I have three sets of solutions:
10, 180
20, 90
30, 60
xy = 1800
2006-12-28 08:34:48
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answer #1
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answered by icehoundxx 6
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GCF = 10 = 2 * 5
=> x,y have 2,5 as their common factors(1)
LCM = 180 = 2 * 3^2 * 5
=> either x or y is a multiple of 3(2)
since(1) and (2), only x or only y is a multiple of 3, otherwise they must have 3 as their common factor
thus x = 2 * 5 ; y = 2 * 5 * 3^2 , x * y = 1800 ( or x = 2 * 5 * 3^2 and y = 2 * 5)
2006-12-28 10:09:13
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answer #2
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answered by James Chan 4
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Use prime factorizations.
The GCF of 10 means that the list of factors for each number will include a 2 and a 5, but no other common factors.
The LCM of 180 means that the total list of factors (excluding duplicates) will be 2, 2, 3, 3, 5
Since you know that you've excluded a 2 and a 5 (because they are the only duplicates), the factors for the product of x and y are 2, 2, 2, 3, 3, 5, 5; which multiply to 1800.
2006-12-28 08:46:45
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answer #3
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answered by Tim P. 5
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If you multiply the less common multiple and the greatest common factor, you will have xy.
Ana
2006-12-28 08:39:08
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answer #4
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answered by MathTutor 6
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gcf(x,y) . lcm(x,y) = 10 . 180 = 1800
we have a theorem that says that x.y = 1800 too.
The we don't have a unique solution for x and y. Any numbers x and y integer positive that x.y = 1800.
For example: (90,20), (18,100), (30,60) ...
2006-12-28 09:29:17
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answer #5
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answered by Jacaré 2
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Remember that you will have 2 min but you can think more of a due question if you think less of another one.
Remember this:
ab = mD
Ana
2006-12-28 08:34:13
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answer #6
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answered by Ilusion 4
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x = 10
y = 180
xy = 1800
2006-12-28 08:34:37
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answer #7
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answered by Dave 6
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xy = 1800
There is not a unique solution for x and y, but their product is unique.
Possile values for x and y are
(10,180)
(20,90)
2006-12-28 08:30:50
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answer #8
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answered by Northstar 7
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