In the “expression” each of the twelve @ symbols is replaced with either $\times$ or $÷$ such that the value of the resulting expression is an integer.
1@2@3@4@5@6@7@8@9@10@11@12@13
Find the greatest common factor of all such integer values.
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1ya sorry its my first time using this site. Sorry Ill change it – Pinak Paliwal Jan 12 '21 at 01:55
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2None of 7,11,13 can be preceded by a $\div$ sign. The situation with 3,9 is more complicated, because of the presence of 12 and 6. Similarly, the situation with 5 is more complicated because of the 10. The only approach that comes to my mind is to compute the prime factorization of all 12 numbers $> 1$, and then consider that if (for example) $\alpha$ is the largest exponent such that $p^\alpha$ divides the result, where $p$ is a prime, then $\alpha$ must be non-negative, so that the result is an integer. – user2661923 Jan 12 '21 at 02:05
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First every such integer is divisible by $7, 11, 13$.
Highest power of $2$ of $13!$ is even, so you need to find an integer product such that it's odd. This can be easily done.
Then the highest power of $3$ of $13!$ is odd, so you need to find an integer product such that it's divisible by $3$ but not by $9$. This is easy too.
The highest power of $5$ of $13!$ is $2$. It's easy to find one integer product that's not divisible by $5$.
Therefore the $\gcd$ of all integer products is $3\times 7 \times 11 \times 13 = 3003$.
Neat Math
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2@bajun65537 One of the integers is $10/5$ times everything else, which is not divisible by $5$. – Neat Math Jan 13 '21 at 18:43
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