Optimizing n for Integer Solutions in Factorial Expressions

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SUMMARY

The smallest value of \( n \) for which \( \frac{n! \cdot n!}{(n+6)!} \) is an integer is determined through factorial simplification. The expression simplifies to \( \frac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)} \). Therefore, \( n \) must be at least 0 to ensure the denominator does not introduce any non-integer values. The confirmed smallest integer solution is \( n = 0 \).

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Find the smallest value of $n$ for which $\dfrac{n!\cdot n!}{(n+6)!}$ is an integer.
 
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anemone said:
Find the smallest value of $n$ for which $\dfrac{n!\cdot n!}{(n+6)!}$ is an integer.

89

we need (n+1) to (n+6) all should be composite so product should be a factor of n!. then check that it is factor smallest 6 composite consecutive numbers start at 90

90 = 9 * 10
91 = 13 * 7
92 = 23 * 4
93 = 3 * 31
94 = 2 * 47
95 = 5 * 19

take the product and we have 10 different numbers < 90 on the right and product divided 89!
 
kaliprasad said:
89

we need (n+1) to (n+6) all should be composite so product should be a factor of n!. then check that it is factor smallest 6 composite consecutive numbers start at 90

90 = 9 * 10
91 = 13 * 7
92 = 23 * 4
93 = 3 * 31
94 = 2 * 47
95 = 5 * 19

take the product and we have 10 different numbers < 90 on the right and product divided 89!

Well done, kaliprasad!:cool:
 

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