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Greatest Common Factor and Least Common Multiple


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A while ago, I gave my Pre-algebra class a puzzle problem: "Given two natural numbers, m and n. If their GCD is G=6 and their LCM is L=72, what are the numbers?"
a) What were all possible (m, n) for G=6 and L=72 (m<=n)?
b) What's the smallest sum, m+n, for any (m, n) pair that share the same G=gcd and L=lcm (with another m<=n; G>=2)?
c) Find the (G, L) pair with the most solutions (m, n) for the same G=gcd(m,n) and L=lcm(m,n) (G>1, L < 1001).

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a) (18, 24) and (6, 72)

b) If G=2 and L=2, then m+n=2+2= 4. I'm not sure if that's what you were asking. :wacko:

c) I think (2, 420), (3, 630), and (4, 840) all work...

Sorry if I misunderstood the question. :(

for B, of pairs of whole numbers, what is the smallest m+n where (m,n) and (p,q) have the same lcd and gcf?

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a) (18, 24) and (6, 72)

b) If G=2 and L=2, then m+n=2+2= 4. I'm not sure if that's what you were asking. :wacko:

c) I think (2, 420), (3, 630), and (4, 840) all work...

Sorry if I misunderstood the question. :(

for B, of pairs of whole numbers, what is the smallest m+n where (m,n) and (p,q) have the same lcd and gcf?

Oh, I see. So DeGe pretty much had it.

G =

2

L = 6

Pairs are (2,6) and (2,3)

Smallest sum then is 2+3 = 5

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a) (18, 24) and (6, 72)

b) If G=2 and L=2, then m+n=2+2= 4. I'm not sure if that's what you were asking. :wacko:

c) I think (2, 420), (3, 630), and (4, 840) all work...

Sorry if I misunderstood the question. :(

for B, of pairs of whole numbers, what is the smallest m+n where (m,n) and (p,q) have the same lcd and gcf?

Oh, I see. So DeGe pretty much had it.

G =

2

L = 6

Pairs are (2,6) and (2,3)

Smallest sum then is 2+3 = 5

M, n, p, and q are distinct

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Wow. I made a really stupid mistake. That's embarrassing.

G = 2

L = 12

The pairs are (2, 12) and (4, 6).

The smallest sum is then 4+6=10.

a) (18, 24) and (6, 72)

b) If G=2 and L=2, then m+n=2+2= 4. I'm not sure if that's what you were asking. :wacko:

c) I think (2, 420), (3, 630), and (4, 840) all work...

Sorry if I misunderstood the question. :(

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