KCET · Maths · Basic of Mathematics
The number of positive divisors of 252 is
- A 9
- B 5
- C 18
- D 10
Answer & Solution
Correct Answer
(C) 18
Step-by-step Solution
Detailed explanation
We know that, if \(a=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \ldots\)
Then, the total number of positive divisors of \(a\) is
\(T(a)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots \ldots .\)
Given, \(\quad 252=2^{2} \times 3^{2} \times 7^{1}\)
Here, \(\quad \alpha_{1}=2, \alpha_{2}=2, \alpha_{3}=1\)
\(\therefore \quad T(a)=(2+1)(2+1)(1+1)\)
\(=3 \cdot 3 \cdot 2\)
\(=18\)
Then, the total number of positive divisors of \(a\) is
\(T(a)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots \ldots .\)
Given, \(\quad 252=2^{2} \times 3^{2} \times 7^{1}\)
Here, \(\quad \alpha_{1}=2, \alpha_{2}=2, \alpha_{3}=1\)
\(\therefore \quad T(a)=(2+1)(2+1)(1+1)\)
\(=3 \cdot 3 \cdot 2\)
\(=18\)
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