Question

remainder when 2^(32^32) is divided by 7

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

To find the remainder when $2^{32^{32}}$ is divided by 7, we use properties of modular arithmetic.

1. Analyze the base modulo the divisor:
   We need to find the remainder of $2^X \pmod{7}$, where $X = 32^{32}$.
   Let's look at the 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 $2^k \pmod{7}$ depends on $k \pmod{3}$. Specifically, if $k \equiv r \pmod{3}$, then $2^k \equiv 2^r \pmod{7}$ (for $r \neq 0$). If $k \equiv 0 \pmod 3$, then $2^k \equiv 2^3 \equiv 1 \pmod 7$.

2. Evaluate the exponent modulo the cycle length:
   The exponent is $X = 32^{32}$. We need to find $X \pmod{3}$.
   First, reduce the base of the exponent (32) modulo 3:
   $32 = 10 \times 3 + 2$, so $32 \equiv 2 \pmod{3}$.
   Alternatively, $32 \equiv -1 \pmod{3}$.
   Now, substitute this into the exponent:
   $X = 32^{32} \equiv (-1)^{32} \pmod{3}$.
   Since 32 is an even number, $(-1)^{32} = 1$.
   Therefore, $X = 32^{32} \equiv 1 \pmod{3}$.

3. Combine the results:
   Since $32^{32} \equiv 1 \pmod{3}$, we can write $32^{32}$ in the form $3k+1$ for some non-negative integer $k$.
   Now substitute this back into the original expression:
   $2^{32^{32}} = 2^{3k+1}$.
   We want to find $2^{3k+1} \pmod{7}$.
   $2^{3k+1} = 2^{3k} \cdot 2^1 = (2^3)^k \cdot 2 \pmod{7}$.
   From step 1, we know that $2^3 \equiv 1 \pmod{7}$.
   So, $(2^3)^k \cdot 2 \equiv 1^k \cdot 2 \pmod{7}$.
   $1^k \cdot 2 \equiv 1 \cdot 2 \pmod{7}$.
   $1 \cdot 2 \equiv 2 \pmod{7}$.

Thus, the remainder when $2^{32^{32}}$ is divided by 7 is 2.