Question

If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + .... + ^{2017}C_{1008} = 2^\lambda$, then the remainder when $\lambda$ is divided by 33 is _____.

Answer 1

3

Answer 2

The full sum of binomial coefficients for $n = 2017$ is:

$$\sum_{k=0}^{2017} \binom{2017}{k} = 2^{2017}.$$

Since $2017$ is odd and the binomial distribution is symmetric, we have:

$$\binom{2017}{0} + \binom{2017}{1} + \cdots + \binom{2017}{1008} = \frac{2^{2017}}{2} = 2^{2016}.$$

Given that this sum equals $2^\lambda$, it follows that $\lambda = 2016$.

Now, to find the remainder when $\lambda = 2016$ is divided by $33$:

$$2016 \div 33 = 61 \text{ remainder } 3.$$