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Find the number of positive integers $n$ less than 2017 such that\[1 n \frac{n^2}{2!} \frac{n^3}{3!} \frac{n^4}{4!} \frac{n^5}{5!} \frac{n^6}{6!}\]is an integer.

Respuesta :

Answer:

Step-by-step explanation :

n.n²/2.n³/3.n⁴/4.n⁵/5.n⁶/6 = k [ k is an integer ]

⇒ n²¹/720 = k

⇒ n²¹=720k

this way you can find the integer ( hit & trail rule )

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