The average (arithmetic mean) of the positive integers x, y, and z is 3
We can write: \(\frac{x+y+z}{3}=3\)

Multiply both sides of the equation by \(3\) to get: \(x+y+z=9\)

At this point we need to find the greatest possible value of \(z\)
Important: Keep in mind that the three numbers are DIFFERENT POSITIVE INTEGERS

Since we know that \(x+y+z=9\), we can MAXIMIZE the value of \(z\) by MINIMIZING the values of \(x\) and \(y\)
We know that \(x\) is the smallest value.
Since \(x\) must be a positive integer, the smallest possible value of \(x\) is \(1\)
Since \(y\) must be different from \(x\), the smallest possible value of \(y\) is \(2\)
At this point we have maximized the value of \(z\)

If \(x=1\) and \(y=2\), then \(z=6\)

Answer: B

Cheers,
Brent

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The smallest positive integer is 1 and 2nd smallest 2.
So, 3*3 = 3+x.
Hence, x= 6. B is the correct choice.

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Solution

Given:

• The average (arithmetic mean) of the positive integers x, y, and z is 3
• Also, x < y < z

To find:

• The greatest possible value of z

Approach and Working:
As the average is 3,

• Then sum of all 3 = 3 * 3 = 9

Also, x < y < z and each are positive integers.

To maximise z, we should take minimum value of x and y

• Minimum possible x = 1
• Minimum possible y = 2
• Therefore, maximum possible z = 9 – (1 + 2) = 6

Hence, the correct answer is option B.

Answer: B
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Given

\(\frac{x + y + z}{3} = 3\)

x + y+ z =9

In order to maximize z we must minimize x and y. Note: x<y<z and all are positive integers. So the minimum value we have is 1.

1 + 2 + 6 = 9

The best answer is B.

\(x + y + z = 9\)

We have to find the max value of z is minimizing the value of x & y

Since, \(x < y < z\), minimizing x & y will be -

x < y < z = 1 < 2 < 6 where x + y + z = 9, Thus Answer must be (B)
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Bunuel so zero isn't counted as an integer? This always confuses me.

0 is an integer but it's neither positive nor negative.


ZERO.

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check for more below threads:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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[quote="dvishal387"]The smallest positive integer is 1 and 2nd smallest 2.
So, 3*3 = 3+x.
Hence, x= 6. B is the correct choice.


Why does everyone automatically assume that X and Y are different integers? I know that OG solution says that, but here every solution implies this by default. Is this a rule for this type of question - assume X and Y are different unless specified otherwise?

\(\frac{x+y+z}{3} = 3\)

\(x+y+z = 9\)

\(z = 9-(x+y)\)

Since we need the greatest possible value of z, we have to minimize x and y

\(x_{min} = 1\); \(y_{min} = 2\)

\(z_{max}= 9-(1+2) = 6 \)

Given: The average (arithmetic mean) of the positive integers x, y, and z is 3.

Asked: If x < y < z, what is the greatest possible value of z ?

x + y + z = 9 & x < y < z
For z to be greatest, x & y have to be least.
Since x, y & z are positive integers
x = 1; y = 2; z = 6

IMO B
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In order to increase the value of Z to its maximum we will need to reduce the values of x and y to its lowest.

Also as given in the question:

(x + y + z) / 3 = 3

The minimum value for x and y can be 1 and 2, hence

(1 + 2 + z)= 9
z= 9 - 3 = 6

We can create the equation for the average of the three numbers:

(x + y + z)/3 = 3

x + y + z = 9

Since z is the largest integer and we want it to be as large as possible, we can let x = 1 and y = 2, and thus the largest value z could be is 6.

Answer: B
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The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

x + y + z = 9

To find the greatest possible value of z, we need to minimize x and y. Remember that 0 is not a positive integer. Let x = 1 and y = 2. Thus, z = 6.

Answer is B.
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