Key Concepts Used

This problem primarily relies on Modular Arithmetic (also known as congruence relation) and the properties derived from the Binomial Theorem for finding remainders of large powers.

Modular Arithmetic: The notation $a \equiv b \pmod{m}$ means that $a$ and $b$ have the same remainder when divided by $m$. This is equivalent to stating that $a-b$ is divisible by $m$. Key properties used here are:

• If $a \equiv b \pmod{m}$, then $a^n \equiv b^n \pmod{m}$ for any positive integer $n$.
• If $a \equiv b \pmod{m}$ and $c \equiv d \pmod{m}$, then $a+c \equiv b+d \pmod{m}$.

Binomial Theorem for Remainders: For an expression of the form $(km+r)^n$, where $km$ is a multiple of $m$, its expansion using the Binomial Theorem will be: $(km+r)^n = \binom{n}{0}(km)^0 r^n + \binom{n}{1}(km)^1 r^{n-1} + \dots + \binom{n}{n}(km)^n r^0$ All terms except the first one (containing $r^n$) will have a factor of $km$, and thus be divisible by $m$. Therefore, $(km+r)^n \equiv r^n \pmod{m}$.

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Step-by-step Solution

1. Finding $\alpha$: Remainder when $(22)^{2022}+(2022)^{22}$ is divided by 3

Goal: Determine $\alpha$ such that $(22)^{2022}+(2022)^{22} \equiv \alpha \pmod{3}$, where $0 \le \alpha < 3$.

• Step 1.1: Simplify the base numbers modulo 3.

  • For the term $(22)^{2022}$: We find the remainder of $22$ when divided by $3$. $22 = 3 \times 7 + 1$ Therefore, $22 \equiv 1 \pmod{3}$.
  • For the term $(2022)^{22}$: We find the remainder of $2022$ when divided by $3$. A number is divisible by 3 if the sum of its digits is divisible by 3. Sum of digits of $2022 = 2+0+2+2 = 6$. Since $6$ is divisible by $3$, $2022$ is divisible by $3$. Therefore, $2022 \equiv 0 \pmod{3}$.

• Step 1.2: Apply modular arithmetic properties to the powers.

  • Since $22 \equiv 1 \pmod{3}$, we can raise both sides to the power of $2022$: $(22)^{2022} \equiv (1)^{2022} \pmod{3}$ $(22)^{2022} \equiv 1 \pmod{3}$ Explanation: If a number leaves a remainder of $1$ when divided by $m$, any positive integer power of that number will also leave a remainder of $1$ when divided by $m$.
  • Since $2022 \equiv 0 \pmod{3}$, we can raise both sides to the power of $22$: $(2022)^{22} \equiv (0)^{22} \pmod{3}$ $(2022)^{22} \equiv 0 \pmod{3}$ Explanation: If a number is divisible by $m$ (remainder $0$), any positive integer power of that number will also be divisible by $m$.

• Step 1.3: Combine the remainders for the sum. We add the congruences obtained in the previous step: $(22)^{2022} + (2022)^{22} \equiv 1 + 0 \pmod{3}$ $(22)^{2022} + (2022)^{22} \equiv 1 \pmod{3}$ Explanation: The remainder of a sum is the sum of the remainders (modulo the divisor).

• Conclusion for $\alpha$: The remainder when $(22)^{2022}+(2022)^{22}$ is divided by 3 is $\alpha = 1$.

2. Finding $\beta$: Remainder when $(22)^{2022}+(2022)^{22}$ is divided by 7

Goal: Determine $\beta$ such that $(22)^{2022}+(2022)^{22} \equiv \beta \pmod{7}$, where $0 \le \beta < 7$.

• Step 2.1: Simplify the base numbers modulo 7.

  • For the term $(22)^{2022}$: We find the remainder of $22$ when divided by $7$. $22 = 3 \times 7 + 1$ Therefore, $22 \equiv 1 \pmod{7}$.
  • For the term $(2022)^{22}$: We find the remainder of $2022$ when divided by $7$. $2022 \div 7$ $2022 = 7 \times 288 + 6$ So, $2022 \equiv 6 \pmod{7}$. Alternatively, and often more conveniently, we can express $6 \pmod{7}$ as $-1 \pmod{7}$, since $6+1=7$. Thus, $2022 \equiv -1 \pmod{7}$. Explanation: Using a negative remainder like $-1$ can simplify calculations, especially when dealing with large powers, as it's easier to calculate powers of $-1$.

• Step 2.2: Apply modular arithmetic properties to the powers.

  • Since $22 \equiv 1 \pmod{7}$, we raise both sides to the power of $2022$: $(22)^{2022} \equiv (1)^{2022} \pmod{7}$ $(22)^{2022} \equiv 1 \pmod{7}$ Explanation: Similar to Step 1.2, any positive integer power of a number that leaves a remainder of $1$ will also leave a remainder of $1$.
  • Since $2022 \equiv -1 \pmod{7}$, we raise both sides to the power of $22$: $(2022)^{22} \equiv (-1)^{22} \pmod{7}$ $(2022)^{22} \equiv 1 \pmod{7}$ Explanation: An even power of $-1$ is always $1$. If the exponent were odd, the remainder would be $-1$, which would then be converted to a positive remainder ($7-1=6$).

• Step 2.3: Combine the remainders for the sum. We add the congruences obtained: $(22)^{2022} + (2022)^{22} \equiv 1 + 1 \pmod{7}$ $(22)^{2022} + (2022)^{22} \equiv 2 \pmod{7}$ Explanation: As before, the remainder of a sum is the sum of the remainders (modulo the divisor).

• Conclusion for $\beta$: The remainder when $(22)^{2022}+(2022)^{22}$ is divided by 7 is $\beta = 2$.

3. Calculating $\alpha^2 + \beta^2$

Goal: Compute the final value using the determined $\alpha$ and $\beta$.

• We found $\alpha = 1$ and $\beta = 2$.
• Substitute these values into the expression $\alpha^2 + \beta^2$: $\alpha^2 + \beta^2 = (1)^2 + (2)^2$ $\alpha^2 + \beta^2 = 1 + 4$ $\alpha^2 + \beta^2 = 5$

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Tips and Common Mistakes to Avoid

• Always simplify the base first: Before dealing with large exponents, always reduce the base number modulo the divisor. For example, instead of $(22)^{2022} \pmod{3}$, simplify $22 \pmod{3}$ to get $1 \pmod{3}$, making the calculation $(1)^{2022} \pmod{3}$.
• Utilize negative remainders: When the remainder is $m-1$ (e.g., $6 \pmod{7}$), it's often beneficial to use $-1 \pmod{m}$ instead. This dramatically simplifies calculations involving powers, as $(-1)^n$ is either $1$ or $-1$. Remember to convert a final negative remainder back to a positive one by adding the modulus (e.g., $-1 \pmod{7} \equiv 6 \pmod{7}$).
• Binomial Expansion Shortcut: Recognize that for $(km+r)^n \pmod{m}$, the remainder is simply $r^n \pmod{m}$. This is a direct consequence of the Binomial Theorem and avoids explicit expansion.
• Watch for large exponents: For extremely large exponents where Euler's Totient Theorem or Fermat's Little Theorem might be applicable, remember to reduce the exponent modulo $\phi(m)$ (or $m-1$ for prime $m$ not dividing the base). This problem, however, only required direct application of basic modular arithmetic due to the bases simplifying to $1$ or $-1$.

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Summary and Key Takeaway

This problem is a classic application of modular arithmetic to simplify complex expressions involving large powers. By understanding that $(A+B) \pmod{M} \equiv (A \pmod{M} + B \pmod{M}) \pmod{M}$ and that $(X^N) \pmod{M} \equiv (X \pmod{M})^N \pmod{M}$, we can break down the problem into manageable steps. The strategic use of remainders, including negative remainders, is crucial for efficiency and accuracy. The final calculation of $\alpha^2 + \beta^2$ then becomes a straightforward substitution.