The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
To determine the largest \( n \in \mathbb{N} \) such that \( 3^n \) divides \( 50! \), we will use the method of finding the highest power of a prime \( p \) dividing \( n! \). The formula is:
\(\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\)
For \( n = 50 \) and \( p = 3 \), the expression becomes:
\(\left\lfloor \frac{50}{3} \right\rfloor + \left\lfloor \frac{50}{9} \right\rfloor + \left\lfloor \frac{50}{27} \right\rfloor + \left\lfloor \frac{50}{81} \right\rfloor + \ldots\)
Calculating each term step by step:
Since all further terms will also be zero, the sum is:
\(16 + 5 + 1 = 22\)
Thus, the largest \( n \) such that \( 3^n \) divides \( 50! \) is 22.
To determine the largest \( n \in \mathbb{N} \) such that \( 3^n \) divides \( 50! \), we need to calculate the number of times the prime number 3 appears in the factorization of \( 50! \). This can be done using Legendre's formula, which states that the exponent of a prime \( p \) in \( n! \) is given by:
\[\text{Exponent of } p = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\]
In this case, \( n = 50 \) and \( p = 3 \). We calculate:
\[\left\lfloor \frac{50}{3} \right\rfloor = 16\]
\[\left\lfloor \frac{50}{9} \right\rfloor = 5\]
\[\left\lfloor \frac{50}{27} \right\rfloor = 1\]
\[\left\lfloor \frac{50}{81} \right\rfloor = 0\]
For higher powers of 3, the floor function results in zero.
Adding these values gives the largest \( n \) for which \( 3^n \) divides \( 50! \):
\[16 + 5 + 1 + 0 = 22\]
Therefore, the largest \( n \) is 22.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.