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The total number of factors of $3528$ greater than $1$ but less than $3528$ is

  1. $35$
  2. $36$
  3. $34$
  4. None of these
in Numerical Ability by Boss (16.8k points)
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2 Answers

+2 votes
$3528=2^{3}\times 3^{2}\times 7^{2}$

Number of proper factors $=(3 + 1) (2 + 1) (2 + 1) - 2 \text{(for 1 and  3528)}=34$
by Junior (761 points)
edited by
+1 vote

Answer: $\mathbf C$

The total number of factors are: $\mathbf{34}$

Explanation:

$2,3,4,6,7,8,9,12,14,18,21,24,28,36,42,49,56,63,72,84,98,126,147,168,196,252,294,392,441,504,588,882,1176,1764\;\;\;\text{[Excluding 1 and 3528]}$


OR (Shortcut)

$3528 = 2^3\times3^2\times7^2$

Using the formula:

$\color {red}{\mathbf {\mathrm{(p+1)(q+1)(r+1)}}}- 2(\text{Exclude 1  and 3528}) = (3+1)(2+1)(2+1)-2=34$ 

where $\color{red} {\mathrm {p, q, r}}$ are the  powers of the prime factors.

$\therefore \mathbf C$ is the correct option.


For more such shortcuts: https://www.integers.co/questions-answers/what-are-the-factor-combinations-of-the-number-3528.html

by Boss (14k points)
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