Question

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

Answer 1

4

Answer 2

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

1. Determine the cycle of powers of 2 modulo 7: We examine the first few 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 powers of 2 modulo 7 repeat with a cycle of length 3. This means $2^k \pmod{7}$ depends on the value of $k \pmod{3}$.

2. Calculate the exponent modulo the cycle length (3) using the binomial theorem: The exponent is $E = 23^{23}$. We need to find $E \pmod{3}$. We can express the base of the exponent, 23, as $23 = 3 \times 7 + 2$. Using the binomial theorem for $(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^{n-i} b^i$, let $a = 3 \times 7$, $b = 2$, and $n = 23$. $E = (3 \times 7 + 2)^{23} = \sum_{i=0}^{23} \binom{23}{i} (3 \times 7)^{23-i} 2^i$. When we consider this expression modulo 3, all terms where $(3 \times 7)$ is raised to a power of 1 or greater will be divisible by 3. The only term that might not be divisible by 3 is the one where the power of $(3 \times 7)$ is 0, which occurs when $i = 23$. So, $E \equiv \binom{23}{23} (3 \times 7)^{23-23} 2^{23} \pmod{3}$. $E \equiv 1 \cdot (3 \times 7)^0 \cdot 2^{23} \pmod{3}$. $E \equiv 1 \cdot 1 \cdot 2^{23} \pmod{3}$. $E \equiv 2^{23} \pmod{3}$. Now, we evaluate $2^{23} \pmod{3}$. Since $2 \equiv -1 \pmod{3}$: $E \equiv (-1)^{23} \pmod{3}$. Since 23 is an odd exponent, $(-1)^{23} = -1$. Therefore, $E \equiv -1 \equiv 2 \pmod{3}$. This means $23^{23} \equiv 2 \pmod{3}$.

3. Calculate the final remainder: We found that the exponent $23^{23}$ has a remainder of 2 when divided by 3. So we can write $23^{23} = 3m + 2$ for some integer $m$. Substitute this back into the original expression: $2^{23^{23}} = 2^{3m+2}$. Now, we find the remainder modulo 7: $2^{3m+2} = (2^3)^m \cdot 2^2$. Using the cycle we found in step 1 ($2^3 \equiv 1 \pmod{7}$): $(2^3)^m \cdot 2^2 \equiv 1^m \cdot 2^2 \pmod{7}$. $1^m \cdot 2^2 \equiv 1 \cdot 4 \pmod{7}$. $1 \cdot 4 \equiv 4 \pmod{7}$.

The remainder is 4.