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

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-----Original Message-----

From: Marcel Martin [mailto:

znz@...]

Sent: 04 October 2001 23:46

Cc: Prime Numbers

Subject: Re: [PrimeNumbers] Brain Teaser

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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