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Consider the following statements
I : The number of non-trivial even divisors of the number \(2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}\), is \(\left(\alpha_1+1\right)\left(\alpha_2+1\right)\left(\alpha_3+1\right)\left(\alpha_4+1\right)\left(\alpha_5+1\right)-2\)
II : The number of non-trivial odd divisors of the number \(2^{\alpha_1} \cdot 3^{\alpha_2} \cdot 4^{\alpha_3} \cdot 5^{\alpha_4} \cdot 6^{\alpha_5}\), is \(\alpha_2+\alpha_4+\alpha_5+\alpha_2 \alpha_4+\alpha_4 \alpha_5\). Then
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I : The number of non-trivial even divisors of the number \(2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}\), is \(\left(\alpha_1+1\right)\left(\alpha_2+1\right)\left(\alpha_3+1\right)\left(\alpha_4+1\right)\left(\alpha_5+1\right)-2\)
II : The number of non-trivial odd divisors of the number \(2^{\alpha_1} \cdot 3^{\alpha_2} \cdot 4^{\alpha_3} \cdot 5^{\alpha_4} \cdot 6^{\alpha_5}\), is \(\alpha_2+\alpha_4+\alpha_5+\alpha_2 \alpha_4+\alpha_4 \alpha_5\). Then
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Verified Answer
The correct answer is:
I is false and II is true
Given number \(=2^{\alpha_1} 3^{\alpha_2} 4^{\alpha_3} 5^{\alpha_4} 6^{\alpha_5}\)
\(=2^{\alpha_1+2 \alpha_3+\alpha_5 3^{\alpha_2}+\alpha_5 5^{\alpha_4}}\)
Number of non-trivial even divisors
\(=\left(\alpha_1+2 \alpha_3+\alpha_5\right)\left(\alpha_2+\alpha_5+1\right)\left(\alpha_4+1\right)-1\)
And, number of non-trivial odd divisors
\(\begin{aligned}
& =\left(\alpha_2+\alpha_5+1\right)\left(\alpha_4+1\right)-1 \\
& =\alpha_2+\alpha_4+\alpha_5+\alpha_2 \alpha_4+\alpha_4 \alpha_5+1-1
\end{aligned}\)
\(\therefore\) Statement \(\mathrm{I}\) is false and \(\mathrm{II}\) is true.
\(=2^{\alpha_1+2 \alpha_3+\alpha_5 3^{\alpha_2}+\alpha_5 5^{\alpha_4}}\)
Number of non-trivial even divisors
\(=\left(\alpha_1+2 \alpha_3+\alpha_5\right)\left(\alpha_2+\alpha_5+1\right)\left(\alpha_4+1\right)-1\)
And, number of non-trivial odd divisors
\(\begin{aligned}
& =\left(\alpha_2+\alpha_5+1\right)\left(\alpha_4+1\right)-1 \\
& =\alpha_2+\alpha_4+\alpha_5+\alpha_2 \alpha_4+\alpha_4 \alpha_5+1-1
\end{aligned}\)
\(\therefore\) Statement \(\mathrm{I}\) is false and \(\mathrm{II}\) is true.
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