A number when divided by 7 leaves a remainder of 4. What is the remainder when its square is divided by 7?
4
- Step 1: Express the number in modular form - Let $n = 7k + 4$ for some integer $k$.
- Step 2: Square the expression - \[ n^2 = (7k + 4)^2 = 49k^2 + 56k + 16 \]
- Step 3: Take modulo 7 - Since $49k^2$ and $56k$ are multiples of 7, they leave remainder 0. Thus: \[ n^2 \equiv 16 \pmod{7} \]
- Step 4: Simplify remainder - $16 \div 7 = 2$ remainder $2$, so: \[ n^2 \equiv 2 \ (\text{mod } 7) \]
- Step 5: Conclusion - The remainder is $2$, matching option (2).
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: