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# x+y+z

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Manager
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Kudos [?]: 41 [0], given: 2

x+y+z [#permalink]  02 May 2011, 12:35
00:00

Difficulty:

(N/A)

Question Stats:

36% (02:21) correct 64% (00:51) wrong based on 12 sessions

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Kudos [?]: 8493 [1] , given: 194

Re: x+y+z [#permalink]  02 May 2011, 17:19
1
KUDOS
Expert's post
MSDHONI wrote:

When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'.

$$3x = 4y = 7z = k$$

$$x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7}$$
Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.

$$x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61$$
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Kudos [?]: 46 [0], given: 27

Re: x+y+z [#permalink]  02 May 2011, 13:15
I believe it's D. The solution is the lowest common multiplier. Both 3 and 7 are prime numbers and odd, while 4 is a multiple of 2, which doesn't help with the other two. So if you multiply 3, 4, and 7, you get 84. If you deduce further, x = 28, y = 21, z = 12, just the product of the 2 numbers left after the reference. So the sum is 61.

If there's any hole in my reasoning, I welcome comments. As I can benefit from it too
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Kudos [?]: 385 [0], given: 36

Re: x+y+z [#permalink]  02 May 2011, 17:17
x + y + z = 3x/4 + x + 3x/7

= (21 + 28 + 12 )x/28 = 61x/28

61 is not divisible by 28(it's a prime #), so for least value, x = 28

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Re: x+y+z [#permalink]  03 May 2011, 00:53
x= (4/3)y = (7/3)z
x+y+z = (61/28)x. Since 61 is a Prime number,x has to be 28.
thus 61.
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Re: x+y+z   [#permalink] 03 May 2011, 00:53
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