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# What is the greatest common factor of "a" and "b"? 1) The

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Manager
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What is the greatest common factor of "a" and "b"? 1) The [#permalink]  15 May 2003, 05:38
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What is the greatest common factor of "a" and "b"?

1) The greatest common factor of a/2 and b/2 is 5
2) The greatest common factor of 2a and 2b is 10
SVP
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An interesting one.
I think the problem works for the least common multiple as well.
Manager
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ya, I think it works for LCM also

Can we generalise

that if GCF of "x*a" and "x*b" is Z, then GCF of "a" and "b" will be x*Z,

where x is a positive integer or fraction.

similarly for LCM
SVP
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You mean
If GCD of AX and BX is ZX, then GCD of A and B is Z, and vice versa.
I think it should be the same for LCM.
Looks like a new theorem.
Senior Manager
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This one is for Stolyar or brstorewala [#permalink]  06 Feb 2004, 08:54
brstorewala wrote:
ya, I think it works for LCM also

Can we generalise

that if GCF of "x*a" and "x*b" is Z, then GCF of "a" and "b" will be x*Z,

where x is a positive integer or fraction.

similarly for LCM

Applying this rule, statement (1) and (2) will give different values for "GCF of a and b". Are you sure about this?
Director
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Re: This one is for Stolyar or brstorewala [#permalink]  06 Feb 2004, 15:46
gmatblast wrote:
Applying this rule, statement (1) and (2) will give different values for "GCF of a and b". Are you sure about this?

Yes ! The answers is different for (1) & (2). But, the concept, IMO,
should be OK!
Manager
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Probably the following might be helpful. I hope you won't say 'what are you trying to demonstrate here' for adding my reply.

let a = p1^k1 * p2^k2 * ... pn^kn
b = p2^m1 * p2^m2 * ... pn^mn

gcd(a,b) = p1^min(k1, m1) * p2^min(k2, m2) * ... pn^min(kn, mn)
lcm(a,b) = p1^max(k1, m1) * p2^max(k2, m2) * ... pn^max(kn, mn)

min(c, d)+max(c,d) = c +d

hence gcd(a,b)*lcm(a,b) = a*b
Director
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Zhung Gazi wrote:

min(c, d)+max(c,d) = c +d

I've lost you at this step. Would you elaborate?
Manager
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if c < d then min(c,d) = c , max(c,d) =d min(c,d)+max(c,d) = c+d
if c >= d then min(c,d) = d, max(c, d) = c and min(c,d)+max(c,d) = c+d

max(c,d) - min(c,d) = |c-d|

Further you can have algebraic equivalents as follows
max(c,d) = (1/2)(c+d+|c-d|)
min(c,d) = (1/2)(c+d-|c-d|)
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