Question

If a + b + c + ab + bc + ca + abc = 2005 where a, b, c are distinct natural numbers such that a < b < c then which of the following options are true?
A. a = 1
B. b = 15
C. c = 58
D. a + b + c = 75

Answer 1

Hint: Since, a + b + c + ab + bc + ca + abc = 2005, try to write this whole expression (in terms of a, b, c) as a product of three different terms and compare both sides to find a, b & c.

Complete step-by-step answer:
It is given that,
a + b + c + ab + bc + ca + abc = 2005
We will add 1 on both sides:-
1 + a + b + c + ab + bc + ca + abc = 2005 + 1
By rearranging:-
1 + c + a + ca + b + bc + ab + abc = 2006
1(1 + c) + a(1 + c) + b(1 + c) + ab(1 + c) = 2006
⇒ (1 + c) + (1 + a + b + ab) = 2006
Simplifying further we get,
(1 + c)[1(1 + a) + b(1 + a)] = 2006
⇒ (1 + c) [(1 + b)(1 + a)] = 2006
⇒ (1 + a)(1 + b)(1 + c) = 2006
We will now try to write 2006 as a product of three numbers
i.e 2006 = 2 × 17 × 59
which nothing but prime factorization of 2006
⇒ (a + 1)(b + 1)(c + 1) = 2 × 17 × 59
(a + 1)(b + 1)(c + 1) = (1 + 1)(16 + 1)(58 + 1)
Comparing terms on both sides we will get,
a = 1, b = 16, c = 58
also, a + b + c = 1 + 16 + 58
a + b + c = 75

∴ Correct options are A, C, D

Note: This was a multiple choice question with more than one answer correct. Here you need to remember that:-
(a + 1)(b + 1)(c + 1) = a + b + c + ab + bc + ca + 1 + abc.
This can come handy in some situations. We can also check the answer as since we got a = 1, b = 16 & c = 58 given condition is verified i.e. a < b < c.