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805+ (Hard)|   Number Properties|   Roots|         
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Veerenk
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 17:41
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Answer can be A or B or C or D

It may needs to be rephrased, which can not be x + y + z?
Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more

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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 18:52
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Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more
Write the given fractions with same base i. e 3/5, 0.6

(0.6)^(3/9x) < (0.6)^(2/4y) < (0.6)^z

(0.6)^(1/3x) < (0.6)^(1/2y) < (0.6)^z

Since for proper fractions positive powers make them smaller relation between them will be:

1/3x > 1/2y > z

z > 2y > 3x

Given x, y, z are single digit non primes which are to be chosen from {1,4,6,8,9}

z=9, y=4, x=1 satisfies this equation

9 > 2*4 > 3*1

9+4+1 = 14

Correct answer B
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 20:02
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(27/125)^(1/9x) = (3/5)^(1/3x)
(9/25)^(1/4y) = (3/5)^(1/2y)
(3/5)^(1/z)

As 3/5<1 then the inequility is true when

1/3x > 1/2y > 1/z

which is true when (as x, y and z are positive)

3x < 2y < z

single-digit non-prime positive integers: 1, 4, 6, 8, 9

x must 1 (if x is greater than 1, for example 4, then it is impossible that 3x < z with the given values)
y must be 4 (if y is greater than 4, for example 6, then it is impossible that 2y < z with the given values)
z must be 9

x + y + z = 1 + 4 + 9 = 14

IMO B
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 20:18
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The expression can be simplified because the 9x root of something is equal to the 3x root of the cube root and the 4y root of something is equal to the 2y root of the square root.

(0.6)^(1/3x) < (0.6)^(1/2y) < (0.6)^(1/z)

They have the same base and it is less than 1:

(1/3x) > (1/2y) > (1/z)

It occurs only when:

3x < 2y < z

Possible values for x, y and z are 1, 4, 6, 8 and 9

The only combination that matches is x=1, y=4 and z=9

x+y+x=14

Answer B
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 20:45
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27=3^3 and 125=5^3 -> 9x root of 27/125 = 3x root of 3/5
9=3^2 and 25=5^2 -> 4y root of 9/25 = 2y root of 3/5
z root of 3/5

a^(1/x) is a increasing function when a<1

3x<2y<z

The single-digit non-prime positive integers are 1,4,6,8,9
There is only one solution for (x,y,z)=(1,4,9) whose sum is 14

The answer is B
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 21:05
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Taking the cube root in the first term and the square root in the second term, the expression is equivalent to:

3x root (3/5) < 2y root (3/5) < z root (3/5)

which is equivalent to:

1/3x > 1/2y > 1/z

which is equivalent to (all the numbers are positive):

3x < 2y < z

Values for x, y and z: 1,4,6,8,9

The answer must be x = 1, y = 4 and z = 9 and their sum is 14

The correct answer is B
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Answer C simplify the roots as if they were the equation and make the contents on the square root the same.
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 22:18
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Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more
Non prime ={1, 4, 6 , 8,9}
(3/5)^3*1/9x < (3/5)^2*1/4y < (3/5)^1/z
(3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z

3x<2y<z

when x= 1 y=4 z=9
3<8<9
true

when x= 4 y=6 z=8
12<12<8 false
therefore x+y+z = 1+4+9 =14
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 22:28
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Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more
x, y, z are distinct single-digit non-prime positive integers
Set: {1,4,6,8,9}
[9x][/27/125] < [4y][9/25] < [z][/3/5]

Rewrite everything with the same base:
(27/125) = (3/5)^3, (9/25) = (3/5)^2
(3/5)^3/9x < (3/5)^2/4y < (3/5)^1/z
Sine (3/5)<1, inequality sign reverses
1/3x > 1/2y > 1/z
1/3x > 1/2y = 2y > 3x
1/2y > 1/z = z> 2y
3x < 2y < z
Hit & Trial:
x = 1; y = 4; z = 9
3 < 8 < 9 (Satisfied)

Hence, x+y+z = 14

Note: You can try other numbers from the set too, but the condition matched on this triplet only.
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Let us rewrite the terms as follows:
(3/5)^(3/9x) < (3/5)^(2/4y) < (3/5)^(1/z)
this implifies to
(3/5)^(1/3x) < (3/5)^(1/2y) < (3/5)^(1/z)
Since 3/5 is less than one, the greater the denominator in the power (the greater the root), the greater the number.
So we need to find numbers that satisfy 3x<2y<z
Considering all non prime positive single digit numbers: 1 4 6 8 9
We can see that only the combination x=1. y=4, z=9 satisfies the inequality.
Hence x+y+z=1+4+9=14
Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 22:53
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(27/125)^1/9x can be readjusted as (3/5)^3*1/9x = (3/5)^1/3x
Similarly (9/25)^1/4y can be readjusted as (3/5)^2*1/4y= (3/5) ^1/2y
And lastly (3/5)^1/z

These satisfy the inequality (3/5)^1/z>(3/5)^1/2y>(3/5)^1/3x
Since 3/5 is a number less than 1 & we know that the lesser the powers to which 3/5 is raised the greater the resulting number, it follows that
1/z<1/2y<1/3x - Condition 1

Now we know that all of the numbers need to be distinct single digit non prime positive numbers which leaves us with the following options (1,4,6,8,9). If we put the corresponding values for x,y & z we would find that if z = 9 it would assume its lowest possible value. So if z=9, then y could be either 4 or 1 since anything above that would make it lower than 1/z. If we put 1 then 1/3x would be bigger hence y =4. Now that z=9 & y=4, to satisfy condition 1, x can only assume the value 1.

Hence z=9,y=4 & x= 1 & their sum comes out to be 14, which is option (B)
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 22:58
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Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more
Lets rewrite,

27/125 = (3/5)^3
9/25 = (3/5)^2
3/5 = (3/5)^1

Then,

=> (27/125)^1/9x < (9/25)^1/4y < (3/5)^1/z
=> (3/5)^3/9x < (3/5)^2/4y < (3/5)^1/z
=> (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z

Since 0 < 3/5 < 1 => The higher the power for 3/5, the lesser the value,

=> 1/3x > 1/2y > 1/z
=> 3x < 2y < z

Given,
x, y, z are distinct single digit positive non-primes. i.e., possibilities : {1, 4, 6, 8, 9}

We maximum value we can assign for y is 4, since 2y = 2*(4) = 8 and there's only one value greater than 8 which we can assign to z
so, z = 8
and x = 1

Then,
3x (3) < 2y (8) < z (9)

Therefore, x + y + z = 1 + 4 + 8 = 13. Answer -> A) 13
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 23:14
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(27/125)^1/9x < (9/25)^1/4y< (3/5)^1/z can be written as (3/5)^1/3x<(3/5)^1/2y<(3/5)^1/z

as 0<3/5<1 ; 1/3x>1/2y>1/z
x,y,z is a non prime positive single digit integer i.e 1,4,6,8,9.
The only combination that can satisfy is z=9, y=4, z=1
Sum=9+4+1=14
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 15 Dec 2025, 23:47
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The answer is choice : B since 3x<2y<z and and the smallest single digit positive non-prime numbers that can satisfy this inequality is 3*1<2*4<9=>x=1,y=4 and z=9. So, the total of x+y+z=14
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Please check answer in the screenshot attached
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Non-prime single digit numbers are (1,4,6,8,9)
9x✓(27/125) < 4y✓(9/25) < z✓(3/5)
Can be written as
(27/125)^1/9x < (9/25)^1/4y < (3/5)^1/z
=> (3/5)^3/9x < (3/5)^2/4y <(3/5)^1/z
=> (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z
Since the value lie between 0 and 1
Considering the power as
1/3x > 1/2y > 1/z
Taking reciprocal
3x<2y<z

Now consider x= 1 and highest value to z = 9
Then y can only be 4 to make the power condition true
Thus 1+4+9 = 14
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If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17

convert the power to base
(3^3/5^3) ^x <(3^2/5^2)^y <(3/5)^z
base is same equality reverses
3x>2y>z

x,y,z are single digit non prime integers
x=6 , y= 8, z =1 ; 18>16>1
sum is 15
OPTION C; 15
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 16 Dec 2025, 02:24
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Bunuel
If x, y, and z are distinct single-digit non-prime positive integers, and \(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\), what is the value of x + y + z?

A. 13
B. 14
C. 15
D. 16
E. 17
Show more

\(\sqrt[9x]{\frac{27}{125}}\) = \(\sqrt[3x]{\frac{3}{5}}\)

\(\sqrt[4y]{\frac{9}{25}}\) = \(\sqrt[2y]{\frac{3}{5}}\)

The third equation does not need to be reduced

All these fractions are less than 1 hence positive roots of these numbers are greater than the original base. Hence, we can figure out that 3x < 2y < z

Now we know that x, y and, z belong to set [1, 4, 6, 8, 9]

since z is the largest, we can see from the set above that z can have a maximum value of 9.
However, we also see that 3x < 9 and 2y < 9
=> x < 3 and y < 4.5
Hence x < 3
Since we also have to satisfy and these are distinct numbers, x can only be 1 and y can only be 4 since 3x < 2y => 3*1 < 2*4 (satisfying condition)

Now since z > 2y => z > 8 the only option is z = 9

Hence x + y + z = 9 + 4 + 1 = 14
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 16 Dec 2025, 02:50
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The only single-digit non-prime positive integers are 1, 4, 6, 8, 9

Simplifying the inequality, we get:

\(3x\sqrt{\frac{3}{5}}<2y\sqrt{\frac{3}{5}}<z\sqrt{\frac{3}{5}}\)

Since the base \(\frac{3}{5}\) is between 0 and 1, the inequality for the exponents is reversed:

\(\frac{1}{3x}>\frac{1}{2y}>\frac{1}{z}\). And that means 3x < 2y < z

The only combination that fits this is x=1, y=4 and z=9

x+y+z=14
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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z 16 Dec 2025, 03:40
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Remember, 0.5^(1/2) =0.7, 0.5^(1/3) =0.794

Simplify the equation, you will get :

(3/5)^(1/3x) < (3/5)^(1/2y) < (3/5)^(1/z)

=>1/3x < 1/2y < 1/z [Taking the reference of 0.5 values above]

That means only one combination possible from {1,4,6,8,9} and that is z=9, y=4, x=1.

Answer : x+y+z = 14
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