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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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05 Oct 2012, 08:14

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kingb wrote:

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

Notice that since 3, 4, and 7 are co-prime (they don't share any common factor but 1), then:

x must be a multiple of 4 and 7, so the least value of x is 28 (LCM of 4 and 7); y must be a multiple of 3 and 7, so the least value of y is 21; z must be a multiple of 3 and 4, so the least value of z is 12;

So, the least possible value of x + y + z is 28+21+12=61.

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

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02 May 2011, 18:19

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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.

\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\) _________________

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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15 Jun 2013, 13:47

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kingb wrote:

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

We know x,y and z are postive integers so based on given information, we can say x= 4/3y and z= 4/7y

then x+y+z= 4/3y+y+4/7y-------> (4/3y+4/7y)+ y ------> 40y/21 +y -----> 61 y/21

61y/21= Sum of positive integers x,y and z. This sum also is a postive integer. Let us say it as Integer "a"

therefoee 61y/21= a now this implies y= a*21/61 ----For y to be an integer, we see that "a" has to be multiple of 61 and the smallest multiple of 61 is 1

Hence ans should D i.e 61 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

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02 May 2011, 14:15

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 (Thanks in advance.) _________________

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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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29 Jul 2014, 12:37

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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]

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01 Feb 2016, 07:00

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If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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16 Mar 2016, 09:34

VeritasPrepKarishma wrote:

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.

\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\)

Excellent Approach i actually did it the other way Big takeaway from your Solution => IF YOU SEE ANY THREE OR MORE TERMS EQUATED => MAKE THEM EQUAL TO K . Thanks

gmatclubot

If x, y, and z are positive integers and 3x = 4y = 7z, then
[#permalink]
16 Mar 2016, 09:34

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