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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
A. 13
B. 14
C. 15
D. 16
E. 17
12 Jan 2026, 05:40
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Official Solution:
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
Notice that \(\frac{27}{125} = (\frac{3}{5})^3\) and \(\frac{9}{25} = (\frac{3}{5})^2\), so:
\(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[(3*3x)]{\frac{27}{125}} < \sqrt[(2*2y)]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[3x]{(\frac{27}{125})^{\frac{1}{3}}} < \sqrt[2y]{(\frac{9}{25} )^{\frac{1}{2}}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}\)
Here’s the key idea: when you take a positive integer root of a number between \(0\) and \(1\), the result becomes larger than the original number, and the higher the root, the closer the value moves toward \(1\).
For example, \(\sqrt{\frac{1}{2}} < \sqrt[3]{\frac{1}{2}} < \sqrt[4]{\frac{1}{2}} < \sqrt[5]{\frac{1}{2}}...\)
Thus, \(\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}\) implies:
\(3x < 2y < z\)
Since \(x\), \(y\), and \(z\) are distinct single-digit non-prime positive numbers, they can be selected only from \(\{1, 4, 6, 8, 9\}\).
Now, \(x\) cannot be \(4\) or greater, because in that case \(3x\) becomes \(12\) or greater and cannot be less than \(z\), which is at most \(9\). Thus, \(x = 1\).
Similarly, \(y\) cannot be \(6\) or greater, so \(y = 4\).
And finally, this forces \(z\) to be \(9\).
Thus: \(x = 1\), \(y = 4\), and \(z = 9\).
\((3 * 1) < (2 * 4) < 9\)
Therefore, \(x + y + z = 1 + 4 + 9 = 14\).
Answer: B
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
Notice that \(\frac{27}{125} = (\frac{3}{5})^3\) and \(\frac{9}{25} = (\frac{3}{5})^2\), so:
\(\sqrt[9x]{\frac{27}{125}} < \sqrt[4y]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[(3*3x)]{\frac{27}{125}} < \sqrt[(2*2y)]{\frac{9}{25}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[3x]{(\frac{27}{125})^{\frac{1}{3}}} < \sqrt[2y]{(\frac{9}{25} )^{\frac{1}{2}}} < \sqrt[z]{\frac{3}{5}}\)
\(\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}\)
Here’s the key idea: when you take a positive integer root of a number between \(0\) and \(1\), the result becomes larger than the original number, and the higher the root, the closer the value moves toward \(1\).
For example, \(\sqrt{\frac{1}{2}} < \sqrt[3]{\frac{1}{2}} < \sqrt[4]{\frac{1}{2}} < \sqrt[5]{\frac{1}{2}}...\)
Thus, \(\sqrt[3x]{\frac{3}{5}} < \sqrt[2y]{\frac{3}{5}} < \sqrt[z]{\frac{3}{5}}\) implies:
\(3x < 2y < z\)
Since \(x\), \(y\), and \(z\) are distinct single-digit non-prime positive numbers, they can be selected only from \(\{1, 4, 6, 8, 9\}\).
Now, \(x\) cannot be \(4\) or greater, because in that case \(3x\) becomes \(12\) or greater and cannot be less than \(z\), which is at most \(9\). Thus, \(x = 1\).
Similarly, \(y\) cannot be \(6\) or greater, so \(y = 4\).
And finally, this forces \(z\) to be \(9\).
Thus: \(x = 1\), \(y = 4\), and \(z = 9\).
\((3 * 1) < (2 * 4) < 9\)
Therefore, \(x + y + z = 1 + 4 + 9 = 14\).
Answer: B
