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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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02 May 2011, 12:35
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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

[Reveal] Spoiler:
[Reveal] Spoiler: OA

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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, 07:14
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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.

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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, 17:19
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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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Kudos [?]: 211 [2] , given: 2

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

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05 Oct 2012, 06:53
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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, 07:14, 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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05 Oct 2012, 07:40
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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 [#permalink]

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15 Jun 2013, 12:47
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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 [#permalink]

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31 Jul 2014, 20:05
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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

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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, 13:15
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
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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, 17:17
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

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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, 00:53
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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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then [#permalink]

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05 Oct 2012, 07:37
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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14 Jun 2013, 04:26
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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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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, 06:21
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.
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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, 16:14
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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Re: If x, y, and z are positive integers and 3x = 4y = 7z, then   [#permalink] 18 Dec 2016, 16:14
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