This problem asks us to find the remainder of a sum of two very large numbers when divided by 9. This type of problem is best solved using modular arithmetic, which allows us to work directly with remainders.

Key Concepts: Modular Arithmetic

The notation $a \equiv b \pmod m$ means that $a$ and $b$ have the same remainder when divided by $m$. Here are the essential properties we'll use:

1. Congruence of Sums: If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a+c \equiv b+d \pmod m$. This means we can find the remainder of each term separately and then sum their remainders.
2. Congruence of Powers: If $a \equiv b \pmod m$, then $a^n \equiv b^n \pmod m$ for any positive integer $n$. This allows us to simplify the base of a power before dealing with the exponent.
3. Divisibility Rule for 9: A number is congruent to the sum of its digits modulo 9. This is a quick way to find the remainder of a large number when divided by 9.

Our goal is to find the value of $(11)^{1011}+(1011)^{11} \pmod 9$. We will calculate each term's remainder individually and then sum them.

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Step 1: Simplify the base of each term modulo 9

The first step in dealing with large powers in modular arithmetic is to reduce the base to its remainder modulo the divisor.

• For the first term, $(11)^{1011}$: We find the remainder of the base, 11, when divided by 9. $11 = 1 \times 9 + 2$ So, $11 \equiv 2 \pmod 9$. Why this step? According to the congruence of powers property, if $11 \equiv 2 \pmod 9$, then $(11)^{1011} \equiv (2)^{1011} \pmod 9$. This simplifies the base from 11 to 2, making subsequent calculations easier.

• For the second term, $(1011)^{11}$: We find the remainder of the base, 1011, when divided by 9. We can efficiently use the divisibility rule for 9 (sum of digits). Sum of digits of $1011 = 1+0+1+1 = 3$. So, $1011 \equiv 3 \pmod 9$. Why this step? Similarly, this simplifies the base from 1011 to 3, so $(1011)^{11} \equiv (3)^{11} \pmod 9$.

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Step 2: Evaluate the first term: $(11)^{1011} \pmod 9$

From Step 1, we know this is equivalent to finding $2^{1011} \pmod 9$. To evaluate a large power modulo a number, we look for a repeating pattern (cycle) in the powers of the base modulo the divisor.

Let's list the powers of 2 modulo 9:

• $2^1 \equiv 2 \pmod 9$
• $2^2 \equiv 4 \pmod 9$
• $2^3 \equiv 8 \pmod 9$
• $2^4 \equiv 16 \equiv 7 \pmod 9$
• $2^5 \equiv 14 \equiv 5 \pmod 9$
• $2^6 \equiv 10 \equiv 1 \pmod 9$ Why this step? We observe that $2^6 \equiv 1 \pmod 9$. This means the remainders repeat every 6 powers. The cycle length (or order) of 2 modulo 9 is 6. This property is crucial because it allows us to reduce the large exponent $1011$ to a much smaller one. Specifically, $2^k \equiv 2^{k \pmod 6} \pmod 9$ (unless $k \pmod 6 = 0$, in which case $2^k \equiv 1 \pmod 9$).

Now, we need to find the remainder of the exponent $1011$ when divided by the cycle length, 6. $1011 \div 6$ $1011 = 6 \times 168 + 3$ So, $1011 \equiv 3 \pmod 6$. Why this step? By finding $1011 \pmod 6$, we can determine which element in the cycle $2^1, 2^2, ..., 2^6$ corresponds to $2^{1011}$.

Therefore, $2^{1011} \equiv 2^3 \pmod 9$. Calculating $2^3$: $2^3 = 8$. So, $(11)^{1011} \equiv 8 \pmod 9$.

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Step 3: Evaluate the second term: $(1011)^{11} \pmod 9$

From Step 1, this is equivalent to finding $3^{11} \pmod 9$. Let's list the powers of 3 modulo 9:

• $3^1 \equiv 3 \pmod 9$
• $3^2 = 9 \equiv 0 \pmod 9$ Why this step? We notice that $3^2$ is a multiple of 9. Once a power of the base becomes 0 modulo $m$, all subsequent higher powers will also be 0 modulo $m$. This is because for any $k \ge 2$, $3^k = 3^2 \cdot 3^{k-2} \equiv 0 \cdot 3^{k-2} \equiv 0 \pmod 9$.

Since the exponent is $11$, and $11 \ge 2$, we can directly conclude: $3^{11} \equiv 0 \pmod 9$. So, $(1011)^{11} \equiv 0 \pmod 9$.

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Step 4: Combine the remainders

We have found the remainder for each term:

• $(11)^{1011} \equiv 8 \pmod 9$
• $(1011)^{11} \equiv 0 \pmod 9$

Using the property of congruence of sums ($a+c \equiv b+d \pmod m$ if $a \equiv b \pmod m$ and $c \equiv d \pmod m$): $(11)^{1011}+(1011)^{11} \equiv 8 + 0 \pmod 9$ $(11)^{1011}+(1011)^{11} \equiv 8 \pmod 9$ Why this step? This is the final step to find the remainder of the entire expression. We simply add the individual remainders and take the result modulo 9. Since $8$ is already less than 9, it is the final remainder.

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Tips for Success & Common Mistakes to Avoid

• Simplify the Base First: Always reduce the base of any power modulo the divisor before attempting to work with the exponent. This prevents calculations with unnecessarily large numbers.
• Identify Cyclic Patterns: For powers $a^n \pmod m$ where $a$ and $m$ are coprime, look for the smallest $k$ such that $a^k \equiv 1 \pmod m$. This is the order of $a$ modulo $m$. Then reduce the exponent $n$ modulo $k$. Euler's Totient Theorem states $a^{\phi(m)} \equiv 1 \pmod m$ for coprime $a,m$, where $\phi(m)$ is Euler's totient function. For $m=9$, $\phi(9) = 9(1-1/3) = 6$, which matches our cycle length for 2.
• Watch Out for Common Factors: If the base and the modulus share common factors (like 3 and 9 here), the pattern might quickly become 0 (as with $3^2 \pmod 9$) or might not have a cycle that ends in 1. Be alert for these "zero-out" scenarios.
• Divisibility Rules are Your Friend: For common moduli like 3, 9, 10, 11, etc., quickly apply their divisibility rules to simplify bases.
• Final Remainder: The remainder must always be a non-negative integer strictly less than the divisor. If you get a negative remainder (e.g., $-1 \pmod 9$), add the divisor to get the positive remainder (e.g., $-1+9 = 8 \pmod 9$).

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Summary/Key Takeaway

This problem perfectly illustrates how modular arithmetic simplifies complex number theory problems. By systematically simplifying the bases, identifying cyclic patterns (or "zero-out" conditions for non-coprime bases), and then combining the remainders, we can efficiently find the remainder of even very large expressions. Precision in modular reductions and exponent handling is key.

The remainder when $(11)^{1011}+(1011)^{11}$ is divided by 9 is 8.