Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2006-09-10 08:24:54

numen
Member
Registered: 2006-05-03
Posts: 115

### determinant

I need some help with this problem, I'm not sure how to approach this.

The numbers 13413, 39865, 58752, 77639 and 96577 are divisible by 17. Show that the determinant

is also divisible by 17.

Bang postponed. Not big enough. Reboot.

Offline

## #2 2006-09-11 01:17:18

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: determinant

numen wrote:

I need some help with this problem, I'm not sure how to approach this.

The numbers 13413, 39865, 58752, 77639 and 96577 are divisible by 17. Show that the determinant

is also divisible by 17.

Are you meant to show that that monstrosity is divisible by 17 simply by working out what it is, or are there some fancy proof tricks you can use that I don't know about?

If it's a case of just working it out, it's fairly simple to do, but it's a ton of work.

You just need to keep dividing that determinant into smaller ones.

becomes

And then you'd need to split each of those determinants into terms containing four determinants of 3*3 matrices, and then those 3*3 determinants into terms containing three 2*2 determinants, and then you could finally give it a value.

So it's possible, but it's a lot of work. Usually you'd get a computer to do that kind of thing.

Why did the vector cross the road?
It wanted to be normal.

Offline

## #3 2006-09-11 06:31:26

numen
Member
Registered: 2006-05-03
Posts: 115

### Re: determinant

I can see where you're going, but I was hoping for a better way of doing it. Working with 60 2*2 determinants doesn't sound like a lot of fun =P

Bang postponed. Not big enough. Reboot.

Offline

## #4 2006-09-11 07:20:05

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: determinant

It seems to be that the if a, b, c..., n are all divisible by x, and:

The digits of a are a0, a1, ..., an
...of b are b0, b1, ..., bn
...
the digits of n are n0, n1, ..., nn

Then the determinate of:

``````a0 a1 ... an
b0 b1 ... bn
...
n0 n1 ... nn``````

Is divisble by x.

At least I can't find any counter examples.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Offline

## #5 2006-09-11 07:30:21

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: determinant

I believe there exists some relation to multiplying numbers and the determinate of their digits, although I can't find exactly what that is.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Offline

## #6 2006-09-11 08:33:31

slfer100
Member
Registered: 2006-09-11
Posts: 106

### Re: determinant

...............

naver give up on the war of love unless if you want to!

Offline