## The comeback of brainteasers

Anything goes, but keep it seemly...
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

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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
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A bit of math

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
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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.

This can be easily demonstrated by brute force but I assume that JPF is looking for an elegant proof. The game is afoot!
Bigtone53

Posts: 413
Joined: 19 September 2005

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
Code: Select all
`913286457`

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

C
coloin

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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).

RW
RW
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Bigtone53 wrote:This can be easily demonstrated by brute force but I assume that JPF is looking for an elegant proof.
right

coloin wrote:Initial thoughts...
final thoughts are welcome

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).
Excellent RW, neat and smart proof.

JPF
JPF
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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

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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
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Okay RW, points accepted. I was confused by the term "minimum product"...

Thinking about your mysterious sequence... When you say "next digit" do you mean it must be a single-digit number?
udosuk

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udosuk wrote:Thinking about your mysterious sequence... When you say "next digit" do you mean it must be a single-digit number?

Sorry, my mistake. I'll edit it to say number (even though it might of course also be a single-digit number...)
RW
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Triple click to see the first hint I wrote:Look for a clue in the past, look for the answer in the near past.

RW
RW
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I think your intended answer for now is 128, but that's subject to change...

And I strongly disagree the omission of 5 in the sequence...
udosuk

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

RW
RW
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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...