Find Vishnoo's positive integers

Adding 1 both sides to the given equation yields: $a+b+c+ab+bc+ca+abc+1 = 1001$
On factorising both sides, we get $(a+1)(b+1)(c+1) = 7 \times 11 \times 13 \times 1$ .
Now clearly as $a, b, c$ are positive integers, therefore none of three factors on L.H.S. can be 1.
Hence, Without Loss of generality, A set of solution for $a, b$ and $c$ is : $a+1=7$ and $b+1=11$ and $c+1 =13$ , i.e.
$a = 6$ and $b = 10$ and $c = 12$
$\Rightarrow a+b+c =28$

The problem states that a, b, and c are positive integers. It is also given that a + b + c + ab + bc + ca + abc = 1000. We are asked to find the value of a + b + c.

We start working from our given equation which is a + b + c + ab + bc + ca + abc =1000

Basically, we can just simplify the LHS to (a+1)(b+1)(c+1) - 1 but of course, this is hard to see at first glance, so below is a step by step process on how to arrive at that.

So once again, we look at our original equation which is a + b + c + ab + bc + ca + abc = 1000

Rearranging the terms, a + ac + b + bc + ab + abc +c = 1000

Factoring out the common factors, a(c+1) + b(c+1) + ab(c+1) +c =1000

Factoring out (c+1), (c+1) (a+b+ab) + c =1000

Then, we can express (a + b + ab) as (a + b + ab +1) - 1 which also equals (a+1)(b+1) - 1

Going back to where we were previously, (c+1) (a+b+ab) + c = 1000

Substituting (a+b+ab) for (a+1)(b+1) - 1, this becomes (c+1) [(a+1)(b+1)-1] +c = 1000

Expanding, (c+1)(a+1)(b+1) - 1(c+1) + c =1000

Then, (a+1)(b+1)(c+1) - c - 1 + c =1000

Cancelling out -c and +c, (a+1)(b+1)(c+1) - 1 =1000

Transposing the 1, (a+1)(b+1)(c+1) = 1001

Since it is given in the problem that a, b, and c are positive integers, (a+1), (b+1), and (c+1) must also be positive integers. Furthermore, it is obvious that (a+1), (b+1), and (c+1) are all greater than one.

From above, (a+1)(b+1)(c+1) = 1001 And 1001 = 7 x 11 x 13

Since none of (a+1), (b+1), and (c+1) can be one, then (a+1), (b+1) (c+1) must be 7, 11, and 13 (not necessarily in that order).

That means a,b, and c must be 6, 10, and 12. (also not necessarily in that order).

Their sum a+b+c must then equal 6 + 10 + 12 = 28

Therefore, a + b + c = 28.

We have

1001 = abc + ab + ac + bc + a + b + c + 1 = (a+1)(b+1)(c+1)

and 1001 has a prime factor decomposition of 1001 = (7)(11)(13), so we have a+b+c = 6+10+12 = 28.

(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc+1=1001

1001=7 * 11 * 13

a+b+c+3=31

a+b+c=28

a + b + c + ab + bc + ca + abc = 1000, We add 1 to the both sides, a + b + c + ab + bc + ca + abc + 1 = 1000 + 1, The left hand side can be factorized as follow: (a + 1)(b + 1)(c + 1) = 1001; 1001 = 7 * 11 * 13; The solution of this equation is (6, 10, 12) So, (a + b + c) = 6 + 10 + 12 = 28

So we can rewrite this equation $a+b+c+ab+bc+ca+abc=1000$ as

$(a+1)(b+1)(c+1) - 1 = 1000$

or, $(a+1)(b+1)(c+1) = 1001$

which can be re written as $(a+1)(b+1)(c+1) = 7\times11\times13$

Now, $7, 11$ and $13$ are prime numbers and can be rewritten as $(6+1),(10+1),(12+1)$

So, $a+b+c = 28$ .

a + b + c + ab + bc + ca + abc = 1000

Factorizes to give:

( a + 1)( b + 1)( c + 1) - 1 = 1000

so

( a + 1)( b + 1)( c + 1) = 1001

which gives

( a + 1)( b + 1)( c + 1) = 7 * 11 * 13

Comparing factors we get:

a + 1 = 7, so a = 6

b + 1 = 11, so b = 10

c + 1 = 13, so c = 12

Therefore, a + b + c = 28

We have-

a + b + c + ab + bc + ca + abc = 1000

Adding 1 to both sides, we get-

1 + a + b + c + ab + bc + ca + abc = 1001

Now, by applying the identity x^3 + (a+b+c) x^2 + (ab+ac+bc) x + abc, and taking x=1, we get-

1 + a + b + c + ab + bc + ca + abc = (1+a)(1+b)(1+c) = 1001

Now, taking out the prime factors of 1001, we get the factors as- 7, 11 and 13

Therefore, by comparing, we get-

1+a = 7, 1+b = 11 and 1+c = 13

Therefore a = 6, b = 10 and c = 12.

And a + b + c = 6 + 10 + 12 = 28.

:)