If x, y, and z are distinct single-digit non-prime positive integers, and \sqrt {\frac{27}{125}} < \sqrt {\frac{9}{25}} < \sqrt {\frac{3}{5}} , what is the value of x + y + z? A. 13 B. 14 C. 15 D. 16 E. 17 M46-01

Possible values for x, y, z are: {1,4,6,8,9} Simplifying the given inequality, and, using the rule \sqrt {a^m} = a^{m/k} 27/125 = (3/5)^3 9/25 = (3/5)^2 3/5 = (3/5)^1 Left Term: \sqrt {(3/5)^3} = (3/5)^{3/9x} = (3/5)^{1/3x} Middle Term: \sqrt {(3/5)^2} = (3/5)^{2/4y} = (3/5)^{1/2y} Right Term: \sqrt {3/5} = (3/5)^{1/z} Inequality now is, (3/5)^{1/3x} 1/2y > 1/z Implies, 3x< 2y < z Trying cases from {1,4,6,8,9} to satisfy this simplified, (1,4,9) fits. 1 +4 + 9 = 14 Option B

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12 Days of Christmas GMAT Competition 2025 - Day 1: If x, y, and z

1 2 3 4

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The given inequality can be rewritten as (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z

Since 3/5 is between 0 and 1, any number raised to a higher power will reduce the value.

Therefore, 1/z < 1/2y < 1/3x

Rearranging the above, z>2y>3x

x,y, and z can be 1,4,6,8,9

z can only be 9 as 2y has to be minimum 8(when y=4)

x can only be 1 in order to satisfy the inequality.

Therefore, 9+4+1 =14.

Option B

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I'm not at all sure about this one, but here goes my Hail Mary:

First thing I see is that the three root numbers are the same as 3 / 5, just squared. So, I wrote 27 / 125 as 3^3 / 5^3; 9 / 25 as 3^2 / 5^2. Among these, if 27 / 125 needs to be the smallest number, and it is under-root 9x, by taking 3^3 / 5^3, so it would be 9x - 2 (to reduce 3^3 / 5^3 by two powers, down to 5/3). The smallest x I can take here that is non-zero, positive, and non-prime, is one. So I'm at 9*1 - 2 (x is taken as 1). For y, it will be 4*3 - 1 (we now minus 1, as it is 3^2 / 5^3 now), and to find a number bigger than 9, it will need to be 4*3 - 1 or 11. So y = 3. Finally, z. For this, we simply take 12, as it is the first number bigger than 11. 12 + 3 + 1 = 16.

I'm pretty doubtful about this one, but here goes an attempt either way.

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

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15 Dec 2025, 10:33

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If all terms are reduced to it's simplest exponent form , then the equality will become (3/5)^1/3x <(3/5)^1/2y < (3/5)^1/z . Now it can further be deduced that since base of all the terms are same and less than 1 then denominator of the fraction of the exponent should be as large as possible to the maximum value out of the term. In other ways , the equality will be simplified as 3x<2y<z. Now x ,y ,z will lie amongst single non prime positive integers which are 1,4,6,8,9. Only solution set will satisfy this that is x= 1 ,y=4 ,z=9. Therefore , x+y+z = 14

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Below is the solution.

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First thing to clarify is that in the question, they are not multiples like $9x * \sqrt{\frac{27}{125}}$ ... instead they are like $9x^{th}$ root of $\frac{27}{125}$, $4y^{th}$ root of $\frac{9}{25}$ and $z^{th}$ root of $\frac{3}{5}.$

as in $(\frac{27}{125})^{\frac{1}{9x}} , (\frac{9}{25})^{\frac{1}{4y}} , (\frac{3}{5})^{\frac{1}{z}}$

Single non-prime distinct digits => {$1, 4, 6, 8, 9$}
(1 is also neither prime nor composite, so we have to include 1 as well. In fact if you exclude 1, the least possible sum is 4+6+8 => 18, which is not even in the solution)

taking the given equation:
$(\frac{27}{125})^{\frac{1}{9x}} < \frac{9}{25}^{\frac{1}{4y}} < \frac{3}{5}^{\frac{1}{z}}$

$(\frac{3}{5})^{\frac{3}{9x}} < (\frac{3}{5})^{\frac{2}{4y}} < (\frac{3}{5})^{\frac{1}{z}}$

$(\frac{3}{5})^{\frac{1}{3x}} < (\frac{3}{5})^{\frac{1}{2y}} < (\frac{3}{5})^{\frac{1}{z}}$ --------- (1)

Note that $\frac{3}{5}$ is a value between 0 and 1 (0.6 to be precise)

Remember that when $0<a<1$,
$a < \sqrt{a} < \sqrt[3]{a}$
i.e.
$a < a^{\frac{1}{2}} < a^{\frac{1}{3}}$

let $a = \frac{3}{5}$, we have to choose $x, y, z$ such that the eq (1) is satisfied
$a^{\frac{1}{3x}} < a^{\frac{1}{2y}} < a^{\frac{1}{z}}$

so now we need to choose $x, y, z$ such that
$3x<2y<z$

we have 5 distinct options of {$1, 4, 6, 8, 9$} for each of $x, y, z$,
we can start with basic permutations where we assign smallest value to $x$, followed by $y$ and then followed by $z$ so that the chance of hitting the answer soon is maximum:

if $x=1, y=4, z=6$,
then
$1*3 < 2*4 > 6$
=> doesn't satisfy the constraint.

if $x=1, y=4, z=8$,
then
$1*3 < 2*4 = 8$
=> doesn't satisfy the constraint

if $x=1, y=4, z=9$,
then
$1*3 < 2*4 < 9$

$3 < 8 < 9$

=> satisfies the constraint.

So sum of 1, 4 and 9 => 14, which is option B

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My attempt:

Possible values for x, y and z are limited to: 0, 1, 4, 6, 8, 9.

Notice all of the numbers in the fractions can be rewritten under bases of 3 and 5.

Rewrite the cube roots as fractional exponents and simplify.

Remember that when we take the square root of x where 0 < x < 1, the square root of x will actually increase. Does this hold true as we increase the number of the root we are taking? Yes, it does. So as we increase the degree of the root we are taking (from square root to cube root, for example), the number should get even larger. Therefore, when written as fractional exponents, the denominators of the terms with the variables should be increasing from left to right. 3x < 2y < z .

Which numbers from the given possible values satisfy the inequality? Pick x = 1 y = 4, x = 9.

Should be 14.

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IMO Option B
As per Q, x,y, z could be : 0,1,4,6,8,9

and simplying the eq, 1/3x > 1/2y > 1/z (so, 0 cant be the value)
which implies: x+y+z could be : 1+4+9 = 14

Bunuel

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I have to just check whether by removing the root equation is satsified, accoringly take value of x, y and z to satisfy and get the answer by addition of those

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$/sqrt[9x]{27/125}$ < $/sqrt[4y]{9/25}$ < /sqrt[z]{3/5}

Can be written as (3/5)^3/9x < (3/5)^2/4y < (3/5)^1/z ==> (3/5)^1/3x < (3/5)^1/2y < (3/5)^1/z

As 3/5<1 => Greater power means smaller result. Hence,

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

x,y,z are one digit , non-prime ..... (2)

Only possible solution -> x=1 , y=4, z=9

Hence total - x+y+z = 14 => B is correct

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15 Dec 2025, 11:16

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Possible values for x, y, z are: {1,4,6,8,9}

Simplifying the given inequality, and, using the rule \sqrt[k]{a^m} = a^{m/k}

27/125 = (3/5)^3

9/25 = (3/5)^2

3/5 = (3/5)^1

Left Term: \sqrt[9x]{(3/5)^3} = (3/5)^{3/9x} = (3/5)^{1/3x}

Middle Term: \sqrt[4y]{(3/5)^2} = (3/5)^{2/4y} = (3/5)^{1/2y}

Right Term: \sqrt[z]{3/5} = (3/5)^{1/z}

Inequality now is,

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

Base = 3/5 <1 ; therefore the inequality is reveresed

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

Implies, 3x< 2y < z

Trying cases from {1,4,6,8,9} to satisfy this simplified, (1,4,9) fits.

1 +4 + 9 = 14

Option B

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15 Dec 2025, 11:31

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Let's Simplify ...we get {(3/5)^3}1/9x < {(3/5)^2}1/4y < (3/5)1^z-----Simplify.....3/5^1/3x < 3/5^1/2y < 3/5^1/z
Now remember for any number between 0 and 1 (Exclusive) ....the square is smaller than the number and the root is greater than the number.....and bigger the under root bigger is the number ......This means.....1/z < 1/2y < 1/3x .....Now since numerator is same...Denominator must be become larger to make the number smaller.....So z > 2y > 3x .......
Rewrite 3x < 2y < z

Now comes the challenge.....the number are single digit non prime positive integers only...so they can be 1,4,6,8,9
Now......We have to find the value of x + y + z ....and 3x < 2y < z...Lets do some hit and trial ...x be = 1 ..3x will be 3...then y cant be 1(because then 2y will be less than 3x) then it can be 4...2y becomes 8....z has to be greater than this 9 is the only option.....we get .... `1 + 4 + 9....= 14 ..We got in the first one here...i would stop here.....Wont waste any more time...although I feel to try out other options too...

Bunuel

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15 Dec 2025, 11:41

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Available integers 1, 4, 6 ,8 , 9

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

Now 3/5 is less than 1.
So value of z must be maximum z=9
Now, 2y<z, possible only for y=4
& 3x<2y iff x=1

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

Ans is 14

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There are two key inferences to be made here
First one is regarding the numbers
Distinct, single digit, non-prime integers
There are only five
1 , 4 , 6, 8, 9
One is a trap here since we consider 1 as , nor composite nor prime, Some might get confused, but it is non-prime.

No, the second inference to make is we need to equalize the base to make a comparison.

9xth root of 27/125 can be written as (27/125)^ 1/9x = (3/5)^ 3/9x= (3/5) 1/3x
the other numbers become (3/5)^1/2y and (3/5)^z
3/5 is a fraction so it becomes bigger every time you take a route of it or cubit or so on.
So now, as the base is common, we get

3x<2y<z
only combination that works is
x=1 y= 4 and z=9
x+y+z= 14

Quote:

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15 Dec 2025, 12:11

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B. 14. 1,4,9 are those non-prime numbers

Bunuel

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Bunuel

Given that x, y, z are DISTINCT , SINGLE DIGIT, NON PRIME positive integers.

So, x,y,z has to be greater than 0 and less than 10.

The possible values of x,y,z are : (1,4,6,8,9)

9x th root of (27/125):

27 = 3^3 , and 125 = 5^3

(27/125)^(1/9x) = [(3^3)/ (5^3)]^{1/9x} = (3/5)^ (1/3x)

4y th root of (9/25) :

9 = 3^2 , and 25 = 5^2

(9/25)^(1/4y) = [(3^2)/(5^2)]^{1/4y} = (3/5)^(1/2y)

The third term = (3/5)^(1/z)

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

For this condition to be satisfied, the powers have to be greater.

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

Multiply by 6xyz, we get

3x < 2y < z

We know the values of x,y,z should be amongst (1,4,6,8,9)

If we put x =4, then y has to be 6. Both terms will be equal.

So, x = 1 , y = 4 and z = 9

3< 8<9

Satisfied.

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

Option B

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15 Dec 2025, 12:58

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cORRECT Answer: 14

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15 Dec 2025, 12:59

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Given

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

condition => x, y, z - distinct positive signle digit non prime integers

((3/5)^3)^9x/2 < ((3/5)^2)^4y/2 < (3/5)^z/2

(3/5)^27x/2 < (3/5)^8y/2 <(3/5)^z/2

here base = 3/5, where 3/5 lies between 0 and 1 i.e., 0< 3/5 < 1, hence powers are written as following

27x/2 > 8y/2 > z/2

cancel denominator, we have, 27x > 8y > z

now we have distinct single positive non prime integers = 1, 4, 6, 8, 9 and x <> y <> z (<> - not equal)

if we consider x = 4 , then 4 * 27 = 108 > 8y > z

if we can try considering y = 9 or 8 or 6 or 1, where in almost all cases, 108 > 8y [ 72 or 64 or 48 or 4 ]

when x = 4 and y = 9 , then z = 1 based on options x + y +z = 14

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Bunuel

To make 9x a single digit unit means x has to be 1, z can take 9 while can either take 4 or 8 but only 4 works given that 8 would take the sum to 18 which is not among the answers hence 1+9+4= 14

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15 Dec 2025, 13:27

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The possible values of x and y given the conditions (distinct, single digit, non-prime, positive integer) are - 1,4,6,8,9

Inequality is
(3/5)^(3/9x) < (3/5)^(2/4y) < (3/5)^(z)

As 3/5 <1, larger exponent would mean smaller value.

Therefore as per the inequality provided, (3/9x) > (2/4y) > z
On simplifying, (1/3x) > (1/2y) > z
On inversing, 3x < 2y < z

If we try possible values listed above, only one of them will satisfy this condition, which is present in the option between 13 to 17. And that is (1,4,9)

1+4+9 = 14
The answer is (B)

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15 Dec 2025, 15:48

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Simplify the expression:

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

9x[27][/125] = (3/5)^1/3x
4y[9][/125] = (3/5)^1/2y
z[3][/5] = (3/5)^1/z

Since 3/5 is less than 1, the inequality reverses when comparing the exponents

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

3x<2y<z

if x=1, 3x=3,
if y=4, 2y=8
=> z>8 so z=9

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

Any other values will exceed 17 and hence 14 is the only possible choice.

Bunuel