How many factors does $36^2$ have

Question

How many factors does $36^2$ have

$(A)2 \\ (B)8 \\ (C)24 \\ (D)25 \\ (E)26$

---

$36^2=2^4\times3^4$
How do i count the number of factors?I don't know.

Answer 1

The counting of factors is to be done in the following way:

As you have written, $$36^2=2^4 \times 3^4$$ Hence you have $4$ $2$'s and $4$ $3$'s as prime factors. So all factors of $36^2$ possible are different combinations of these $8$ factors.
To form the factors, you can choose one $2$, or two $2$'s, or three $2$'s, or four $2$'s, or none of them.
So you have $4+1=5$ choices. Similarly, you can choose one $3$, and so on.
In this way, you have $(4+1)(4+1)$ choices for the $8$ prime factors.
Thus the total number of factors = $(4+1)(4+1)=25$

Answer 2

It can be written as $2^4.3^4$ so now total factors are $(p_1+1)(p_2+1)$ where $p_i$ represents powers of prime thus total factors $(4+1)(4+1)=25$ thus option D

Answer 3

The question is ambiguous - do you mean prime factors or divisors?

---

If you mean prime factors, then $36^2$ has the same amount as $36$ (that is, $2$ prime factors).

---

If you mean divisors, then each of the $2$ prime factors can appear in each divisor between $0$ and $4$ times, hence there are $5^2=25$ divisors:

• $2^0\cdot3^0$
• $2^0\cdot3^1$
• $2^0\cdot3^2$
• $2^0\cdot3^3$
• $2^0\cdot3^4$
• $2^1\cdot3^0$
• $2^1\cdot3^1$
• $2^1\cdot3^2$
• $2^1\cdot3^3$
• $2^1\cdot3^4$
• $2^2\cdot3^0$
• $2^2\cdot3^1$
• $2^2\cdot3^2$
• $2^2\cdot3^3$
• $2^2\cdot3^4$
• $2^3\cdot3^0$
• $2^3\cdot3^1$
• $2^3\cdot3^2$
• $2^3\cdot3^3$
• $2^3\cdot3^4$
• $2^4\cdot3^0$
• $2^4\cdot3^1$
• $2^4\cdot3^2$
• $2^4\cdot3^3$
• $2^4\cdot3^4$

Answer 4

Considering the prime decomposition of the number,

$$36^2=2^4\cdot3^4$$

you can form distinct factors by combining all possible multiplicities of the primes, including $0$, $2^0\cdot3^0,2^1\cdot3^0,\cdots2^3\cdot3^2\cdots2^4\cdot3^4$. There are $5\cdot5=25$ such combinations (counting $1$ and $36^2$).

More generally, there will be $(m_1+1)(m_2+1)\cdots(m_n+1)$ distinct factors for the number $$p_1^{m_1}\cdot p_2^{m_2}\cdots p_n^{m_n}.$$

Answer 5

Each factor of $36^2 = 2^43^4$ is of the form $2^{\alpha}3^{\beta}$, where $\alpha$ and $\beta$ are integers satisfying the inequalities $0 \leq \alpha \leq 4$ and $0 \leq \beta \leq 4$. Since there are five choices for each exponent, the number of factors of $1296$ is $5 \cdot 5 = 25$.

In general, if $n = p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k} = \prod_{j = 1}^{k} p_j^{a_j}$, then each of its factors will be of the form $p_1^{b_1}p_2^{b_2} \cdots p_k^{b_k}$, where $b_j$ is an integer satisfying the inequalities $0 \leq b_j \leq a_j$ for $1 \leq j \leq k$. Hence, $n$ will have $$\prod_{j = 1}^{k} (a_j + 1)$$ factors.