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If x, y, and z are positive integers and 3x = 4y = 7z, then

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

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

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New post 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 the least [#permalink]

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New post 02 May 2011, 18:17
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

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

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New post 03 May 2011, 01:53
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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If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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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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Last edited by Bunuel on 05 Oct 2012, 08:14, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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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.

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

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New post 05 Oct 2012, 08:37
i did wrong..:(..

i was luking at the choice that must be multiple of all 3 4 and 7.. so i tuk 84 ..

Thanks bunuel :)..

bunuel..is there any other way too to do this question ?
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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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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


The LCM of 3, 4, and 7 = 84. (Easy to figure out - since 3, 4, and 7 are prime, the LCM is 3*4*7)

3(x) = 84 -> x = 28
4(y) = 84 -> y = 21
7(z) = 84 -> z = 12

28+21+12 = 61.

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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 [#permalink]

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

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LCM of 3, 4 & 7 is 84

Least value of x+y+z

\(= \frac{84}{3} + \frac{84}{4} + \frac{84}{7}\)

= 28 + 21 + 12 = 61

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

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New post 16 Mar 2016, 09:34
VeritasPrepKarishma wrote:
MSDHONI wrote:
Image


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