Do you want to count to three?

Suppose without loss of generality that the 3 positive integers are $a \le b \le c$ . Then

$abc = a + b + c \le 3c$ , and so $ab \le 3$ . This implies that the only possibilities for $(a,b)$ are

$(1,1), (1,2)$ and $(1,3)$ . Testing out each of these in turn, we find that for

• $(1,1) \Longrightarrow c = 2 + c \Longrightarrow 0 = 2$ , discard,

• $(1,2) \Longrightarrow 2c = 3 + c \Longrightarrow c = 3$ , valid solution,

• $(1,3) \Longrightarrow 3c = 4 + c \Longrightarrow c = 2$ , repeat of previous solution.

Thus $(a,b,c) = (1,2,3)$ is the only suitable triple, giving $k = a + b + c = abc = \boxed{6}$ .

Comment: Using a similar method, if we are to find 4 positive integers $a \le b \le c \le d$ such that

$a + b + c + d = abcd$ then the only valid quadruple is $(1,1,2,4)$ , giving a sum/product of $8$ .

$k =$ $6$

$1 + 2 + 3 = 6$

$1 * 2 * 3 = 6$

We immediately know that 1 is one of the values because 2x2x2 > 2+2+2, guess and check 1,2,3 yep it works cool the answer is 6

Let three no.s be a-d,a,a+d According to 1st condition: (a-d)+a+(a+d)=K According to 2nd condition: (a-d) a (a+d)=K Therefore on comparing 1st and 2nd eqn we get 3a=a*(a^2-d^2)

-> a^2-d^2=3 Assuming d=1 we get a=2

Therefore the no.s are. 2-1 , 2 , 2+1 i.e. 1,2,3 K=1+2+3=1 2 3=6

First,

1+2+3=6

1•2•3 is also 6