Question

The remainder obtained when \( (2m + 1)^{2n} \), \( m, n \in \mathbb{N} \) is divided by 8 is

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

Squares of all odd numbers are congruent to 1 modulo 8. This is useful in modular arithmetic problems.

Answer 1

The Correct Option is A

Step 1: Consider the general term: \( (2m + 1)^{2n} \). This is an odd number raised to an even power.
Step 2: Let \( a = 2m + 1 \). Since \( m \in \mathbb{N} \), \( a \) is always odd.
Step 3: Try small values of \( a \):
\[ \begin{aligned} &\text{If } a = 1, && 1^{2n} = 1 \mod 8 = 1
&\text{If } a = 3, && 3^2 = 9 \mod 8 = 1
&\text{If } a = 5, && 5^2 = 25 \mod 8 = 1
&\text{If } a = 7, && 7^2 = 49 \mod 8 = 1
\end{aligned} \]
Step 4: Observe that any odd number squared gives 1 modulo 8, hence: \[ (2m + 1)^{2n} \mod 8 = 1 \] \[ \boxed{\text{Remainder is } 1} \]