In how many ways can a number be written as a product of two different factors?
Let Number=x
Let Its prime factors are:
2a*3b*5c…………….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=225331
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=245231
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=35
Number of factors=(5+1)=6
Required number of ways=6/2=3
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