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

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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, 13:35
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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 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.

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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New post 02 May 2011, 18:19
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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 16 May 2011, 17:51
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given 3x=4y=7z

x+y+z in terms of x

= x+(3x/4)+(3x/7) = 61x/28

now checking with each of the answers and see which value gives a minimum integer value.

A x = 28*33/61 , not an integer

B,C can be ruled out similarly.

D is minimum value as x = 61*28/61 = 28

Answer is 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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New post 16 May 2011, 20:30
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3x = 4y = 7z

LCM = 3 * 4 * 7 = 84
84 / 3 = 28
84 / 4 = 21
84 / 7 = 12

28 + 21 + 12 = 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 05 Oct 2012, 08:40
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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.

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

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New post 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
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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 31 Jul 2014, 21:05
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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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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least  [#permalink]

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New post 09 Oct 2019, 09:50
MSDHONI 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


1) We should minimize z while keeping 7z divisible by a multiple of 3 and 4 (which means z must be even, since 7 is odd and we need divisibility by an even number)

2) 7z must be divisible by 12 because 3x and 4y will have 12 as their LCM. First multiple of 7 divisible by 12 is 84, which is trap AC E) --> we need x+y+z NOT the LCM of all 3

3) 84/3 = 28, 84/4 = 21, 84/7 = 12
28 + 21 + 12 = 61, D)
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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least   [#permalink] 09 Oct 2019, 09:50
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