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If x, y, and z are positive integers and 3x = 4y = 7z, then the least
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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
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05 Oct 2012, 08:14
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 coprime (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
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02 May 2011, 18:19
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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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least
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16 May 2011, 17:51
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
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16 May 2011, 20:30
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
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05 Oct 2012, 08:40
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 = 284(y) = 84 > y = 217(z) = 84 > z = 1228+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
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14 Jun 2013, 05:26
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then the least
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15 Jun 2013, 13:47
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
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31 Jul 2014, 21:05
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
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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)




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