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Integers $1, 2, ..., 20$ are partitioned into $2$ groups. Sum of all integers in one group is equal to $n$ and the product of all integers in another group is also equal to $n$. Find the maximal $n$.

Since $1 + 2 + ... + 20 = 210$, so the product of all integers in another group is less than $210$.

Please suggest how to proceed.

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    It's not $209(=11\cdot19)$ or $208(=13\cdot16=13\cdot2\cdot8)$ either. And it's not $207$ or $206$ or $205$ either, because each one of them has a prime factor larger than $20$. Perhaps you can continue eliminating your options in a similar manner. – 2017-02-23
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    It's not $204=12\cdot17=3\cdot4\cdot17=2\cdot6\cdot17$ either. And again, it's not $203$ or $202$ or $201$ either, because each one of them has a prime factor larger than $20$. So you can start your investigation from $200$. – 2017-02-23
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    Oops, each one of those products in my previous comments can be also made with a "$1$" It doesn't make any difference with regards to the conclusion, but still (FYI). – 2017-02-23
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    The best I've found is $180=20\cdot 9\cdot 1 = 210-(20+9+1)$ – 2017-02-23
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    @JaroslawMatlak It is $192=4*8*6=210-(4+8+6)$ – 2017-02-23
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    and it is none of $193$ ... $200$ for reasons that @barakmanos described – 2017-02-23

1 Answers 1

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Here's a answer by trial and error.

Notice that the product of the smallest $5$ integers is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ and the product of $6$ integers is bigger then $210$.

Now the sum will be bigger if we choose smaller numbers,lets try to make the sum the max possible value $<210$ by changing the digit $5$,testing $8$ is the biggest digit so the sum doesn't exceed $210$.We have that $1\cdot 2\cdot 3\cdot 4\cdot 8=192$ and the sum is $210-1-2-3-4-8=192$ so we have that $192$ can be $n$.

Now lets try to prove that $n\leq 192$ first $193$ is a prime $194=97\cdot 2$ where $97$ is prime so those are not solutions (because of the product).Now lets try numbers $\geq 195$ then sum of those numbers is $\leq 15$ checking products of $2$ numbers such that their sum is $\leq 15$ the greatest such number is $8\cdot 7=56$, now checking product of $3$ numbers the biggest is $5\cdot 4\cdot 6=120$ since the closer the $3$ numbers the bigger the product is,checking $4$ numbers we get that the biggest product is $2\cdot 3\cdot 4\cdot 5=120$ and at last checking five numbers it's $1\cdot 2\cdot 3\cdot 4\cdot 5=120$ this isn't the answer either so we concluded that $n=192$.

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    I got it. Thank you, everyone. – 2017-02-23