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• To: Prime Numbers From: Jon Perry Date sent: Thu, 4 Oct 2001 22:51:33
Message 1 of 13 , Oct 5, 2001
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From: "Jon Perry" <perry@...>
Date sent: Thu, 4 Oct 2001 22:51:33 +0100

> Should be quite simple this one.
>
> What is the highest power of two that divides ab+bc+ca?
>
> e.g.a=1, b=2, c=3, ab+bc+ca=2+6+3=11, therefore 0.
>
> a=2,b=2,c=3 ab+bc+ca=4+6+6=16, therefore 4.

Here's my (version 1) "solution":

Let a, b, c be 2^i.s, 2^j.t, 2^k.u, where s,t,u are odd, and i <= j <= k

1) If j < k or i=j=k:

2) If i<j=k:
Let t+u = 2^m.v, v is odd

2.1) If i+m > j:

2.2) If i+m < j:

2.3) If i+m = j:
Let sv+tu = 2^n.w, w is odd.

Messy... :-(

Michael Hartley : Michael.Hartley@...
Sepang Institute of Technology
+---Q-u-o-t-a-b-l-e---Q-u-o-t-e----------------------------------
"There was a young man from Peru,
Whose limericks stopped at line two."
-- Spike Milligan
• If you consider 2,5,10, i.e. A=2^0*5 B=2^1*1 C=2^1*5 ... (r+s) in this case translates to (s+t), or (1+5)=6, therefore k=1. ... (c+k) becomes a+k = 1. a
Message 2 of 13 , Oct 6, 2001
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If you consider 2,5,10, i.e.

A=2^0*5
B=2^1*1
C=2^1*5

b and c are equal. Hence:

>3) If 2 only are equal, say a = b, it comes

>S = 2^(a+a)rs + 2^(a+c)rt + 2^(a+c)st
>S = 2^(2a)rs + 2^(a+c+k)t(r+s)/2^k
>where 2^k is the highest power of 2 that divides (r+s),

(r+s) in this case translates to (s+t), or (1+5)=6, therefore k=1.

>Once more, there are 3 cases:
>1) if (c+k) > a, the result is 2^(2a)
>2) if (c+k) < a, the result is 2^(a+c+k)
>3) if (c+k) = a, the result is 2^(2a+h) where 2^h is the greatest
> power of 2 that divides rs + t(r+s)/2^k

(c+k) becomes a+k = 1. a becomes what?

Say a becomes b=1. a+k=b, therefore 2^(0+1+1)=4
Say a becomes c=1. a+k=c, therefore 4 again.

Only the answer to ab+bc+ca is 80.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Marcel Martin [mailto:znz@...]
Sent: 04 October 2001 23:46
Cc: Prime Numbers

I assume the numbers A, B and C are positive and different from 0.

A = 2^a * r (r odd)
B = 2^b * s (s odd)
C = 2^c * t (t odd)
sorted such that a <= b <= c

S = AB + AC + BC = 2^(a+b)rs + 2^(a+c)rt + 2^(b+c)st
= 2^(a+b) (rs + 2^(c-b)rt + 2^(c-a)st)

1) If a < b < c, the highest power of 2 that divides S is 2^(a+b)

2) If a = b = c, the highest power of 2 that divides S is 2^(2a)

3) If 2 only are equal, say a = b, it comes

S = 2^(a+a)rs + 2^(a+c)rt + 2^(a+c)st
S = 2^(2a)rs + 2^(a+c+k)t(r+s)/2^k
where 2^k is the highest power of 2 that divides (r+s),

Once more, there are 3 cases:
1) if (c+k) > a, the result is 2^(2a)
2) if (c+k) < a, the result is 2^(a+c+k)
3) if (c+k) = a, the result is 2^(2a+h) where 2^h is the greatest
power of 2 that divides rs + t(r+s)/2^k

Marcel Martin

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• Hey, you know you are right! Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript
Message 3 of 13 , Oct 6, 2001
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Hey, you know you are right!

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Marcel Martin [mailto:znz@...]
Sent: 06 October 2001 09:30
Cc: Prime Numbers

>(c+k) becomes a+k = 1. a becomes what?
>Say a becomes b=1. a+k=b, therefore 2^(0+1+1)=4
>Say a becomes c=1. a+k=c, therefore 4 again.

Are you serious?. To find 'what a becomes', it is sufficient to

------------------------------------------
S = 2^(a+b) rs + 2^(a+c) rt + 2^(b+c) st
------------------------------------------

We have a < b and b = c
S = 2^(a+b)rs + 2^(a+b)rt + 2^(2b)st
S = 2^(a+b+k)r(s+t)/2^k + 2^(2b)st
where 2^k is the highest power of 2 that divides (s+t)

There are 3 cases:
1) if (a+k) > b, the result is 2^(2b)
2) if (a+k) < b, the result is 2^(a+b+k)
3) if (a+k) = b, the result is 2^(2b+h) where 2^h is the greatest
power of 2 that divides r(s+t)/2^k + st.

Now, let's apply this to your numbers.

A=2^0*5 -> a=0, r=5
B=2^1*1 -> b=1, s=1
C=2^1*5 -> c=1, t=5

S = 2^(0+1)5 + 2^(0+1)25 + 2^2 5
S = 2^(1+k)5(1+5)/2^k + 2^2 5 -> thus k = 1
we are in the case a+k = b so the answer is 2^(2+h) with
2^h dividing r(s+t)/2^k + st = 5(1+5)/2 + 5 = 20 -> 2^h = 4 -> h = 2

The answer is 2^(2+h) = 2^4

Marcel Martin

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• You have missed i=j
Message 4 of 13 , Oct 6, 2001
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You have missed i=j<k.

1) is true - I checked it in JScript (I'm working on the others).

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: MICHAEL HARTLEY [mailto:Michael.Hartley@...]
Sent: 06 October 2001 04:26
To: Jon Perry

From: "Jon Perry" <perry@...>
Date sent: Thu, 4 Oct 2001 22:51:33 +0100

> Should be quite simple this one.
>
> What is the highest power of two that divides ab+bc+ca?
>
> e.g.a=1, b=2, c=3, ab+bc+ca=2+6+3=11, therefore 0.
>
> a=2,b=2,c=3 ab+bc+ca=4+6+6=16, therefore 4.

Here's my (version 1) "solution":

Let a, b, c be 2^i.s, 2^j.t, 2^k.u, where s,t,u are odd, and i <= j <= k

1) If j < k or i=j=k:

2) If i<j=k:
Let t+u = 2^m.v, v is odd

2.1) If i+m > j:

2.2) If i+m < j:

2.3) If i+m = j:
Let sv+tu = 2^n.w, w is odd.

Messy... :-(

Michael Hartley : Michael.Hartley@...
Sepang Institute of Technology
+---Q-u-o-t-a-b-l-e---Q-u-o-t-e----------------------------------
"There was a young man from Peru,
Whose limericks stopped at line two."
-- Spike Milligan
• Sorry - you haven t. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry BrainBench MVP for HTML and JavaScript http://www.brainbench.com
Message 5 of 13 , Oct 6, 2001
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Sorry - you haven't.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: 06 October 2001 12:43
To: MICHAEL HARTLEY

You have missed i=j<k.

1) is true - I checked it in JScript (I'm working on the others).

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: MICHAEL HARTLEY [mailto:Michael.Hartley@...]
Sent: 06 October 2001 04:26
To: Jon Perry

From: "Jon Perry" <perry@...>
Date sent: Thu, 4 Oct 2001 22:51:33 +0100

> Should be quite simple this one.
>
> What is the highest power of two that divides ab+bc+ca?
>
> e.g.a=1, b=2, c=3, ab+bc+ca=2+6+3=11, therefore 0.
>
> a=2,b=2,c=3 ab+bc+ca=4+6+6=16, therefore 4.

Here's my (version 1) "solution":

Let a, b, c be 2^i.s, 2^j.t, 2^k.u, where s,t,u are odd, and i <= j <= k

1) If j < k or i=j=k:

2) If i<j=k:
Let t+u = 2^m.v, v is odd

2.1) If i+m > j:

2.2) If i+m < j:

2.3) If i+m = j:
Let sv+tu = 2^n.w, w is odd.

Messy... :-(

Michael Hartley : Michael.Hartley@...
Sepang Institute of Technology
+---Q-u-o-t-a-b-l-e---Q-u-o-t-e----------------------------------
"There was a young man from Peru,
Whose limericks stopped at line two."
-- Spike Milligan

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... Who is you ? ? uoy si ohW Haven t what? ?tahw t nevaH Please don t top post. .tsop pot t nod esaelP Phil
Message 6 of 13 , Oct 6, 2001
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On Sat, 06 October 2001, "Jon Perry" wrote:
>
> Sorry - you haven't.
>
> Jon Perry
> perry@...
> http://www.users.globalnet.co.uk/~perry
> BrainBench MVP for HTML and JavaScript
> http://www.brainbench.com

Who is 'you'? ?'uoy' si ohW
Haven't what? ?tahw t'nevaH
Please don't top post. .tsop pot t'nod esaelP

Phil lihP

-- --

Phil lihP

Please don't top post. .tsop pot t'nod esaelP
Haven't what? ?tahw t'nevaH
Who is 'you'? ?'uoy' si ohW

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