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If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink ]
19 Oct 2006, 09:33

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If x, y, and z are positive integers and 3x = 4y = 7z, then the least possible value of x + y + z is
(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

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The answer should be a (D).The LCM of (3,4,7) will be the least number when all three will be equal which happens when

x=28

y=21

z=12

The sum of all three is 61.

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3x = 4y = 7z
Henec y=3*x/4 , z= 3*x/7
Hence x+y+z = 61*x/28.Hence answer has to be a multiple of 61.Which is (D)

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sangarelli wrote:

The answer should be a (D).The LCM of (3,4,7) will be the least number when all three will be equal which happens when x=28 y=21 z=12 The sum of all three is 61.

and how do you come up with those numbers? pick numbers? It would be time consuming....

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No you just find the least common multiple of 3, 4, and 7 (which is 28) and then solve each equation and add.

Director

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Got it at last

I am slow

Had to clarify what the Least Common Multiple means in my language... Another gap in my knowledge was repaired , that is good

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burningskulls wrote:

No you just find the least common multiple of 3, 4, and 7 (which is 28) and then solve each equation and add.

Least common multiple cannot be 28...as 3 is not divisible by 28.

You need to find the least common multiple of 3, 4, 7 whch is 84 and then you get

x=28

y=21

z=12

= 61

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Easier method to solve this [#permalink ]
20 Oct 2006, 11:27

3x = 4y = 7z
Thus will only work when 3 numbers are not divisioble by each other
x will be min 4*7=28
y will be 3*7=21
z will be 3*4=12
Add them up.
If the numbers where divisble by even 1 other number
LCM would be better way to go

Easier method to solve this
[#permalink ]
20 Oct 2006, 11:27

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