**In how many ways can a number be written as a product of two different factors?**

Let Number=x

Let Its prime factors are:

2^{a}*3^{b}*5^{c}…………….and so on

Number of factors

=(a+1)(b+1)(c+1)………… and so on

Required number of ways

=[(a+1)(b+1)(c+1)………… and so on]/2

If Required number of ways result to a.b then Required number of ways=a+1

**Example 1:**

**In how many ways can 1500 be resolved into two factors?**

(a)12

(b)18

(c)20

(d)10

**Answer: **12

**Explanation:**

1500=2*750=2*2*3*5*5*5=2^{2}5^{3}3^{1}

Number of factors=(2+1)(3+1)(1+1)=3*4*2=24

Required number of ways=16/2=8

**Example 2:**

**In how many ways can 1200 be resolved into two factors?**

(a)12

(b)18

(c)20

(d)10

**Answer: **12

**Explanation:**

1200=12*100=2*2*3*5*2*5*2=2^{4}5^{2}3^{1}

Number of factors=(4+1)(2+1)(1+1)=5*3*2=30

Required number of ways=30/2=15

**Example 3:**

**In how many ways can the number 243 be resolved into two factors?**

(a)12

(b)18

(c)20

(d)10

**Answer: **12

**Explanation:**

243=3*3*3*3*3=3^{5}

Number of factors=(5+1)=6

Required number of ways=6/2=3
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