:Obviously

BTW....... do you know then how that geezer worked out the 13th root of a 200 digit number ...in his head ???

http://news.bbc.co.uk/1/hi/england/london/7138252.stm

Not a riddle, just a question.

C

JPF wrote:Obviously

BTW....... do you know then how that geezer worked out the 13th root of a 200 digit number ...in his head ???

http://news.bbc.co.uk/1/hi/england/london/7138252.stm

Not a riddle, just a question.

C

- coloin
**Posts:**1737**Joined:**05 May 2005

coloin wrote:BTW....... do you know then how that geezer worked out the 13th root of a 200 digit number ...in his head ???

I don't know how he does it in his head...

But one way to do it with a computer is to use a Newton method based algorithm.

There are many derivations of this algorithm.

For instance to calculate PI with this formula :

PI^26=27! / [(76977927) x 2^24] x [1^(-26)+2^(-26)+3^(-26)+4^(-26)+...]

JPF

- JPF
- 2017 Supporter
**Posts:**3754**Joined:**06 December 2005**Location:**Paris, France

A bit of math

Let's consider a box of Sudoku , here's one :

The product of row 1 is : 1 x 8 x 9 = 72

The product of column 2 is : 8 x 5 x 4 = 160

Prove that for any box of sudoku, there is at least one row or one column such that its product is greater than or equal to 90.

JPF

Let's consider a box of Sudoku , here's one :

- Code: Select all
`+-------+`

| 1 8 9 |

| 2 5 7 |

| 3 4 6 |

+-------+

The product of row 1 is : 1 x 8 x 9 = 72

The product of column 2 is : 8 x 5 x 4 = 160

Prove that for any box of sudoku, there is at least one row or one column such that its product is greater than or equal to 90.

JPF

- JPF
- 2017 Supporter
**Posts:**3754**Joined:**06 December 2005**Location:**Paris, France

Initial thoughts

6 3-tupels per box

84 essentially different 3-tupels

? how many with product 90 or above ?

Prove that one of these is always in a box

worst case scenario perhaps

9! / [6^2]*[2] essentially different boxes =7!

C

6 3-tupels per box

84 essentially different 3-tupels

? how many with product 90 or above ?

Prove that one of these is always in a box

worst case scenario perhaps

- Code: Select all
`913`

286

457

9! / [6^2]*[2] essentially different boxes =7!

C

- coloin
**Posts:**1737**Joined:**05 May 2005

Here's a proof, don't know if it's elegant enough...

-A column or row that doesn't include any of the digits 123 has a minimum product of 4*5*6=120.

-To find a box where all columns and rows have a smaller product than 120, we must spread digits 123 so that each column and row contains one of them:

-For the row and column including digit 3, one must have a minimum product of 3*4*(insert larger number here) and the other a minimum product of 3*5*(insert larger number here). The latter is always greater than or equal to 90 (3*5*6=90).

RW

-A column or row that doesn't include any of the digits 123 has a minimum product of 4*5*6=120.

-To find a box where all columns and rows have a smaller product than 120, we must spread digits 123 so that each column and row contains one of them:

- Code: Select all
`1..`

.2.

..3

-For the row and column including digit 3, one must have a minimum product of 3*4*(insert larger number here) and the other a minimum product of 3*5*(insert larger number here). The latter is always greater than or equal to 90 (3*5*6=90).

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

rightBigtone53 wrote:This can be easily demonstrated by brute force but I assume that JPF is looking for an elegant proof.

final thoughts are welcomecoloin wrote:Initial thoughts...

Excellent RW, neat and smart proof.RW wrote:Here's a proof, don't know if it's elegant enough...

-A column or row that doesn't include any of the digits 123 has a minimum product of 4*5*6=120.

-To find a box where all columns and rows have a smaller product than 120, we must spread digits 123 so that each column and row contains one of them:

- Code: Select all
`1..`

.2.

..3

-For the row and column including digit 3, one must have a minimum product of 3*4*(insert larger number here) and the other a minimum product of 3*5*(insert larger number here). The latter is always greater than or equal to 90 (3*5*6=90).

Your turn again.

JPF

- JPF
- 2017 Supporter
**Posts:**3754**Joined:**06 December 2005**Location:**Paris, France

RW wrote:-For the row and column including digit 3, one must have a minimum product of 3*4*(insert larger number here) and the other a minimum product of 3*5*(insert larger number here). The latter is always greater than or equal to 90 (3*5*6=90).

Just a minor pick: it should be "... and the other a minimum product of 3*(5|6)*(insert larger number here)."

Looking forward for RW's next riddle!

- udosuk
**Posts:**2698**Joined:**17 July 2005

udosuk wrote:RW wrote:-For the row and column including digit 3, one must have a minimum product of 3*4*(insert larger number here) and the other a minimum product of 3*5*(insert larger number here). The latter is always greater than or equal to 90 (3*5*6=90).

Just a minor pick: it should be "... and the other a minimum product of 3*(5|6)*(insert larger number here)."

Sorry if it was a bit unclear. What I meant was that the first product is always equal to or greater than 3*4*(x) and the second is always equal to or greater than 3*5*(y). If the first is 3*4*5,the second is still equal to or greater than 3*5*(y). Even though the first option isn't possible, the statement should still be true, right?

udosuk wrote:Looking forward for RW's next riddle!

Ok, here we go. To solve this one, you have to think backwards...

riddler wrote:What's the next number in the sequence:

1, 4, 14, 15, 46, ...

RW

Last edited by RW on Sat May 17, 2008 10:08 am, edited 1 time in total.

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

RW wrote:You are on the right track, but the answer is not 128, and 5 should not be in this part of the sequence!

In that case the sequence is probably:

1, 4, 14, 15, 46, 5, 128, ... (Edited: This is wrong!)

Although I really don't know the ordering mechanism yet...

Putting the 1 at the front and the 5 so far back is very puzzling...

Added later:

Got it! The answer must be 4!

- udosuk
**Posts:**2698**Joined:**17 July 2005