Question

If a + b + c +ab +bc + ca + abc = 2005 where a, b, c are distinct natural number such that a < b < c then

• A. a= 1
• B. b = 15
• C. c = 58
• D. a + b+ c = 75

Answer 1

The correct option is D a + b+ c = 75

We have,

If $a+b+c+ab+bc+ca+abc=2005$

We know that,

$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc+1$

$\Rightarrow (a+1)(b+1)(c+1)=2005+1$

$\Rightarrow (a+1)(b+1)(c+1)=2006$

$\Rightarrow (a+1)(b+1)(c+1)=2\times 17\times 59$

$\Rightarrow (a+1)(b+1)(c+1)=(1+1)(16+1)(58+1)$

On comparing that,

$a=1,b=16,c=58$

Then,

$a+b+c=1+16+58$

$a+b+c=75$

Hence, this is the answer.