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2 Answers

Best answer
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9 votes

$7200$ can be written as $2^{5} * 3^{2} * 5^{2}$

Therefore, there are $(5+1) * ( 2+1) * (2+1) = 54$ divisors including $1$, and $7200$.

Hence, the total divisors (excluding $1$ and $7200$) = $54-2 = 52$

Here's some justification to support the above method : 

Let $n\in\mathbb{N}$. Then by fundamental theorem of arithmetic we can write $n\in \mathbb{N}, n\neq 1$ by $n=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ where $p_1,p_2,\dots p_k$ are prime and $a_k\in\mathbb{N}$, $k=1,2,...,k$. Hence, number of divisors of $n$ =  $(a_1+1)(a_2+1)\cdots (a_k+1)$.

More on this here.

Answer ( C )

edited
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0 votes
7200=2^5 *3^2 * 5^2

total divisors=(5+1)*(2+1)*(2+1)=54

but 1 and 7200 excluding

so remaining divisors are 54-2=52

so ans is (C).
Answer:

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