Question

2^(23^23) divided by 7 will give what remainder using binomial theorem

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

To find the remainder when $2^{23^{23}}$ is divided by 7, we can use modular arithmetic and the binomial theorem.

1. Identify the cycle of powers of 2 modulo 7:

   • $2^1 \equiv 2 \pmod{7}$
   • $2^2 \equiv 4 \pmod{7}$
   • $2^3 \equiv 8 \equiv 1 \pmod{7}$ The cycle length of powers of 2 modulo 7 is 3. This means that the remainder of $2^n \pmod{7}$ depends on the value of $n \pmod{3}$.

2. Determine the exponent modulo the cycle length: The exponent is $23^{23}$. We need to find the remainder of $23^{23}$ when divided by 3.

   • First, reduce the base (23) modulo 3: $23 = 7 \times 3 + 2$, so $23 \equiv 2 \pmod{3}$. Alternatively, $23 \equiv -1 \pmod{3}$.
   • Now, substitute this into the exponent: $23^{23} \equiv (-1)^{23} \pmod{3}$.
   • Since 23 is an odd number, $(-1)^{23} = -1$.
   • Therefore, $23^{23} \equiv -1 \equiv 2 \pmod{3}$.

3. Apply the findings to the original expression: Since $23^{23} \equiv 2 \pmod{3}$, we can write $23^{23}$ in the form $3k+2$ for some integer $k$. So, the problem becomes finding $2^{3k+2} \pmod{7}$. $2^{3k+2} = 2^{3k} \cdot 2^2 = (2^3)^k \cdot 4$.

4. Use the binomial theorem (implicitly): We know $2^3 = 8$. We want to find $(2^3)^k \pmod{7}$, which is $8^k \pmod{7}$. We can write $8$ as $7+1$. Using the binomial theorem for $(7+1)^k$: $(7+1)^k = \sum_{i=0}^k \binom{k}{i} 7^i 1^{k-i}$. When we take this modulo 7, all terms where $i \ge 1$ will have a factor of 7, making them 0 modulo 7. So, $(7+1)^k \equiv \binom{k}{0} 7^0 1^k \pmod{7}$. $(7+1)^k \equiv 1 \cdot 1 \cdot 1 \equiv 1 \pmod{7}$. Thus, $(2^3)^k \equiv 1 \pmod{7}$.

5. Final Calculation: Substitute the result from step 4 back into the expression from step 3: $2^{23^{23}} \equiv (2^3)^k \cdot 4 \pmod{7}$ $2^{23^{23}} \equiv 1 \cdot 4 \pmod{7}$ $2^{23^{23}} \equiv 4 \pmod{7}$.

The remainder is 4.