We need to find the remainder of:

$\frac{10^{122} + 12^{122}}{1331}$

Step 1: Factorizing 1331

$1331 = 11^3$ Thus, we need to compute:

$(10^{122} + 12^{122}) \mod 1331$

Step 2: Applying Fermat's Little Theorem

Fermat's theorem states that if $p$ is a prime and $a$ is not divisible by $p$, then:

$a^{p-1} \equiv 1 \mod p$

For $p = 11$, we get:

$10^{10} \equiv 1 \mod 11$ $12^{10} \equiv 1 \mod 11$

Since $122 = 12 \times 10 + 2$, we can use this periodicity:

$10^{122} \equiv 10^2 \equiv 100 \equiv 1 \mod 11$ $12^{122} \equiv 12^2 \equiv 144 \equiv 1 \mod 11$

So,

$10^{122} + 12^{122} \equiv 1 + 1 = 2 \mod 11$

Step 3: Reducing Modulo $11^3 = 1331$

We also check modulo $11^2 = 121$. Using similar calculations,

$10^{122} \equiv 1 \mod 121$ $12^{122} \equiv 1 \mod 121$

Thus,

$10^{122} + 12^{122} \equiv 2 \mod 121$

Using the Chinese Remainder Theorem, since both modulo 11 and 121 give the remainder 2, we conclude:

$10^{122} + 12^{122} \equiv 2 \mod 1331$

Final Answer:

The remainder is 2.

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Do you want a breakdown of any step?

Here are some related questions:

1. How does Fermat’s theorem apply to modular arithmetic?
2. How do we solve similar problems for different bases and exponents?
3. What is the Chinese Remainder Theorem, and how is it applied here?
4. How can we generalize this method for different divisors?
5. How do modular exponentiation and fast calculations help in large exponent problems?

Tip: When dealing with large exponents, look for periodicity using modular properties like Fermat's theorem.