Xjyuk

Let's review a triple of numbers, $1, 2, 3$, it is a curiosity that

$$1+2+3 = 1times2times3 = 6$$

Are there another triples (or not necessary triples) such that their multiple equal to their sum?

And generalised pattern of such identities would be interesting and appreciated.

PS: Conjecture: Reviewing $t$ fold case of such numbers, they are seem to be the integer solutions of the equation
$$n(n+1)(n+2)cdots(n+t) = binomt+11n + binomt+12$$

PSS Integer solution (for consequent integers)
$$prod_k=0^2s (n+k) = sum_k=0^2s (n+k)$$
for $n=-s$. But these sums and products are 0.

Hint: $$a+b+c=abc$$ and $a<b<c$ so we have $$3c>abcimplies ab<3$$

Working with integers
$$n(n+1)(n+2)=3n+3=3(n+1)$$

With $n=-1$, we have $$-1,0,1$$ as a solution

Otherwise

$$n(n+2)=3$$

$$n^2+2n-3=0$$

$$(n+3)(n-1)=0$$

$$n=3,n=1$$
Thus we have $$-3,-2,-1$$ or $$1,2,3$$ as solutions.

"And generalised pattern of such identities would be interesting and appreciated"
Well, while there is nothing particular about the triplet $59,60$ and $61$, we do have $tan59+tan60+tan61=(tan59)(tan60)(tan61)$, the triplet being in degrees. This particular case comes from the identity $tanA+tanB+tanC=(tanA)(tanB)(tanC)$ where $A+B+C=180$.

If we know, that $A=1,B=2,C=3$ is a solution we can look for another solution with larger numbers by
$$(A+a)+(B+b)+(C+c) = (A+a)(B+b)(C+c) \
-----------------------------\
(1+a)+(2+b)+(3+c) = (1+a)(2+b)(3+c)\
6+a+b+c = 6+ 2c+3b+6a+bc+3ab+2ac+abc\
a+b+c = 2c+3b+6a+bc+3ab+2ac+abc\
0 = c+2b+5a+bc+3ab+2ac+abc\
$$
If no number $a,b,c$ is negative, all must be zero.

Considering $(n,n+1,n+2)$ solutions with $nin N$, I have: $3n+3=n(n+1)(n+2) Leftrightarrow 3(n+1)=n(n+1)(n+2) Leftrightarrow 3=n(n+2)$. So, I obtain: $$n^2+2n-3=0$$The equation becomes: $(n-1)(n+3)=0$ so $n=1$ or $n=-3$. The second solution is impossible because $n in N$, the first is the only correct.