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If x, y, and z are positive integers and 3x = 4y = 7z, then the least [#permalink]
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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 [#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 the least [#permalink]
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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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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. \(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\)
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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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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.
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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 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 [#permalink]
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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 = 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 [#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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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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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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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]
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30 Aug 2016, 07:21
l.c.m of 3,4,7 is 84. 84/3,84/4,84/7 are 28,21,12 respectively. sum of 28, 21,12 is 61. therefore D is answer.



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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]
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18 Dec 2016, 17:14
Excellent Question.
Here is what i did in this one =>
3x=4y=7z Hence x must be divisible by both 4 and 7 => least x = LCM(4,7)=28 Hence x=28 => y=21 =>12
Hence x+y+z= 61
Hence D
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If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]
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31 May 2017, 19:57
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
4y = 3x so, y = 3x/4 7z = 3x so, z = 3x/7 x + y + z= x+3x/4+3x/7 =(28x+21x+12x)/28 =61x/28 It’s given that x,y and z are positive integers, so x+y+z must be a positive integer. So, 61x/8 must be a positive integer. 61 and 28 does not have any common divisor, so we cannot break it down further. In that case x must be divisible by 28 and the least value of x could be 28. So, x+y+z = 61x/28 = 61*28/28 = 61 Answer is D




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