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Bunuel
If 6x = 8y = 14z, then what is a possible sum of positive integers x, y, and z?

A. 52
B. 58
C. 84
D. 122
E. 168

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The possible value for which 6x = 8y =14z ,will be LCM of 6,8,14 i.e. 168
Therefore x = 28, y = 21, z = 12
sum = 61.
But we don't have 61 in the ans choices. Only possible values which satisfy the eqn given will be multiples of 168.
We move onto the next multiple(336), which also doubles the value of x, y and z.
Now x+y+z = 122, which is present in the answer choice.
Hence D.

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Bunuel
If 6x = 8y = 14z, then what is a possible sum of positive integers x, y, and z?

A. 52
B. 58
C. 84
D. 122
E. 168

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x : y
8 : 6

y : z
14 : 8

x : y : z
56 : 42 : 24

x+y+z = 122 could be possible sum
IMO D
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Bunuel
If 6x = 8y = 14z, then what is a possible sum of positive integers x, y, and z?

A. 52
B. 58
C. 84
D. 122
E. 168

We can divide the given equation by 2 and we have:

3x = 4y = 7z

In order for each term to be equal, x could be 4(7) = 28, y could be 3(7) = 21, and z could be 3(4) = 12.

Thus, we see that the smallest value of x + y + z is 28 + 21 + 12 = 61. However, since that is not an answer choice, the next highest value is 2 x 61 = 122.

Answer: D
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6x = 8y = 14z → x = 14z/6 = 7z/3 and y = 14z/8 = 7z/4
Sum = x+y+z = 7z/3 + 7z/4 + z → z (7/3 + 7/4 +1) = 61z / 12

Because the sum of integers is also an integer. z must be divisible by 12.
Possible values of z = 12, 24, …

Only option that satisfies is 122 (when z = 24)
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3x=4y=7z=k
Sum=k/7+k/4+k/3= 61k/84

Only option with a multiple of 61 is D)
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