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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 ?

Explanation	Calculations	Help
$We translate the statement about the average into an equation.$	$\frac{x + y + z}{3} = 3$	Theory & Tactics Method Card TRANS1-B Click to view full details
$We solve this equation to find the sum of the three integers.$	$x + y + z = 9$
$We need to maximize z given x < y < z and that x, y, z are positive integers, so we choose the smallest possible values for x and y .$	$x = 1; y = 2$
$We substitute x and y into the sum to find z .$	$z = 9 - 1 - 2 = 6$	Theory & Tactics Method Card EQUA3-A Click to view full details

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6

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 ?

Explanation	Calculations	Help
$We translate the statement about the average into an equation.$	$\frac{x + y + z}{3} = 3$	Theory & Tactics Method Card TRANS1-B Click to view full details
$We solve this equation to find the sum of the three integers.$	$x + y + z = 9$
$We need to maximize z given x < y < z and that x, y, z are positive integers, so we choose the smallest possible values for x and y .$	$x = 1; y = 2$
$We substitute x and y into the sum to find z .$	$z = 9 - 1 - 2 = 6$	Theory & Tactics Method Card EQUA3-A Click to view full details

Scroll horizontally to view all columns

6