Question

The fractional part of the number \(\tfrac{4^{2022}}{15}\) is equal to:

Answer 1

Observe that \[ \left\{\frac{4^{2022}}{15}\right\} = \left\{\frac{2^{4044}}{15}\right\} = \left\{\frac{(1 + 15)^{1011}}{15}\right\}. \] By the binomial theorem, \[ (1 + 15)^{1011} = \sum_{k=0}^{1011} \binom{1011}{k} 1^{1011-k} 15^k = \sum_{k=0}^{1011} \binom{1011}{k} 15^k. \] Expanding the sum, we have \[ (1 + 15)^{1011} = \binom{1011}{0} 15^0 + \binom{1011}{1} 15^1 + \binom{1011}{2} 15^2 + \cdots + \binom{1011}{1011} 15^{1011}. \] \[ (1 + 15)^{1011} = 1 + 1011 \cdot 15 + \binom{1011}{2} 15^2 + \cdots + 15^{1011}. \] Dividing by 15, we get \[ \frac{(1 + 15)^{1011}}{15} = \frac{1}{15} + 1011 + \binom{1011}{2} 15 + \cdots + 15^{1010}. \] All terms except \(\frac{1}{15}\) are integers. Therefore, the fractional part is \[ \left\{\frac{(1 + 15)^{1011}}{15}\right\} = \left\{\frac{1}{15} + \text{integer}\right\} = \left\{\frac{1}{15}\right\} = \frac{1}{15}. \] Therefore, \[ \left\{\frac{4^{2022}}{15}\right\} = \frac{1}{15}. \]