Question

if y is a positive integer, for how many different values of y is \\(\sqrt3{\frac{144}{y}}\\) a whole number?

1
2
6
15

Explanation:

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<reasoning>

Set up the condition for a whole number

Using the Integer Divisibility knowledge point
\[

$$\begin{aligned} &k = \sqrt[3]{\frac{144}{y}} \in \{0, 1, 2, 3, \dots\}\\ &k^3 = \frac{144}{y}\\ &y = \frac{144}{k^3} \end{aligned}$$

\]

Analyze constraints on the integer variables

Since \(y\) must be a positive integer, we have:
\[

$$\begin{aligned} &y > 0 \implies \frac{144}{k^3} > 0 \implies k \ge 1\\ &y \in \mathbb{Z}^+ \implies k^3 \text{ must be a positive divisor of } 144 \end{aligned}$$

\]

Find perfect cube divisors of 144

We find all positive integers \(k\) such that \(k^3\) divides \(144\).
First, find the prime factorization of \(144\):
\[
144 = 12^2 = (2^2 \cdot 3)^2 = 2^4 \cdot 3^2
\]
For \(k^3\) to divide \(2^4 \cdot 3^2\), the prime factorization of \(k^3\) can only contain primes \(2\) and \(3\) with exponents less than or equal to those in \(144\):
\[

$$\begin{aligned} &k^3 = 2^a \cdot 3^b \quad \text{where } a \in \{0, 3\} \text{ and } b = 0\\ &\text{Possible values for } k^3:\\ &k^3 = 2^0 \cdot 3^0 = 1 \implies y = \frac{144}{1} = 144\\ &k^3 = 2^3 \cdot 3^0 = 8 \implies y = \frac{144}{8} = 18 \end{aligned}$$

\]

Count the valid values of y

There are exactly \(2\) positive integer values of \(y\) (\(144\) and \(18\)) that satisfy the condition.
</reasoning>

<answer>
<mcq-option>(A) 1</mcq-option>
<mcq-correct>(B) 2</mcq-correct>
<mcq-option>(C) 6</mcq-option>
<mcq-option>(D) 15</mcq-option>
</answer>

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]
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</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Set up the condition for a whole number

Using the Integer Divisibility knowledge point
\[

$$\begin{aligned} &k = \sqrt[3]{\frac{144}{y}} \in \{0, 1, 2, 3, \dots\}\\ &k^3 = \frac{144}{y}\\ &y = \frac{144}{k^3} \end{aligned}$$

\]

Analyze constraints on the integer variables

Since \(y\) must be a positive integer, we have:
\[

$$\begin{aligned} &y > 0 \implies \frac{144}{k^3} > 0 \implies k \ge 1\\ &y \in \mathbb{Z}^+ \implies k^3 \text{ must be a positive divisor of } 144 \end{aligned}$$

\]

Find perfect cube divisors of 144

We find all positive integers \(k\) such that \(k^3\) divides \(144\).
First, find the prime factorization of \(144\):
\[
144 = 12^2 = (2^2 \cdot 3)^2 = 2^4 \cdot 3^2
\]
For \(k^3\) to divide \(2^4 \cdot 3^2\), the prime factorization of \(k^3\) can only contain primes \(2\) and \(3\) with exponents less than or equal to those in \(144\):
\[

$$\begin{aligned} &k^3 = 2^a \cdot 3^b \quad \text{where } a \in \{0, 3\} \text{ and } b = 0\\ &\text{Possible values for } k^3:\\ &k^3 = 2^0 \cdot 3^0 = 1 \implies y = \frac{144}{1} = 144\\ &k^3 = 2^3 \cdot 3^0 = 8 \implies y = \frac{144}{8} = 18 \end{aligned}$$

\]

Count the valid values of y

There are exactly \(2\) positive integer values of \(y\) (\(144\) and \(18\)) that satisfy the condition.
</reasoning>

<answer>
<mcq-option>(A) 1</mcq-option>
<mcq-correct>(B) 2</mcq-correct>
<mcq-option>(C) 6</mcq-option>
<mcq-option>(D) 15</mcq-option>
</answer>

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"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Number Theory",
"Integer Divisibility"
]
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