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

$$1+2+3 = 1\times2\times3 = 6$$

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 $$n(n+1)(n+2)\cdots(n+t) = \binom{t+1}{1}n + \binom{t+1}{2}$$

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.

PS3: Still we can easily find such combinations using the following pattern: $$\prod_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1 \times \prod_{k=0}^t a_k = \left(\sum_{k=0}^t a_k\right)+\sum_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1$$

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  • 2
    $\begingroup$ We have $2+2=2\cdot 2$ $\endgroup$ – Arthur Aug 18 at 20:26
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    $\begingroup$ $1+1+2+4=1\cdot 1\cdot 2\cdot 4.$ $\endgroup$ – Thomas Andrews Aug 18 at 20:29
  • $\begingroup$ For $t=2$ solutions to $n(n+1)(n+2)=\binom{3}{1}n + \binom{3}{2}$ are $n=1, n=-1, n=3$. For $t=4$ the solution in integers is $n=-2$. For $t=6$ solution in integers is $n=-3$. $\endgroup$ – Petro Kolosov Aug 18 at 21:27
  • $\begingroup$ For $t=8$ the solution is $n=-4$. We can assume now that integer solution exists only for even $t$. Yes, for $t=10$ the solution is $n=-5$. $\endgroup$ – Petro Kolosov Aug 18 at 21:33
  • $\begingroup$ $1+1+6+1+1+2=1\cdot 1\cdot 1\cdot 6 \cdot 1 \cdot 1 \cdot 2$ $\endgroup$ – Petro Kolosov Aug 18 at 23:20
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Hint: $$a+b+c=abc$$ and $a<b<c$ so we have $$3c>abc\implies ab<3$$

Soince $ab >1$ we have $ab=2$ so $a=1$ and $b=2$ and ...

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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.

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  • $\begingroup$ as I see we can even obtain such $t-$fold set of numbers by recucrsion from previous one, i.e for three numbers we have $1+2+3 = 1*2*3$ then we just substitute this result for 4 fold product $n(n+1)(n+2)(n+3)$. Yeah, I know I speak about it unclear but hope it is possible to understand :) $\endgroup$ – Petro Kolosov Aug 18 at 20:38
  • $\begingroup$ Yes trying to find all possible solutions is very interesting $\endgroup$ – Mohammad Riazi-Kermani Aug 18 at 20:48
  • $\begingroup$ try to check my edit in the question body, may be it would be helpful $\endgroup$ – Petro Kolosov Aug 18 at 20:50
  • $\begingroup$ Your formula for the sum is not correct. You should have $t+1$ choose $1$ Instead of $t$ choose $1$ $\endgroup$ – Mohammad Riazi-Kermani Aug 18 at 21:02
  • $\begingroup$ Yes, I've fixed it $\endgroup$ – Petro Kolosov Aug 18 at 21:06
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"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 $A+B+C=180$.

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  • $\begingroup$ But are they all integers ? I'm not sure $\endgroup$ – Petro Kolosov Aug 18 at 20:58
  • $\begingroup$ @PetroKolosov. As far as the general identity is concerned, $A,B$ and $C$ don't have to be integers, but to satisfy your case, you just have to solve $x+(x+1)+(x+2)=180$ to create a set of consecutives that will do the job. If you allow negatives, you can also consider $-59,-60$ and $-61$ $\endgroup$ – imranfat Aug 18 at 21:11
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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.

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

Easy. Just take some random numbers, say $3$ and $4$. We have $3 \cdot 4 = 12$, but $3+4=7$. So, just pad it with $12-7=5$ more $1$'s, and we have:

$1+1+1+1+1+3+4=1\cdot1\cdot1\cdot1\cdot1\cdot3\cdot4=12$

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Considering $(n,n+1,n+2)$ solutions with $n\in N$, 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.

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